Two elements -A- and -B- form compounds having molecular formulae AB2 and AB4 when dissolved in 20-0 g of benzene- 1-0 g of AB2 lowers the freezing pioint by 2-3-C whereas 1-0 g of AB4 lowers the freezing point by 1-3-C- The molal depression constant for benzene is 5-1 K kg mol-1- Calculate the atomic masses of A and B-

Name: JEE - Grade 12 - Chemistry - Stereochemistry and Reaction Mechanism - Session 20 - W05
Uploaded: 2022-07-04T10:36:42+05:30
Description: Freezing point depression of a solution is given by, T_f = K_f × m where, T_f is the depression in freezing point. nbsp; K_f is molal depression constant. m i ...

Freezing point depression of a solution is given by, T_f = K_f × m where, T_f is the depression in freezing point. nbsp; K_f is molal depression constant. m i ...

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Standard XII

Chemistry

Depression in Freezing Point

Two elements ...

Question

Two elements 'A' and 'B' form compounds having molecular formulae $A B_{2}$ and $A B_{4}$ when dissolved in $20.0 g$ of benzene, $1.0 g$ of $A B_{2}$ lowers the freezing pioint by
${2.3}^{\circ} C$ whereas $1.0 g$ of $A B_{4}$ lowers the freezing point by ${1.3}^{\circ} C$ . The molal depression constant for benzene is $5.1 K k g m o l^{- 1}$ . Calculate the atomic masses of A and B.

A = 25.6 g / m o l; B = 42.6 g / m o l

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A = 40 g / m o l; B = 23 g / m o l

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A = 15.2 g / m o l; B = 20.4 g / m o l

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A = 34.3 g / m o l; B = 52.5 g / m o l

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Solution

The correct option is A $A = 25.6 g / m o l; B = 42.6 g / m o l$
Freezing point depression of a solution is given by,
$△ T_{f} = K_{f} \times m$
where,
$△ T_{f}$ is the depression in freezing point.
$K_{f}$ is molal depression constant.
m is molality of the solution.
From the above equation,
We know that,
$M = \frac{1000 K_{f} \times w}{W \times △ T_{f}}$
where,
$w$ is weight of solute
$W$ is weight of solvent
$M$ is molar mass of the solute

1. Molar mass of $A B_{2}$ (from given data) $= \frac{1000 \times 5.1 \times 1}{20 \times 2.3} M_{A B_{2}} = 110.86 g / m o l$

2. Molar mass of $A B_{4}$ (from given data)
$= \frac{1000 \times 5.1 \times 1}{1.3 \times 20} M_{A B_{4}} = 196.15 g / m o l$

Further, $A B_{4} \Rightarrow A + 4 B = 196.15..... e q . (i)$
$A B_{2} \Rightarrow A + 2 B = 110.86..... e q . (i i)$
Subtracting eq. (ii) from (i), we get,
$2 B = 85.29 B = 42.6 g / m o l$
Putting the value of B in eq. (ii)
$A + 85.29 = 110.86$
or $A = (110.86 - 85.29) = 25.57 \approx 25.6 g / m o l$
Thus, the atomic masses of A and B are $25.6 g / m o l$ and $42.6 g / m o l$ respectively.

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Depression in Freezing Point

Standard XII Chemistry

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