Question

Sodium bicarbonate, $NaHC{O}_3$, can be purified by dissolving it in hot water $\left({60}^{\circ }C\right)$, filtering to remove insoluble impurities, cooling to $0^{\circ }C$ to precipitate solid $NaHC{O}_3$, and then filtering to remove the solid, leaving soluble impurities in solution. Any $NaHC{O}_3$ that remains in the solution is not recovered. The solubility of $NaHC{O}_3$ in hot water at ${60}^{\circ }C$ is $164g/litre$ and is $69g/litre$ in cold water at $0^{\circ }C$. What is the percentage yield of $NaHC{O}_3$, when it is purified by this method?

• A. $55.34$%
• B. $42.07$%
• C. $69$%
• D. $31$%

Answer 1

The correct option is A $55.34$%

During crystallization $5\%$ loss in acceptable expected yied $=164-69=95g$

$5\%=\frac{x}{95}\times 100$

$x=\frac{5\times 95}{100}=4.75g$

actual yield will get $=95-4.75=90.25g$

$\%NaHC{O}_3$ recverable $=\frac{90.25}{164}\times 100$

$=55.03\%$ option $\left(a\right)$