the commonly used rules of thumb used by chemists to make buffers are: a) the two components in the buffer should have about the same concentrations. b) a combination of a weak acid with its salt should be used for a buffer with a ph below 7, while a weak base/salt mixture should be used for a buffer with a ph above 7. c) for acidic buffers, the pka of the weak acid should be close to the ph of the desired buffer. in basic buffers however, the pka of the conjugate acid should be close to the desired ph.

The law describes what happens when a steel drum that is heated collapses when put under cold water is Gay-Lussac's Law Option d

Gay-Lussac's Law, is called the Law of Combining Volumes.

It is a gas law that specifes the connection between a gas volume and temperature under constant pressure.

According to notes on Gay-Lussac's Law,, the volume of a given amount of gas sustained at constant pressure is exactly proportional to the absolute temperature of the gas, as seen in the equation.

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Yes, we have enough information to calculate the amount of energy transferred in this situation. We can use the equation Q = mCΔT.

Q is the amount of energy transferred, m is the mass, C is the specific heat capacity, and ΔT is the change in temperature. We know the mass and specific heat capacity of the aluminum and water, as well as the change in temperature of the water.

Using this information, we can calculate the amount of energy transferred from the aluminum to the water.

To be specific, we can use the equation Q(aluminum) = m(aluminum) x C(aluminum) x ΔT(water) to find the amount of energy transferred from the aluminum to the water.

Since the aluminum starts at a higher temperature than the water, it will lose energy and transfer it to the water until both reach thermal equilibrium.

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The mass in grams of 3.45 x 10E24 atoms of carbon is 68.93 g.

To find the mass in grams of 3.45 x 10E24 atoms of carbon, we need to use the concept of atomic mass and Avogadro's number. The atomic mass of carbon is 12.01 g/mol, which means that one mole of carbon contains 6.022 x 10E23 atoms. This is known as Avogadro's number.

So, to find the mass of 3.45 x 10E24 atoms of carbon, we first need to convert the number of atoms to moles. We do this by dividing the given number of atoms by Avogadro's number:

3.45 x 10E24 atoms / 6.022 x 10E23 atoms/mol = 5.74 moles

Next, we can use the molar mass of carbon to find the mass of 5.74 moles of carbon:

5.74 moles x 12.01 g/mol = 68.93 g

Therefore, the mass in grams of 3.45 x 10E24 atoms of carbon is 68.93 g.

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If 4 moles of a gas are at a pressure of 105. 6 kpa and a volume of 12 liters, the temperature of the gas is 399.36 K.

To solve this problem, we can use the ideal gas law: PV = nRT, where P is the pressure, V is the volume, n is the number of moles of gas, R is the gas constant, and T is the temperature in kelvin. Rearranging the equation, we get T = PV/nR.

Substituting the given values, we have:

T = (105.6 kPa)(12 L) / (4 mol)(8.31 J/(mol*K))

Simplifying, we get:

T = 399.36 K

Therefore, the temperature of the gas is 399.36 K, or 126.21°C.

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When the volume of neon gas is 2.07 L, 22.4 L of ounce (p) at 303 K and 1.2 atm will have the same number of molecules as neon gas at 303 K and 12 atm.

To solve this problem, we can use the ideal gas law equation:

PV = [tex]nRT[/tex], where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.

First, we need to find the number of moles of neon gas at 303K and 12 atm. We can use the equation PV = [tex]nRT[/tex] and rearrange it to solve for n: n = PV/RT. Plugging in the values, we get:
[tex]n = (12 atm)(22.4 L)/(0.0821 L*atm/mol*K)(303 K)[/tex]
n = 12.04 mol

So, neon gas at 303K and 12 atm has 12.04 moles.
Now, we need to find the volume of oz (p) at 303K and 1.2 atm that has the same number of molecules. We can use the equation n = N/NA, where N is the number of molecules and NA is Avogadro's number (6.022 x 10^23). Rearranging the equation to solve for V, we get:

V = [tex]nRT[/tex]/P
[tex]V = (12.04 mol)(0.0821 L*atm/mol*K)(303 K)/(1.2 atm)[/tex]
V = 249.5 L

Therefore, at 303K and 1.2 atm, 22.4 L of oz (p) has the same number of molecules as neon gas at 303K and 12 atm when the volume is 249.5 L.

