Question

$2.5\, g$ of the carbonate of a metal was treated with $100 \,ml$ of $1 \,N H _{2} SO _{4}$. After the completion of the reaction, the solution was boiled off to expel $CO _{2}$ and was then titrated against $1 \,N\, NaOH$ solution. The volume of alkali that would be consumed, if the equivalent weight of the metal is $20$

• A. 50 mL
• B. 25 mL
• C. 75 mL
• D. 100 mL

Answer 1

Answer: (A)

Equivalent weight of metal carbonate

$=20+30=50$

$2.5 \,g$ of metal carbonate $=\frac{2.5}{50}=0.05\, eq$

Number of equivalent of $H _{2} SO _{4}$ would have reacted $=0.05$

Number of equivalent of $H _{2} SO _{4}$ taken

$=\frac{100 \times 1}{1000}=0.1$

Number of equivalent of $H _{2} SO _{4}$ remains unreacted

$=0.1-0.05=0.05\, eq$

$\therefore \,\,\,$ Number of equivalent of alkali consumed $=0.05\, eq$

milli eq. $=$ Normality $\times$ Volume in $mL$

$\therefore 1.0 \times V=0.05 \times 1000$

$V=\frac{0.05 \times 1000}{1.0}=50\, ml$