Q71 / 180 · AIIMS 2009 · Chemistry
$2.5 \mathrm{~g}$ of the carbonate of a metal was treated with $100 \mathrm{~mL}$ of $1 \mathrm{~N} \mathrm{H}_2 \mathrm{SO}_4$. After the completion of the reaction, the solution was boiled off to expel $\mathrm{CO}_2$ and was then titrated against $1 \mathrm{~N} \mathrm{NaOH}$ solution. The volume of alkali that would be consumed, if the equivalent weight of the metal is 20
- Correct answerA$50$
- B$25$
- C$75$
- D100.
Solution
Equivalent weight of metal carbonate $=20+30=50$
$2.5 \mathrm{~g}$ of metal carbonate $=\frac{2.5}{50}=0.05 \mathrm{eq}$.
Number of equivalent of $\mathrm{H}_2 \mathrm{SO}_4$ would have reacted $=0.05$
Number of equivalent of $\mathrm{H}_2 \mathrm{SO}_4$ taken
$=\frac{100 \times 1}{1000}=0.1$
Number of equivalent of $\mathrm{H}_2 \mathrm{SO}_4$ remains unreacted
$=0.1-0.05=0.05 \mathrm{eq}$.
$\therefore$ Number of equivalent of alkali consumed $=0.05 \mathrm{eq}$.
milli eq. $=$ Normality $\times$ Volume in $\mathrm{mL}$
$\begin{aligned} & \therefore \quad 1.0 \times V=0.05 \times 1000 \\ & V=\frac{0.05 \times 1000}{1.0}=50 \mathrm{~mL}\end{aligned}$
Hence, the correct answer is option: 'A'.