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Feliz [49]
3 years ago
13

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.
Chemistry
1 answer:
lora16 [44]3 years ago
6 0

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.

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