Question

$1g$ of bivalent metal carbonate on heating gives $0.5g$ metal oxide. The molar mass of metal sulphate would be: (in g/mol).

• A. $124$
• B. $110$
• C. $156$
• D. $210$

Answer 1

The correct option is A $124$

Given,

$1g$ of bivalent metal carbonate gives

$0.5g$ of metal oxides on heating

Let metal carbonate be$MC{O}_3$ & Metal Oxide be $MO$

The reaction involved is

$MC{O}_3⟶MO+C{O}_2$

Let the mass of metal $M^{2+}$ be $Xg/mol$

Molar mass of $C{O}_3^{2-}$ is $60g/mol$

Molar mass of $O^{2-}$ is $16g/mol$

$\Rightarrow$ Molar mass of $MC{O}_3=(x+60)g/mol$

Molar mass of $MO=(x+16)g/mol$

Also, Moles of $MC{O}_3$ given=$\frac{1g}{(x+60)g/mol}⟶\left(1\right)$

Moles of $MO$ given=$\frac{0.5g}{(x+16)g/mol}⟶\left(2\right)$

Now,

According to reaction above

$1$ mole of $MC{O}_3$ produces $1$ mole of $MO$

$\Rightarrow$ Moles of $MC{O}_3$=Moles of $MO$

From $\left(1\right)$ & $\left(2\right)$ $\frac{1}{(x+60)}mol=\frac{0.5}{(x+16)}mol$

$\Rightarrow 2x+32=x+60$

$=x=28$

$\therefore$ Molar mass of metal $M^{2+}$ is $x=28g/mol$

Now, we know

Molar mass of $S{O}_4^{2-}$ is $96g/mol$

$\Rightarrow$ Molar mass of $MS{O}_4=(28+96)g/mol$

$=124g/mol$.