# Question #fb7bc

Jun 8, 2017

Q-1
Molar mass of

${H}_{2} S {O}_{4} = \left(2 \times 1 + 32 + 4 \times 16\right) g \text{/"mol=98g"/} m o l$

So $0.25 M {H}_{2} S {O}_{4} \to 0.25 \times 98 g \text{/"L=24.5g"/} L$

So $2 L \text{ } 0.25 M {H}_{2} S {O}_{4}$ contains $24.5 \times 2 = 49 g {H}_{2} S {O}_{4}$

Now 80% (w/w) ${H}_{2} S {O}_{4}$ has sp.gr$1.75 g \text{/} m L$ So its 100g will have volume $\frac{100}{1.75} m L$. This volume contains 80g ${H}_{2} S {O}_{4}$.

49g ${H}_{2} S {O}_{4}$. will be obtained from $\frac{100}{1.75} \times \frac{49}{80} m L = 35 m L \text{ } {H}_{2} S {O}_{4}$ of 80%(w/w) sample.

Q-2

7% (w/w) ${H}_{2} S {O}_{4}$ contains 7g ${H}_{2} S {O}_{4}$ in 100g solution. This solution has density 1.4g/mL. So 100g solution will have volume $\frac{100}{1.4} m L$

So 200 mL of this solution will contain $\frac{7}{\frac{100}{1.4}} \times 200 g = 19.6 g \text{ } {H}_{2} S {O}_{4}$

Again 100mL 2.45% (w/v) ${H}_{2} S {O}_{4}$ solution will contain 2.45g${H}_{2} S {O}_{4}$

So total mass of ${H}_{2} S {O}_{4}$ present in 300mL mixture is $19.6 + 2.45 = 22.05 g$. Finally mixture is diluted to a volume of 500mL.
So 1L of this diluted solution contains $\frac{22.05}{500} \times 1000 g = 44.1 g {H}_{2} S {O}_{4}$

Hence molarity of the diluted solution $\frac{44.1 \frac{g}{L}}{98 g \text{/} m o l} = 0.45 M$