Question

If 250 mL of water is added to 100 mL of a 9.8% $\left(\frac{w}{w}\right)H_{2}S{O}_4$ solution (d= 1.5 g/mL), the normality of the resulting solution will be __.

• A. 0.380 N
• B. 0.856 N
• C. 0.180 N
• D. 0.280 N

Answer 1

The correct option is B 0.856 N

Let the mass of solution to be 100 g.
9.8% by mass $H_{2}S{O}_4$ means 9.8 g solute in 100 g of the solution.
Density = $\frac{\text{mass}}{\text{volume}}$;
volume of the solution = $\frac{100}{1.5}$ mL
Moles of $H_{2}S{O}_4=\frac{9.8}{98}=0.1mol$

Molarity = $\frac{\text{moles of solute}}{\text{total volume of solution in mL}}\times 1000$

$=\frac{0.1}{\frac{100}{1.5}}\times 1000=1.5M$

$M_1{V}_1=M_2{V}_2$
$(1.5\times 100)=(M\times 350)$
Molarity of the solution is 0.428 M.
Normality = $2\times 0.428$ = 0.856 N