Question

The weight of sodium carbonate required to prepare $500mL$ of a seminormal solution is

• A. 6.125 g
• B. 26.5 g
• C. 53 g
• D. 13.25 g

Answer 1

Answer: (D)

$N=\frac{w\times 1000}{eq.wt.\times \text{volume in ml.}}$

$\left(eq.wt.=\frac{106}{2}=53\right)$

or $w=\frac{0.5\times 53\times 500}{1000}=13.25$.