Question

When 70.4 g of benzamide (C7H7NO) are dissolved in 850. g of a certain mystery liquid X , the freezing point of the solution is 2.7°C lower than the freezing point of pure X. On the other hand, when 70.4 g of ammonium chloride (NH4CI) are dissolved in the same mass of X, the freezing point of the solution is 9.9 °C lower than the freezing point of pure X.
A) Calculate the van't Hoff factor for ammonium chloride in X. Be sure your answer has a unit symbol, if necessary, and round your answer to 2 significant digits.

Answer 1

Answer:

i=1.62 .

Explanation:

Let, i be the Van't Hoff Factor.

Moles of benzamide,=$\dfrac{Mass}{molar\ mass}=\dfrac{70.4}{121.14}=0.58\ mol.$

Molality of solution, m=$\dfrac{moles }{mass\ of\ solvent}=\dfrac{0.58}{0.85}=0.68\ molal.$

Now, we know

Depression in freezing point, $\Delta T=i\times K_f\times m$  .....1

It is given that,

$\Delta T=2.7^o C\\i=1 ( since\ it\ is\ non\ dissociable\ solutes)\\K_f ( freezing\ constant)$

Putting all these values we get,

$K_f=3.949\ C/m.$

Now, moles of ammonium chloride=$\dfrac{70.4}{53.49}=1.316\ mol.$

molality =$\dfrac{1.316}{0.85}=1.54 molal.\\\Delta T=9.9 .$

Putting all these values in eqn 1.

We get,

i=1.62 .

Hence, this is the required solution.