Question

A certain liquid X has a normal freezing point of 7.60 °C and a freezing point depression constant K= 6.90 °C-kg-mol. Calculate the freezing point of a solution made of 7.57g of sodium chloride (NaCl) dissolved in 350. g of X Round your answer to 3 significant digits. lºc X 5

Answer 1

Answer: The freezing point of solution is -5.11°C

Explanation:

Vant hoff factor for ionic solute is the number of ions that are present in a solution. The equation for the ionization of sodium chloride follows:

$NaCl(aq.)\rightarrow Na^{+}(aq.)+Cl^-(aq.)$

The total number of ions present in the solution are 2.

To calculate the molality of solution, we use the equation:

$Molality=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}$

Where,

$m_{solute}$ = Given mass of solute (NaCl) = 7.57 g

$M_{solute}$ = Molar mass of solute (NaCl) = 58.44 g/mol

$W_{solvent}$ = Mass of solvent (liquid X) = 350.0 g

Putting values in above equation, we get:

$\text{Molality of }NaCl=\frac{7.57\times 1000}{58.44\times 350.0}\\\\\text{Molality of }NaCl=0.370m$

To calculate the depression in freezing point, we use the equation:

$\Delta T=iK_fm$

where,

i = Vant hoff factor = 2

$K_f$ = molal freezing point depression constant = 6.90°C/m

m = molality of solution = 0.370 m

Putting values in above equation, we get:

$\Delta T=2\times 6.90^oC/m.g\times 0.370m\\\\\Delta T=5.11^oC$

Depression in freezing point is defined as the difference in the freezing point of water and freezing point of solution.

$\Delta T=\text{freezing point of water}-\text{freezing point of solution}$

$\Delta T$ = 5.11 °C

Freezing point of water = 0°C

Freezing point of solution = ?

Putting values in above equation, we get:

$5.106^oC=0^oC-\text{Freezing point of solution}\\\\\text{Freezing point of solution}=-5.11^oC$

Hence, the freezing point of solution is -5.11°C