The sulphur in 1.0 g sample of steel is burned to sulphur dioxide and absorbed in 50.0 mL of 0.1 M sodium hydroxide solution. The excess sodium hydroxide requires 24.0 mL of 0.15 M hydrochloric acid solution for neutralization. The percentage of sulphur in the sample is

The reaction of $\mathrm{NaOH}$ and ${\mathrm{SO}}_2$ is $2\mathrm{NaOH}+{\mathrm{SO}}_2\to {\mathrm{Na}}_2{\mathrm{SO}}_3+{\mathrm{H}}_2\mathrm{O}$

Amount of $\mathrm{NaOH}$ to start with $=VM=\left(50\times {10}^{-3}\mathrm{L}\right)\left(0.1\mathrm{molL}=5\times {10}^{-3}\mathrm{mol}\right.$

Amount of $\mathrm{NaOH}$ after the absorption of ${\mathrm{SO}}_2=\left(24.0\times {10}^{-3}\mathrm{L}\right)\left(0.15{\mathrm{molL}}^{-1}\right)=3.6\times {10}^{-3}\mathrm{mol}$

Amount of $\mathrm{NaOH}$ used in absorbing 

${\mathrm{SO}}_2=5\times {10}^{-3}\mathrm{mol}-3.6\times {10}^{-3}\mathrm{mol}=1.4\times {10}^{-3}\mathrm{mol}$

Amount of ${\mathrm{SO}}_2$ formed

$=\left(0.5\right)\left(1.4\times {10}^{-3}\mathrm{mol}\right)=7.0\times {10}^{-4}\mathrm{mol}$

Amount of S in the steel $=7.0\times {10}^{-4}\mathrm{mol}$

Mass percent of $\mathrm{S}=\frac{\left(7.0\times {10}^{-4}\mathrm{mol}\right)\left(32\mathrm{g}{\mathrm{mol}}^{-1}\right)}{10\mathrm{g}}\times 100=2.24\mathrm{\%}$