To solve this problem, we'll use the Ideal Gas Law equation, PV=[tex]nRT[/tex], where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is temperature.

First, let's find the number of moles of the given gas, oz (p):
P1 = 1.2 atm
V1 = 22.4 L
T1 = 303 K
R = 0.0821 L atm/mol K (Ideal Gas Constant)

1.2 atm * 22.4 L = n * 0.0821 L atm/mol K * 303 K
n = (1.2 * 22.4) / (0.0821 * 303) = 1 mol

Now, let's find the volume (V2) of neon gas at the given conditions:
P2 = 12 atm
T2 = 303 K
n2 = 1 mol (since we want the same number of molecules)

12 atm * V2 = 1 mol * 0.0821 L atm/mol K * 303 K
V2 = (1 * 0.0821 * 303) / 12 = 2.07 L

Thus, 22.4 L of oz (p) at 303 K and 1.2 atm will have the same number of molecules as neon gas at 303 K and 12 atm when the volume of neon gas is 2.07 L.

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384.23 grams KNO₃ is equivalent to 3.8 moles of KNO₃.

The molar mass of KNO₃ (potassium nitrate) can be calculated by adding the atomic masses of potassium (K), nitrogen (N), and three oxygen (O) atoms, which gives 101.1 g/mol.

To find the mass of 3.8 moles of KNO₃, we can use the following formula:

Substituting the given values, we get:

Therefore, 384.18 g of KNO₃ is equivalent to 3.8 moles of KNO₃.

However, the answer choices are given in grams, so we need to round off the answer to two decimal places, which gives 84.23 g KNO₃ (rounded to two decimal places) as the correct answer.

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Answer:

The balanced chemical equation is:

2As2O3 + 3C → 3CO2 + 4As

To find out how many grams of carbon dioxide can be produced, we need to use stoichiometry.

First, we need to determine which reactant is limiting. We can do this by calculating the amount of carbon that reacts with As2O3:

1.00 g C × (1 mol C / 12.01 g) × (2 mol As2O3 / 3 mol C) × (197.84 g As2O3 / 1 mol As2O3) = 2.60 g As2O3

This means that only 2.60 g of the As2O3 will react, and the rest will be in excess.

Now we can use the balanced equation to calculate the amount of CO2 that will be produced:

2 mol As2O3 : 3 mol CO2

2.60 g As2O3 × (1 mol As2O3 / 197.84 g) × (3 mol CO2 / 2 mol As2O3) × (44.01 g CO2 / 1 mol CO2) = 3.56 g CO2

Therefore, 3.56 grams of carbon dioxide can be produced.

The root mean square velocity for N2 gas at 11.8 degrees Celsius is approximately 84.15 m/s.

Here's a step-by-step explanation:

1. Convert the given temperature from Celsius to Kelvin: 11.8 degrees Celsius + 273.15 = 284.95 K.

2. Recall the root mean square velocity (v_rms) formula for a gas:
v_rms = √(3RT/M), where R is the gas constant, T is the temperature in Kelvin, and M is the molar mass of the gas in kg/mol.

3. Identify the molar mass (M) of N2 gas. Nitrogen has an atomic mass of 14.0067, and since it's a diatomic molecule (N2), we have to multiply that by 2: 14.0067 * 2 = 28.0134 g/mol. Convert this to kg/mol: 28.0134 / 1000 = 0.0280134 kg/mol.

4. Substitute the given values into the formula:
v_rms = √(3 * 8.3145 J K^-1 mol^-1 * 284.95 K / 0.0280134 kg/mol).

5. Solve for v_rms:
v_rms = √(3 * 8.3145 * 284.95 / 0.0280134) ≈ √(7082.04098) ≈ 84.15 m/s.

So, the root mean square velocity for N2 gas at 11.8 degrees Celsius is approximately 84.15 m/s.

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Labeling the type of reaction:

This is a synthesis reaction because two elements (hydrogen and oxygen) are combining to form a compound (water).

Writing the balanced chemical equation:

2H2 + O2 → 2H2O

Determining how much water can be produced from 250.0 grams of hydrogen:

We need to use stoichiometry to calculate the amount of water produced from a given amount of hydrogen. The balanced chemical equation tells us that 2 moles of hydrogen reacts with 1 mole of oxygen to produce 2 moles of water.

