Question

Two elements $A$ and $B$ form compounds having molecular formula $AB_{2}$ and $AB_{4}$. When dissolved in $20.0\, g$ of benzene $(C_{6}H_{6}), 1.0\, g \,AB_{2}$ lowers the freezing point by $2.3^{\circ}C$ whereas $1.0\, g$ of $AB_{4}$ lowers the freezing point by $1.3^{\circ}C$. The molal depression constant for benzene is $5.1\, K\, kg\, mol ^{-1}$. The atomic masses of $A$ and $B$ respectively are

• A. $74.26,36.45$
• B. $42.64,25.57$
• C. $25.00,75.00$
• D. $25.57,42.64$

Answer 1

Answer: (D)

For $AB_{2} \Delta T_{f} = K_{f} \times \frac{w_{2} \times 1000}{M_{2} \times w_{1}}$
or $2.3 = 5.1 \times \frac{1 \times1000}{M_{2} \times20}$
$\therefore M_{2} = \frac{50 \times5.1}{2.3} = 110.86$
Similarly for $AB_{4} 1.3 = 5.1 \times \frac{1 \times 1000}{M_{2} \times 20}$
$\therefore M_{2} = \frac{50 \times 5.1}{1.3} = 196.15$
Now, molecular weight of $AB_{2} = 110.86$,
molecular weight of $AB_{4} = 196 .15$
$AB_{4} = A + 4B = 196.15$ ...(i)
$AB_{2} = A + 2B= 110.86$ ...(ii)
(i) - (ii) gives $2B = 85.29 \therefore B = 42.645$
Putting the value of B in equation (ii),
$A + 2 \times 42.645.= 110.86$
or, $A = 110.86 - 85.29 = 25.57$