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A mineral consists of an equimolar mixture of the carbonates of two bivalent metals. One metal is present to the extent of 12.5% by weight. 2.8 gm of the mineral on heating lost 1.32 gm of ${CO}_{2}$ . What is the % by weight of the other metal?

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23.21

65.11

87.5

40.35

answer is B.

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Detailed Solution

$\begin{array}{lc} a mol & a mol \\ {MCO}_{3}, & {NCO}_{3} \\ ↓ Δ & ↓ Δ \\ M O + {CO}_{2} & NO + {CO}_{2} \end{array}$

$\begin{array}{l} 2 a \times 44 = 1.32 \\ \Rightarrow a = 0.015 mole \end{array}$

Mass of carbonate = 0.015 x 2 x 60 = 1.8g

Mass of metals = 1.0 g

Wt. of M $= \frac{12.5}{100} \times 2.8 = 0.35 g$

Wt. of N = 0.65 g

%N = 0 · 65 x 100 = 23.21 %

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