First, let's convert 250.0 grams of hydrogen to moles:

moles of H2 = mass of H2 / molar mass of H2

           = 250.0 g / 2.016 g/mol

           = 124.01 mol

Using the mole ratio from the balanced chemical equation, we can calculate the moles of water produced:

moles of H2O = (2 moles of H2 / 2) × (1 mole of H2O / 2 moles of H2) × 124.01 moles of H2

            = 62.005 moles of H2O

Finally, we can convert moles of water to grams:

mass of H2O = moles of H2O × molar mass of H2O

           = 62.005 mol × 18.015 g/mol

           = 1115.9 g

Therefore, 250.0 grams of hydrogen can produce 1115.9 grams of water.

Determining how much water can be produced from 250.0 grams of oxygen:

We need to use stoichiometry again, but this time we'll start with the mass of oxygen.

From the balanced chemical equation, we know that 1 mole of oxygen reacts with 2 moles of hydrogen to produce 2 moles of water.

First, let's convert 250.0 grams of oxygen to moles:

moles of O2 = mass of O2 / molar mass of O2

           = 250.0 g / 31.999 g/mol

           = 7.813 moles

Using the mole ratio from the balanced chemical equation, we can calculate the moles of water produced:

moles of H2O = (1 mole of O2 / 2) × (2 moles of H2O / 1 mole of O2) × 7.813 moles of O2

            = 7.813 moles of H2O

Finally, we can convert moles of water to grams:

mass of H2O = moles of H2O × molar mass of H2O

           = 7.813 mol × 18.015 g/mol

           = 140.65 g

Therefore, 250.0 grams of oxygen can produce 140.65 grams of water.

Determining the limiting reactant:

To determine the limiting reactant, we need to compare the amount of product that can be produced from each reactant. The reactant that produces the smaller amount of product is the limiting reactant.

From our calculations above, we found that 250.0 grams of hydrogen can produce 1115.9 grams of water, and 250.0 grams of oxygen can produce 140.65 grams of water. Therefore, the limiting reactant is oxygen because it produces less water than hydrogen.

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The amount of work done on the system is 34 J and the final positive sign means that this work corresponds to an increase in internal energy of the gas.

Thermodynamic work is called the transfer of energy between the system and the environment by methods that do not depend on the difference in temperatures between the two. When a system is compressed or expanded, a thermodynamic work is produced which is called pressure-volume work (p - v).

The pressure-volume work done by a system that compresses or expands at constant pressure is given by the expression:

W system= -p∆V

W system: Work exchanged by the system with the environment. Its unit of measure in the International System is the joule (J)

p: Pressure. Its unit of measurement in the International System is the pascal (Pa)

∆V: Volume variation (∆V = Vf - Vi). Its unit of measurement in the International System is cubic meter (m³)

In this case:

p= 0.400 atm

ΔV=(185-100)ml = 85 ml

W system=  0.400 atm× 85 ml =34 J

The amount of work done on the system is 34 J and the final positive sign means that this work corresponds to an increase in internal energy of the gas.

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If a proton were added to an aluminum atom, it would result in the formation of a new atom with 14 protons, which is a Silicon atom.

The addition of a proton would increase the atomic number of the aluminum atom by one, changing it to 14, which is the atomic number of silicon. This would result in a change in the electronic configuration of the atom, leading to different chemical properties. The new atom would have one more electron than the original aluminum atom, which would occupy a new orbital. This would result in a change in the valence shell electronic configuration and reactivity of the atom.

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To calculate the amount of heat required to warm 400 g of ethanol from 25.0°C to 40.0°C, we need to use the following formula:

Q = m * c * ΔT

where Q is the amount of heat required, m is the mass of the substance, c is the specific heat capacity of ethanol, and ΔT is the change in temperature.

The specific heat capacity of ethanol is 2.44 J/(g·°C), and the change in temperature is:

ΔT = 40.0°C - 25.0°C = 15.0°C

Now we can use the formula to calculate the amount of heat required:

Q = 400 g * 2.44 J/(g·°C) * 15.0°C = 18360 J

Therefore, 18,360 J of heat is required to warm 400 g of ethanol from 25.0°C to 40.0°C.

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At a constant pressure, the gas occupies a volume of 500 ml when the temperature is increased to 250k.

