Question

Two elements $A$ and $B$ form compounds having molecular formulae $A{B}_2$ and $A{B}_4.$ When dissolved in $20.0$ g of benzene, $1.0$ g of $A{B}_2$ lowers the freezing point by ${2.3}^{o}C$, whereas $1.0$ g of $A{B}_4$ lowers the freezing point by ${1.3}^{o}C$. The molal depression constant for benzene in $1000$ g is $5.1$. Calculate the atomic masses of $A$ and $B$.

Answer 1

We know that,
$m=\frac{1000K_f\times w}{W\times △T}$
Molecular mass of $A{B}_2$ (from given data) = $\frac{1000\times 5.1\times 1}{20\times 2.3}$
$=110.86$
and Molecular mass of $A{B}_4$ (from given data) $=\frac{100\times 5.1\times 1}{1.3\times 20}$
$=196.15$
Further, $A{B}_4=A+4B=196.15$ ...(i)
$A{B}_2=A+2B=110.86$ ...(ii)
Subtracting eq. (ii) from (i),
$2B=85.29$
$B=42.645$
Putting the value of B in eq. (ii),
$A+85.29=110.86$
or $A=110.86-85.29=25.57$
Thus, the atomic masses of $A$ and $B$ are $25.57$ and $42.645$ respectively.