Question

Two elements $A$ and $B$ form compounds of formula $A{B}_2$ and $A{B}_4$. When dissolved in $20.0g$ of benzene $1.0g$ of $A{B}_2$ lowers f. pt. by 2.3${}^{\circ }$C whereas $1.0g$ of $A{B}_4$ lowers f. pt. by 1.3${}^{\circ }$C. The $K_f$ for benzene is $5.1$. The atomic masses of $A$ and $B$ are

• A. 25, 42
• B. 42, 25
• C. 52, 48
• D. 48, 52

Answer 1

Answer: (A)

Let the masses of A and B be a and b. The mass of AB₂ will be (a + 2b) g mol^-1 and AB₄ will be (a + 4b) g mol^-1.
$\Delta T_f=\frac{K_f\times W_B\times 1000}{M_B\times W_A}$
For AB${}_2,2.3=\frac{5.1\times 1\times 1000}{\left(a+2b\right)\times 20}...\left(i\right)$
For AB${}_4,1.3=\frac{5.1\times 1\times 1000}{\left(a+4b\right)\times 20}...\left(ii\right)$
On solving (i) and (ii), we get
a = 25.59 and b = 42.64