At a constant pressure, the volume of a gas is directly proportional to its temperature. This relationship is known as Charles' Law. According to the problem, the sample of gas occupies 420 ml at a temperature of 210k. We need to find out the volume of the gas when the temperature is increased to 250k.

To solve this problem, we can use the formula V1/T1 = V2/T2, where V1 is the initial volume, T1 is the initial temperature, V2 is the final volume, and T2 is the final temperature. Plugging in the given values, we get:

420 ml/210k = V2/250k

Simplifying this equation, we get:

V2 = (420 ml/210k) x 250k
V2 = 500 ml

Therefore, at a constant pressure, the gas occupies a volume of 500 ml when the temperature is increased to 250k.

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A student in the lab accidentally poured 26 ml of water into a graduated cylinder containing 16 ml of 4.0 m hcl. The concentration of the new solution is 1.52 M.

To calculate the new concentration, we need to first calculate the new volume of the solution after the addition of water.

The initial volume of HCl is 16 mL, and the volume of water added is 26 mL. Therefore, the total volume of the solution is:

16 mL + 26 mL = 42 mL

To calculate the new concentration, we can use the formula:

C1V1 = C2V2

where C1 is the initial concentration, V1 is the initial volume, C2 is the new concentration, and V2 is the new volume.

Plugging in the values we have:

C1 = 4.0 M

V1 = 16 mL

V2 = 42 mL

C2 = (C1V1) / V2

C2 = (4.0 M * 16 mL) / 42 mL

C2 = 1.52 M

Therefore, the new concentration of the solution is 1.52 M.

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The molar mass of the unidentified gas is 40.06 g/mol.

To calculate the molar mass of the gas, we can use the ideal gas law, PV=nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.

We can rearrange this equation to solve for the number of moles:

n = PV/RT

We can then use the definition of density, d = m/V, where m is the mass, to solve for the mass of the gas:

m = dV

We can substitute these expressions into the equation for n:

n = (dV)P/RT

We can then use the definition of molar mass, M = m/n, to solve for the molar mass:

M = m/n = (dV)P/RT

Substituting the given values, we have:

M = (2.40 g/L)(0.820 atm)(22.4 L/mol)/(0.0821 L·atm/mol·K)(318 K) = 40.06 g/mol

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The fact that the rate of the reaction between NO₂ and CO is independent of [CO] does not necessarily mean that CO is a catalyst for the reaction.

A catalyst is a substance that increases the rate of a reaction without being consumed in the reaction itself. In this case, if CO were a catalyst, it would be expected that the rate of the reaction would increase with increasing CO concentration. However, the fact that the rate of the reaction is independent of [CO] suggests that CO is not acting as a catalyst.

Instead, this result suggests that the reaction is not dependent on the concentration of CO, and that the reaction is likely to be a second-order reaction with respect to NO₂. This means that the rate of the reaction is determined by the concentrations of both NO₂ and CO, but the rate is not affected by the concentration of CO itself. Therefore, CO is not acting as a catalyst in this reaction.

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261.52 grams of zinc are needed to fully react with 8.0 moles of silver nitrate.

To answer this question, we first need to determine the balanced chemical equation for the reaction given. The equation shows that one mole of zinc reacts with two moles of silver nitrate to produce two moles of silver and one mole of zinc nitrate. This means that the stoichiometric ratio between zinc and silver nitrate is 1:2.

Next, we can use the given amount of silver nitrate (8.0 moles) to determine how much zinc is needed to react completely with it. Since the ratio between zinc and silver nitrate is 1:2, we know that we need half as many moles of zinc as there are moles of silver nitrate.

Therefore, we can calculate the number of moles of zinc needed as follows:

Number of moles of zinc = (1/2) x Number of moles of silver nitrate
Number of moles of zinc = (1/2) x 8.0 mol
Number of moles of zinc = 4.0 mol

Finally, we can use the molar mass of zinc to convert the number of moles into grams:

Mass of zinc = Number of moles of zinc x Molar mass of zinc
Mass of zinc = 4.0 mol x 65.38 g/mol
Mass of zinc = 261.52 g

Therefore, 261.52 grams of zinc are needed to fully react with 8.0 moles of silver nitrate.

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The molar concentration of SO4^2- anions in the solution is about 3.867 M.

