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Chemistry 3 Marks Question

Solution · Rajasthan Board (RBSE) STD 12 Science (Rajasthan - English Medium). 6 questions.

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Question 13 Marks
Calculate the mass of ascorbic acid (Vitamin C, $ C_{6}H_{8}O_{6} $) to be dissolved in 75 g of acetic acid to lower its melting point by $ 1.5^{\circ}C $. ($ K_{f} $ for $ CH_{3}COOH = 3.9 K kg mol^{-1} $)
Answer
Lowering in melting point $\left(\Delta T _f\right)=1.5^{\circ}$
Mass of solvent $\left( CH _3 COOH \right) W _1=75 g$
Molar mass of solute $\left( C _6 H _8 O _6\right), M _2=176 g mol ^{-1}$
$K _f=3.9 K kg mol ^{-1}$
$M _2=\frac{1000 K_f W_2}{W_1 \Delta T_f}$
or $\quad$$W _2=\frac{ M _2 \times W _1 \times \Delta T _f}{1000 \times K _f}$
$\therefore \quad=\frac{\left(176 g mol ^{-1}\right)(75 g)(1.5 K)}{\left(1000 g kg ^{-1}\right)\left(3.9 K kg mol ^{-1}\right)}=5.077 g$.
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Question 23 Marks
State Raoult's Law for solution containing volatile solutes. Write two Characteristics of the solutions which obeys Raoult's law at all concentrations.
Answer
When a volatile liquid is dissolved into a solvent then the sum of vapour pressures of both, solute and solvent will be equal to the total pressure of the solution. When two volatile liquids A and B with mole fraction $ X_{A} $ and $ X_{B} $, Partial vapour pressures $ P_{A} $ and $ P_{B} $ vapour pressure of their pure state $ P_{A}^{0} $ and $ P_{B}^{0} $. So according to Raoult's law:
$ P_{A} = P_{A}^{0} \cdot X_{A} $ and $ P_{B} = P_{B}^{0} \cdot X_{B} $.
Total pressure $ P = P_{A} + P_{B} = [P_{A}^{0} \cdot X_{A} + P_{B}^{0} \cdot X_{B}] $.
Characteristics of the solutions which obeys Raoult's Law at all Concentrations
(i) $ \Delta_{mix}H = 0 $
(ii) $ \Delta_{mix}V = 0 $
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Question 33 Marks
A 4% solution (w/w) of sucrose (M = 342 g $ mol^{-1} $) in water has a freezing point of 271.15 K. Calculate the freezing point of 5% glucose (M = 180 g $ mol^{-1} $) in water. (Given: Freezing point of pure water = 273.15 K)
Answer
We know $ \Delta T_{f} = K_{f} \times M $
$ M = \frac{W_{2} \times 1000}{M_{2} \times M_{1}} $
For Sucrose solution:
$ 273.15 - 271.15 = \frac{K_{f} \times 4 \times 1000}{342 \times 96} $
$ K_{f} = \frac{2 \times 96 \times 342}{4 \times 1000} $ ...(1)
For glucose solution:
$ K_{f} = \frac{(273.15 - x) \times 95 \times 180}{5 \times 1000} $ ...(2)
By Equating (1) and (2):
$ \frac{2 \times 96 \times 342}{4 \times 1000} = \frac{(273.15 - x) \times 180 \times 95}{5 \times 1000} $
x = 268.35 K
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Question 43 Marks
Calculate the freezing point of a solution containing 60 g of glucose (Molar mass = 180 g $ mol^{-1} $) in 250 g of water. ($ K_{f} $ of water = 1.86 K kg $ mol^{-1} $)
Answer
$\Delta T _f= K _f \times M$
$M =\frac{60 / 180 mol}{0.25 kg}$
$=1.33 mol / kg$
$\Delta T _f=1.86 \times 1.33$
$=2.4738 K$
$T _f^0- T _f=2.4738 K$
$T _f= T _f^0-2.4738$
$=273-2.4738$
$T _f=270.5262 K$
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Question 53 Marks
(a) What do you mean by the term 'Depression of freezing point'?
(b) State Raoult's Law of depression of freezing point? How is it useful in determining the molecular weight of non-volatile and non-electrolyte.
Answer
(a) Freezing point of a substance is the temperature at which the solid and the liquids forms the substance are in equilibrium i.e. the solid and the liquid forms of substance have the same vapour pressure. The freezing point of the solution is always power than that of the pure advent. The decrease is called the depression in freezing point.
$\Delta T _f= T _f^{\prime}- T _f$
Molal depression constant may be defined as the depression in freezing point when the moldity of solution is unity i.e. one mole of the solute is dissolved in 1000 g of solvent.
(b) Freezing point : It is that temperature of a liquid at which it begins to solidify. At this temperature vapour pressure becomes zero.
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Question 63 Marks
An anti-freeze solution is prepared from 222.6 g of ethylene glycol $ [C_{2}H_{4}(OH)_{2}] $, and 200 g of water. Calculate the molality of the solution. If the density of the solution is 1.072 g $ mL^{-1} $, then what shall be the molarity of the solution ?
Answer
Molality of ethylene glycol
= $ \frac{222.6}{62 \times \frac{200}{1000}} = 17.95~m $
Wt. of solution = Wt. of glycol + Wt. of water
= 222.6 + 200 = 422.6 g
Volume of solution = $ \frac{422.6}{1.072} $ mL
Molarity of ethylene glycol = $ \frac{222.6}{62 \times \frac{422.6}{1.072 \times 1000}} = 9.11~M $
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