Question

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

• A. $53.54$%
• B. $42.07$%
• C. $69$%
• D. $31$%

Answer 1

The correct option is A $53.54$%

During crystallization, 5% loss in the acceptable expected field =164 -69 = 95g

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

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

$\Rightarrow x=4.75$g

Actual yield =95-4.75 =90.25g

Percentage $NaHC{O}_{3}recoverable=\frac{90.25}{164}\times 100$ = 55.03 %

Hence, correct option is A.