Question

A mixture contains equimolar quantities of carbonates of two bivalent metals. One metal is present to the extent of $13.5\%$ by weight in the mixture and $2.58$g of the mixture on heating leaves a residue of $1.35$g. What percentage by weight of the other metal is there?

• A. $31.6$
• B. $21.3$
• C. $19.2$
• D. $17.8$

Answer 1

The correct option is B $21.3$

Let equimolar metal carbonates have mole be $n$

Let Metal be $M_1$ and $M_2$

As $M_1{CO}_3\underrightarrow{\Delta }{M}_{1}O+{CO}_{2_{\left(g\right)}}\uparrow$

$M_2{CO}_3\underrightarrow{\Delta }{M}_{2}O+{CO}_{2_{\left(g\right)}}\uparrow$

Let molecular wt. of $M_1\to w_1$

molecular wt. of $M_2\to w_2$

for % composition of metal $M_1=\frac{n{w}_1}{n(w_1+60)+n(w_2+60)}=0.135-\left(1\right)$

$\Rightarrow$ we have $n(w_1+60)+n(w_2+60)=2.58gm-\left(2\right)$

$C{O}_2$ librated$=2.58-1.35=1.23gm$

$\therefore$ wt. of $C{O}_2$ in mixture$=n\left(44\right)+n\left(44\right)$

$=88ngm$

$88ngm=1.23gm$

$n=\frac{1.23}{88}=0.01397$

Put $n=0.01397$ mol in $\left(1\right)$ we get

$\frac{0.01397w_1}{2.58}=0.135$

$[w_1=24.93gm]$

From $\left(2\right)$

$n(w_2+w_1+120)=2.58$

$0.01397(24.93+w_2+120)=2.58$

$2.0246+0.01397w_2=2.58$

$[w_2=39.75gm]$

% composition of $M_2=\frac{n{w}_2}{2.58}=\frac{0.01397\times 39.75}{2.58}=0.215$

$=21.5$%