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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.

Respuesta :

Answer:

i=1.62 .

Explanation:

Let, i be the Van't Hoff Factor.

Moles of benzamide,=[tex]\dfrac{Mass}{molar\ mass}=\dfrac{70.4}{121.14}=0.58\ mol.[/tex]

Molality of solution, m=[tex]\dfrac{moles  }{mass\ of\ solvent}=\dfrac{0.58}{0.85}=0.68\ molal.[/tex]

Now, we know

Depression in freezing point, [tex]\Delta T=i\times K_f\times m[/tex]  .....1

It is given that,

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

Putting all these values we get,

[tex]K_f=3.949\ C/m.[/tex]

Now, moles of ammonium chloride=[tex]\dfrac{70.4}{53.49}=1.316\ mol.[/tex]

molality =[tex]\dfrac{1.316}{0.85}=1.54 molal.\\\Delta T=9.9 .[/tex]

Putting all these values in eqn 1.

We get,

i=1.62 .

Hence, this is the required solution.

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