Question

Two elements form compounds having molecular formula $A{B}_2$ and $A{B}_4$. When dissolved in 20 g of benzene, 1.0 g of $A{B}_4$ lowers it by 1.3 K. The molar depression constant for benzene is 5.1 K kg $mo{l}^{-1}$. Calculate the atomic mass of $A$ and $B$.

Answer 1

Given, $K_f=5.1$ K kg ${mol}^{-1}$

$W_1=20$ $W_2=1$

We know, ${\Delta T}_f=K_{f}m$

$m=\frac{W_2\times 1000}{M_2\times W_1}$

$\therefore$ $M_2=\frac{1000\times K_f\times W_2}{W_1\times {\Delta T}_f}$

for ${AB}_4,{\Delta T}_f=1.3$ K

$M_2=\frac{1000\times 5.1\times 1}{1.3\times 20}$

$M_2=196.15$ g ${mol}^{-1}$

let mass of $A=xg$

mass of $B=yg$

for ${AB}_4,x+4y=196.15g\to \left(1\right)$

for ${AB}_2,x+2y=M_2$

$M_2=\frac{1000\times 5.1\times 1}{2.3\times 20}$

$M_2=110.87g=x+2y\to \left(2\right)$

solving $\left(1\right)$ and $\left(2\right)$

$x+4y=196.15g$

$x+2y=11.87g$

mass of $A=x=25.5g$

mass of $B=y=42.6g$