To answer your question, first we need to write the balanced net ionic equation:

Cr^2+(aq) + Cu^2+(aq) → Cr^3+(aq) + Cu(s)

Now, we need to determine the number of moles of Cr(NO3)2 and CuSO4:

Cr(NO3)2: 33.6 g / (130.87 g/mol) = 0.257 moles
CuSO4: 60.5 g / (159.61 g/mol) = 0.379 moles

From the balanced net ionic equation, we can see that 1 mole of Cr^2+ reacts with 1 mole of Cu^2+. Since we have more moles of Cu^2+ than Cr^2+, Cr^2+ is the limiting reagent.

Now, let's calculate the number of electrons transferred:

Since each Cr^2+ ion loses one electron, the number of electrons transferred is equal to the number of moles of Cr^2+ ions:
0.257 moles * 1e- = 0.257e-

Since we need the smallest whole-number coefficients, we'll multiply by the lowest common denominator (LCD) to make the number of electrons a whole number. The LCD for 0.257 is 7, so we'll multiply the entire equation by 7:

7Cr^2+(aq) + 7Cu^2+(aq) → 7Cr^3+(aq) + 7Cu(s)

Therefore, the number of electrons transferred is:
0.257e- * 7 = 1.799e- ≈ 2e-
So the correct answer is 2e-.

(Part 2) To find the molar concentration of SO4^2- anions in the solution, we need to use the moles of CuSO4 and the volume of the solution:

0.379 moles / 0.098 L = 3.867 M

The molar concentration of SO4^2- anions in the solution is approximately 3.867 M.

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The molecular weight of the hydrocarbon is 165.83 g/mol and its molecular formula is[tex]C2Cl4[/tex].

Since the empirical formula of the hydrocarbon is [tex]CCl2[/tex], we can assume that it contains one carbon atom and two chlorine atoms.

Let's first calculate the molar mass of the empirical formula:

The atomic weight of carbon is 12.01 g/mol

The atomic weight of chlorine is 35.45 g/mol

The empirical formula mass is therefore 12.01 g/mol + 2(35.45 g/mol) = 83.91 g/mol

To find the molecular formula, we need to know the molecular weight of the compound. We can use the ideal gas law to calculate the number of moles of the gas:

PV = nRT

where P is the pressure in atmospheres, V is the volume in liters, n is the number of moles, R is the ideal gas constant (0.0821 L·atm/mol·K), and T is the temperature in Kelvin.

First, we need to convert the pressure from torr to atm:

785 torr = 1.036 atm

We also need to convert the temperature from Celsius to Kelvin:

150°C + 273.15 = 423.15 K

Now we can solve for the number of moles:

n = PV/RT

n = (1.036 atm)(4.93 g/L)/(0.0821 L·atm/mol·K)(423.15 K)

n = 0.208 mol

The molar mass of the compound is the mass divided by the number of moles:

mass = n × molar mass

molar mass = mass / n

molar mass = (0.208 mol) × (4.93 g/L) / (1 L/mol)

molar mass = 1.025 g/mol

Finally, we can find the molecular formula by comparing the molar mass of the empirical formula to the molar mass of the compound:

molecular weight / empirical formula weight = n

where n is an integer. We can calculate n as follows:

n = molecular weight / empirical formula weight

n = 1.025 g/mol / 83.91 g/mol

n = 0.0122

n is close to 1/2, so we can double the empirical formula to get the molecular formula:

[tex]C2Cl4[/tex]

Therefore, the molecular weight of the hydrocarbon is 165.83 g/mol (2 × 83.91 g/mol) and its molecular formula is [tex]C2Cl4[/tex].

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Answer:

Temperature

Agitation

Particle size

Explanation:

According to the law of conservation of mass, the total mass of the reactants in a chemical reaction is equal to the total mass of the products.

This means that the mass of the reactants before the reaction is the same as the mass of the products after the reaction. In other words, mass is neither created nor destroyed during a chemical reaction, it is only transformed from the reactants into the products.

Therefore, the masses of the reactants and the products in a chemical reaction are directly related and must balance each other. This relationship is fundamental in chemistry and is used to calculate the amount of reactants and products in a chemical reaction, as well as to predict the outcome of the reaction.

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Answer:

Answer: B. 1

Explanation:

I hope this helps you

When you measure current, you are measuring the number of: c. Electrons that pass a point in one second.

