Answer to Question #139289 in General Chemistry for jane

Question

What will be the freezing point of an aqueous solution containing 80.0 g of NaCl in 3.4 L of H2O?

Answer 1

Q139289

What will be the freezing point of an aqueous solution containing 80.0 g of NaCl in 3.4 L of

H₂O?

Solution :

Depression in freezing point: The Freezing point of the solvent is lowered on the addition of a non-volatile compound.

It is a colligative property that is it depends on the number of the solute particles present and not on their identity.

The formula for depression in freezing point is

ΔT_F = - i * m * K_F, ---------------- Equation 1

where, ΔT_f is the depression in freezing point

i = van’t hoff factor of the solute.

m = molality of the solution.

K_f = Freezing point depression constant.

We will consider that NaCl dissociates completely in water to give Na⁺ (aq) and Cl^- (aq).

Since each NaCl can given two ions so van’t hoff factor, i = 2.

For water, K_F = 1.86 ⁰C/m

Step 1 : To find the moles of NaCl

Convert 80.0g of NaCl to moles using molar mass of NaCl.

Molar mass of NaCl = 1 * atomic mass of Na + 1 * atomic mass of Cl

= 1 * 22.99g/mol + 1* 35.453g/mol

= 58.443g/mol

moles of NaCl = 80.0g of NaCl * 1 mole NaCl/58.443g of NaCl = 1.369 mol NaCl

Step 2: To find the mass of given 3.4L water.

We are given volume of water = 3.4L

Let us assume that the density of water is 1.00 kg/L

Substitute volume and density of water in formula of density and find the mass of water.

Density = mass/ Volume ;

So, mass = Density * volume = 1.00kg/L * 3.4L = 3.4 kg

Step 3: To find the molality of the solution.

Moles of NaCl = 1.369moles and mass of solvent (water ) = 3.4kg

molality = moles of solute/mass of solvent in ‘kg’

molality = 1.369mol/3.4kg = 0.4026m

Step 4: To find the depression in the freezing point

Substitute i = 2, m = 0.4026m and K_F = 1.86 ⁰C/m in the Equation 1, we have

ΔT_F = - i * m * K_F = - 2 * 0.4026m * 1.86 ⁰C/m

ΔT_F = - 1.498 ⁰C

Step 5: To find freezing point of water after addition of NaCl .

We know Freezing point of pure water = 0.00⁰C

ΔT_F = T_solution - T_water

-1.498 ⁰C = T_solution - 0.00⁰C

T_solution = -1.498⁰C

In the question the quantity with the least number of significant figure is 3.4L. It has got 2 significant figures. So we must round up our final answer to 2 significant figures.

Hence the freezing point of the aqueous solution will be -1.5⁰C.