The mass of $\mathrm{Na}_{2} \mathrm{CO}_{3}$ (91.2% purity) required for the neutralisation of $45.6…

The mass of $\mathrm{Na}_{2} \mathrm{CO}_{3}$ (91.2% purity) required for the neutralisation of $45.6 \mathrm{~mL}$ of $0.25 \mathrm{M}~ \mathrm{HC} 1$ solution is
  1. $0.6625 \mathrm{~g}$
  2. $0.4265 \mathrm{~g}$
  3. $0.5765 \mathrm{~g}$
  4. $0.8473 \mathrm{~g}$

Solution

The reaction is $\quad \mathrm{Na}_{2} \mathrm{CO}_{3}+2 \mathrm{HCl} ightarrow 2 \mathrm{NaCl}+\mathrm{H}_{2} \mathrm{CO}_{3}$
Amount of $\mathrm{HCl}$ to be neutralised $=V \times M=\left(45.6 \times 10^{-3} \mathrm{~L}ight)\left(0.25 \mathrm{~mol} \mathrm{~L}^{-1}ight)=11.4 \times 10^{-3} \mathrm{~mol}$
Amount of $\mathrm{Na}_{2} \mathrm{CO}_{3}$ required $=0.5 \times 11.4 \times 10^{-3} \mathrm{~mol}=5.7 \times 10^{-3} \mathrm{~mol}$
Mass of $\mathrm{Na}_{2} \mathrm{CO}_{3}$ required $=\left(5.7 \times 10^{-3} \mathrm{~mol}ight)\left(106 \mathrm{~g} \mathrm{~mol}^{-1}ight)(100 / 91.2)=0.6625 \mathrm{~g}$

Asked in: JEE-TOPICTESTS-CHEMISTRY

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