Two elements $A$ and $B$ form compounds of formula $A B_2$ and $A B_4$. When dissolved in $20.0 \mathrm{~g}$ of benzene $1.0 \mathrm{~g}$ of $A B_2$ lowers f. pt. by $2.3^{\circ} \mathrm{C}$ whereas $1.0 \mathrm{~g}$ of $A B_4$ lowers $\mathrm{f}$. pt. by $1.3^{\circ} \mathrm{C}$. The $K_f$ for benzene is 5.1. The atomic masses of $A$ and $B$ are

• Correct answer
  A

  25,42
• B

  42,25
• C

  52,48
• D

  48,52

Solution

Let the masses of $A$ and $B$ be $a$ and $b$. The mass of $A B_2$ will be $(a+2 b) \mathrm{g} \mathrm{mol}^{-1}$ and $A B_4$ will be $(a+4 b) \mathrm{g} \mathrm{mol}^{-1}$. $\Delta T_f=\frac{K_f \times W_B \times 1000}{M_B \times W_A}$
On solving (i) and (ii), we get $a=25.59$ and $b=42.64$

Hence, the correct answer is option: 'A'.