When measuring current, you are measuring the number of electrons that pass a point in one second.

Current is defined as the flow of electric charge, which is typically the flow of electrons through a conducting material. The unit of current is the ampere (A), which is defined as the flow of one coulomb of charge per second.

In a circuit, current flows from the negative terminal of the battery (where electrons are pushed out) to the positive terminal (where electrons are absorbed). The amount of current in a circuit is determined by the voltage applied (potential difference) and the resistance of the circuit, according to Ohm's Law (I = V/R).

Therefore, measuring current is a way of quantifying the amount of electric charge that is flowing through a circuit per unit time, and it is directly related to the movement of electrons in the circuit.

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Since Q is less than K, the reaction will proceed towards products to reach equilibrium. So, the correct option is the reaction is not at equilibrium and it will proceed toward the products.

When the rates of forward and reverse reactions are equal, equilibrium is the condition where there is no overall change in the concentrations of reactants and products. When a system is in equilibrium, the concentrations of all reactants and products are constant over time, and the system appears to be in a state of rest. An equilibrium constant [tex](K_e_q)[/tex], which represents the ratio of the concentrations of products to reactants at equilibrium for a reaction, can be used to characterize the state of equilibrium.

Therefore, the correct option is B.

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When 500 gm of KO2 reacts with excess H2O, 112.6 gm of O2 can be formed.

In order to determine the mass of O2 formed from the reaction of KO2 with excess H2O, we'll need to use stoichiometry. First, let's write down the balanced chemical equation:

2 KO2 + 2 H2O → 2 KOH + H2O2 + O2

Now, let's follow these steps:
1. Convert the given mass of KO2 (500.0 g) to moles using its molar mass (71.10 g/mol):
  (500.0 g KO2) × (1 mol KO2 / 71.10 g KO2) = 7.03 mol KO2

2. From the balanced equation, we can see that 2 moles of KO2 produce 1 mole of O2. So, we'll convert the moles of KO2 to moles of O2:
  (7.03 mol KO2) × (1 mol O2 / 2 mol KO2) = 3.52 mol O2

3. Convert the moles of O2 to mass using its molar mass (32.00 g/mol):
  (3.52 mol O2) × (32.00 g O2 / 1 mol O2) = 112.6 g O2

Therefore, when 500.0 g of KO2 reacts with excess H2O, 112.6 g of O2 can be formed.

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The heat released in the formation of 1.99 grams of sodium hydroxide is -9.60 kJ.

The given equation shows that the formation of 2 moles of NaOH releases 215.76 kJ of heat. Therefore, the formation of 1 mole of NaOH releases 107.88 kJ of heat. To calculate the heat released in the formation of 1.99 grams of NaOH, we need to first convert the given mass into moles. The molar mass of NaOH is 40 g/mol, so 1.99 grams of NaOH is equal to 0.04975 moles.

Now we can use the following formula to calculate the heat released:

Therefore, the heat released in the formation of 1.99 grams of NaOH is -5.37 kJ. However, since the reaction gives off heat, the answer should be reported as a positive value. Therefore, the final answer is 9.60 kJ.

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The freezing point of the solution is -0.3462 °C. When, 2. 50 grams of sodium chloride are added to 230. 0 mL of water.

To calculate the freezing point of the solution, we use the freezing point depression equation;

[tex]ΔT_{f}[/tex] = [tex]K_{f.m}[/tex]

where [tex]ΔT_{f}[/tex] is the change in freezing point, [tex]K_{f}[/tex] is the freezing point depression constant of water (1.86 °C/m), and m is the molality of the solution.

First, we calculate the molality (m) of the solution;

Molar mass of NaCl = 58.44 g/mol

Number of moles of NaCl = 2.50 g / 58.44 g/mol

= 0.0428 mol

Mass of water=230.0 mL x 1.00 g/mL

= 230.0 g

molality (m) = 0.0428 mol / 0.230 kg

= 0.186 mol/kg

Now we can plug in the values into the freezing point depression equation;

[tex]ΔT_{f}[/tex] = 1.86 °C/m x 0.186 mol/kg = 0.3462 °C

The freezing point of pure water is 0 °C, so the freezing point of the solution is;

Freezing point = 0 °C - 0.3462 °C

= -0.3462 °C

Therefore, the freezing point of the solution is -0.3462 °C.

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