Question

When $4.08$ g of a mixture of $BaO$ and unknown carbonate $MC{O}_3$ was heated strongly, the residue weighed $3.64$ g. This was dissolved in $100$ mL of $1N$ $HCl$. The excess acid required $16$ mL of $2.5N$ $NaOH$ solution for complete neutralisation. Identify the metal $M$.

• A. Chromium
• B. Nickel
• C. Zinc
• D. Calcium
  

Answer 1

The correct option is D Calcium

Weight of $MC{O}_3$ and $BaO=4.08g$ (given)

Weight of residue $=3.64g$ (given)

$\therefore$ Weight of $C{O}_2$ evolved on hating $=(4.08-3.64)g=0.44g$

$=0.44/44=0.01mole$

Number of moles of $MC{O}_3=0.01mole$

$[\because MC{O}_3\stackrel{heat}{\to }MO+C{O}_2]$

volume of $1NHCl$ in which residue is dissolved $=100ml$

volume of $1NHCl$ used for dissolution $=(100-2.5\times 16)ml=60ml$

$=\frac{60}{1000}=0.06$ equivalents

The chemical equation for dissolution can be written as

$\underset{Residue}{\underset{}{BaO+MO}}+4HCl\to BAC{l}_2+MC{l}_2+2H_{2}O$

[Number of moles of $BaO$ and $MO=1+1=2$]

Number of moles of $BaO+$ Number of moles of $MO=\frac{0.06}{2}=0.03$

Number of moles of $BaO=(0.03-0.01)=0.02$ moles

Molecular weight of $BaO=138+16=154$

$\therefore$ Weight of $BaO=(0.02\times 154)g$

$=30.8g$

Weight of $MC{O}_3=(4.08-3.08)=1.0g$

Since weight of $0.01$ mole of $MC{O}_3=1.0g$

$\therefore$ Mol. wt. of $MC{O}_3=\frac{1}{0.01}=100$

Hence atomic weight of unknown $M=(100-60)=40$

The atomic weight of metal is $40$ so the metal $M$ is $Ca$