Question

In transforming $0.01$ mole of $PbS$ to $PbSO _{4}$, the volume of ' $10$ volume' $H _{2} O _{2}$ required will be

• A. $11.2 \,mL$
• B. $22.4\, mL$
• C. $33.6\, mL$
• D. $44.8 \,mL$

Answer 1

The Correct Option is D

$PbS +4 H _{2} O _{2} \longrightarrow PbSO _{4}+4 H _{2} O$
From the above equation
$\because 1$ mole of PbS required $4$ moles of $H _{2} O _{2}$
$\therefore 0.01$ moles of PbS required $0.04$ mole of $H _{2} O _{2}$
Weight of $0.04$ mole $H _{2} O _{2}=1.36\, g$
$10$ volume $H _{2} O _{2}$ means,
$1\, mL$ of such solution of $H _{2} O _{2}$ on decomposition by heat produces $10 mL$ of oxygen at NTP.
$H _{2} O _{2}$ decomposes as,
$2 H _{2} O _{2} \longrightarrow 2 H _{2} O + O _{2}$
Thus, $1\, mL$ of $10$ volume $H _{2} O _{2}$ solution contains
$=\frac{68}{22400} \times 10 g \text { of } H _{2} O _{2}$
$=0.03035\, g H _{2} O _{2}$
$\because 0.03035\, g H _{2} O _{2}$ is present in $1\, mL$ of $10$ volume $H _{2} O _{2}$
$\therefore 1.36\, g\, H _{2} O _{2}$ present in $\frac{1}{0.03035} \times 1.36\, mL$ of 10 volume $H _{2} O _{2}$
$=44.81\, mL$