Question

Two elements $A$ and $B$ form compounds of formula $AB_2$ and $AB_4$. When dissolved in $20.0\, g$ of benzene $1.0 \,g$ of $AB_2$ lowers f. pt. by 2.3$^{\circ}$C whereas $1.0\, g$ of $AB_4$ 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)

25, 42

Answer 2

Let the masses of A and B be a and b. The mass of AB will be (a + 2b) g mol and AB will be (a + 4b) g mol. $\Delta T_{ f} =\frac{K_{ f} \times W_{B}\times1000}{M_{B}\times W_{A}}$ For AB$_{2}, 2.3=\frac{5.1\times1\times1000}{\left(a+2b\right)\times20}\quad\quad\quad\quad\quad...\left(i\right)$ For AB$_{4}, 1.3=\frac{5.1\times1\times1000}{\left(a+4b\right)\times20}\quad \quad \quad \quad \quad ...\left(ii\right)$ On solving (i) and (ii), we get $\quad\quad$ a = 25.59 and b = 42.64