Question

Two elements $A$ and $B$ form compounds having molecular formula $A{B}_2$ and $A{B}_4$. When dissolved in $20.0g$ of benzene $(C_6{H}_6),1.0gA{B}_2$ lowers the freezing point by $2.3^{\circ }C$ whereas $1.0g$ of $A{B}_4$ lowers the freezing point by $1.3^{\circ }C$. The molal depression constant for benzene is $5.1Kkgmo{l}^{-1}$. 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 $A{B}_2\Delta T_f=K_f\times \frac{w_2\times 1000}{M_2\times w_1}$
or $2.3=5.1\times \frac{1\times 1000}{M_2\times 20}$
$\therefore M_2=\frac{50\times 5.1}{2.3}=110.86$
Similarly for $A{B}_{4}1.3=5.1\times \frac{1\times 1000}{M_2\times 20}$
$\therefore M_2=\frac{50\times 5.1}{1.3}=196.15$
Now, molecular weight of $A{B}_2=110.86$,
molecular weight of $A{B}_4=196.15$
$A{B}_4=A+4B=196.15$ ...(i)
$A{B}_2=A+2B=110.86$ ...(ii)
(i) - (ii) gives $2B=85.29\therefore B=42.645$
Putting the value of B in equation (ii),
$A+2\times \mathrm{42.645.}=110.86$
or, $A=110.86-85.29=25.57$