Question

Which among the following has minimum molarity?

• A. 10 g of NaOH in 100 ml solution
• B. 49 g of $H_{2}S{O}_4$ in 500 ml solution
• C. 73 g of HCI in 1000 ml solution
• D. 74 g of $Ca(OH{)}_2$ in 2000 ml solution

Answer 1

The correct option is D 74 g of $Ca(OH{)}_2$ in 2000 ml solution

Molarity of different solutions can be calculated as:
A) 10g of NaOH in 100ml solution:
M = $\frac{10}{40\times 100}\times 1000$ = 2.5M
B) 49g of $H_{2}S{O}_4$ in 500ml solution:
M = $\frac{49}{98\times 500}\times 1000$ = 1M
C) 73g of HCl in 1000ml solution:
M = $\frac{73}{36.5\times 1000}\times 1000$ = 2M
D) 74g of $Ca(OH{)}_2$ in 2000ml solution:
M = $\frac{74}{74\times 2000}\times 1000$ = 0.5M
Therefore option (D) has least molarity