Question

A mineral consists of an equimolar mixture of the carbonates of two bivalent metals. One metal is present to the extent of $15.0\%$ by weight. $3.0$ g of the mineral on heating lost $1.10$ g of $C{O}_2$. The percent by weight of other metal is :

• A. $65$
• B. $25$
• C. $75$
• D. $35$

Answer 1

The correct option is D $35$

$MC{O}_3\to MO+C{O}_2\uparrow$
$M^{\prime }C{O}_3\to M^{\prime }O+C{O}_2\uparrow$
Equivalent of $C{O}_2=$ Equivalent of carbonates of metals
$1$ mol of $C{O}_2\equiv 1$ mol of $C{O}_3^{2-}$
$44$ g of $C{O}_2\equiv$$60$ g of $C{O}_3^{2-}$
$1.10$ g of $C{O}_2\equiv \frac{60}{44}\times 1.10=1.5$ g of $C{O}_3^{2-}$
% of $C{O}_3^{2-}=\frac{1.5\times 100}{3}=50$%
% of one metal $=$ $15$ %
% of another metal $=100-50-15=35$%