Question

Calculate the weight of the precipitate of $BaC{O}_3\left(s\right)$ obtained from $50$ mL of the above test solution.

[Atomic mass of $Ba=137$ g, $C=12$ g and $O=16$ g while molecular weight of $\left(BaC{O}_3\right)=197$ g/mol]

• A. $3.94$ g
• B. $0.394$ g
• C. $1.97$ g
• D. $0.197$ g

Answer 1

The correct option is C $1.97$ g

Milliequivalents of $KOH$ (total) $=500\times 0.1\times 1=50$

Milliequivalents of $HCl$ for $KOH$ ($50$ mL) $=30\times 0.1=3$
MIlliequivalents of excess $KOH$ in $500$ mL $=\frac{3}{50}\times 500=30$

Milliequivalents of $KOH$ reacted $=50-30=20=$ milliequivalents of $C{O}_2$

Millimoles of $C{O}_2=\frac{20}{2}=10=10\times {10}^{-3}$

mol ($1$ mol of $C{O}_2=22.4$ L at STP)

ppm of $C{O}_2=\frac{\left[\right(10\times {10}^{-3}\times 22.4)\times \text{/}{10}^3]\times {10}^6}{224\times \text{/}{10}^3}$ $=1000$

Milliequivalents of $C{O}_2=$ milliequivalents of $C{O}_3^{2-}=$

milliequivalents of $BaC{O}_3=20$

Weight of $BaC{O}_3=20\times {10}^{-3}\times \frac{197}{2}=1.97$ g

Hence, the correct option is $\text{C}$