Question

How many kilograms of wet NaOH containing $10\%$ water are required to prepare 60 litre of a 0.5 M solution?

• A. 1.23 kg
• B. 1.13 kg
• C. 1.33 kg
• D. 1.24 kg

Answer 1

Answer: (C)

1.33 kg

Moles required = $60\times 0.5=30$

30 moles of NaOH = $30\times 40$ = 1200 g of NaOH

but it contains only $90$%$NaOH$

Therefore amount of wet NaOH required =$\frac{100\times 1200}{90}\times {10}^{-3}$ = $1.33kg$