2 Marks Questions

Question 12 Marks

For a 5% solution of urea (molar mass = 60 g/mol), calculate the osmotic pressure at 300 k [R = 0.0821 L atm $ K^{-1} mol^{-1} $]

Answer

V = 100ml = 0.1L, W = 5g, m = 60
$ \pi = \frac{w}{m} \times \frac{R \times T}{V} $
$ = \frac{5}{60} \times \frac{0.0821 \times 300}{0.1} = 20.5~atm $

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Question 22 Marks

Write Raoults law of relative lowering of vapour pressure?

Answer

When a non-volatile solute is dissolved in a solvent, its vapour pressure gets lowered. This lowering of vapour pressure of solvent is termed as relative lowering of vapour pressure. This lowering is governed by Raoult's Law. According to Raoult's law relative lowering in vapour pressure of a solvent is equal to mole fraction of the solute.
Mathematically, this is
$ \frac{P^{0}-P_{s}}{P^{0}}=\frac{n_{1}}{n_{1}+n_{2}} $
It is a colligative properties.

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Question 32 Marks

18 g of glucose ($ C_{6}H_{12}O_{6} $) was added to 1 kg of water at 1.013 bar atmospheric pressure in a vessel. At which temperature will water boil? $ K_{b} $ for $ H_{2}O $ is 0.52 K kg $ mol^{-1} $.

Answer

As we know that
$\Delta T _b=\frac{1000 K_b \cdot W_{ A }}{ W _{ B } M _{ A }}$
$=\frac{1000 \times 0.52 \times 18 \times 10^{-3}}{180 \times 10^{-3} \times 10^{-3}}$
$=\frac{100 \times 0.52}{1000}$$=0.052 K$

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Question 42 Marks

The osmotic pressure of sugar solution is 2.46 atm at 27°C. Calculate the concentration of the solution.

Answer

Given that $ \pi = 2.46 $, T = 27 + 273 = 300 K, R = 0.0821
From the formula, $ \pi = cRT $
$ 2.46 = c \times 0.0821 \times 300 $
$c=\frac{2.46}{0.0821 \times 300}=\frac{2.46}{24.63}$
= 0.0998 mol/litre

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Question 52 Marks

What is Van't Hoff factor? How is it related to degree of dissociation of the electrolyte ?

Answer

Van't Hoff Factor: It is a ratio of experimentally determined value of colligative property and normal value when solute behaves ideally in the solution. Relation between degree of dissociation $ \alpha $ of electrolyte with i (van't Hoff factor)
(A) At the time of association $ \alpha = \frac{i-1}{(1/n)-1} $
(B) At the time of dissociation $ \alpha = \frac{i-1}{n-1} $

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Question 62 Marks

Osmotic pressure of a solution is 0.0821 atm. at a temperature of 400 K. Calculate the concentration of solution in mol/litre. [R=0.0821 L atm $ K^{-1} mol^{-1} $]

Answer

Given that T = 400 K, R = 0.0821, $ \pi = 0.0821 $
From the formula, $ \pi = cR'T $
$\Rightarrow \quad c=\frac{\pi}{ RT }$
$=\frac{0.0821}{0.0821 \times 400}=\frac{1}{400}$
$c=0.0025 mol /litre$

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Question 72 Marks

Explain the following terms with suitable example:
(i) Molarity
(ii) Molality

Answer

(i) Molarity (M) : It is the number of mole of solute dissolved per litre of solution. For example, 58.5 g NaCl when dissolved in one litre of water gives 1 M solution.
(ii) Molality (m) : It is the number of moles of solute dissolved in 1 kg of solvent. For example, 40 g of NaOH when dissolved in 1 kg of water it will give 1 molal solution.

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