Question

If 250 mL of a solution contain 41 g of $H_{3}P{O}_3$ then the molarity and normality respectively are
(Given: Molecular weight of $H_{3}P{O}_3=82g$)

• A. 0.5 M, 1N
• B. 2 M, 4 N
• C. 1 M, 2 N
• D. 2 M, 3 N

Answer 1

Answer: (B)

2 M, 4 N

Moles of solute = $\frac{41}{82}=0.5$
volume of solution = $\frac{250}{1000}=0.25L$
Molality = $\frac{0.5}{0.25}=2M$
Similarly normality = $moles\times valencyfactor=2\times 2=4N$
Because in $H_{3}P{O}_3$ , only two hydrogen ions are available on dissociation because $H_{3}P{O}_3$ is diprotic acid.