Question

What volume of $98$% sulphuric acid (in ml) should be mixed with water to obtain $200mL$ of $15$% solution of sulphuric acid by weight? Given density of $H_{2}O=1.00gc{m}^{-3}$, sulphuric acid $(98$%) $=1.88gc{m}^{-3}$ and sulphuric acid $(15$%) $=1.12gc{m}^{-3}$.

Answer 1

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Correct Answer - V1=18.2mL

1. In the case of dilution, the simplest way is to determine the normality of the 98% and 15% $H{}_{2}SO{}_4$ .

2. For 98% $H{}_{2}SO{}_4$, the volume of $H{}_{2}SO{}_4$ in 100g solution is = $\frac{100}{1.88}cm{}^3$.

3. Let normality of 98% $H{}_{2}SO{}_4$ is N1, and $N{}_1=\frac{1000\times W}{E\times V}$ , where W is the weight of $H{}_{2}SO{}_4$ (which is 98g), E is the equivalent weight of $H{}_{2}SO{}_4$ (which is 49g). V is the volume that we have calculated above that is $\frac{100}{1.88}cm{}^3$

3. $N_1=\frac{1000\times 98}{49\times \frac{100}{1.88}}$ = 37.6N.

4. For 15% $H{}_{2}SO{}_4$, the volume of $H{}_{2}SO{}_4$ in 100g solution is =$\frac{100}{1.12}cm{}^3$

5. Let normality of 15% $H{}_{2}SO{}_4$ is $N{}_2$ and $N_2=\frac{1000\times w}{E{}^1\times v}$ where $w$ is the weight of $H{}_{2}SO{}_4$(which is 98g) , E is the equivalent weight of $H{}_{2}SO{}_4$ which is 49g.

6.$N_2=\frac{1000\times 98}{49\times \frac{100}{1.12}}$ = 3.43N

6. As we know that $N{}_{1}V{}_1=N_2{V}_2$,

7. $\left(37.6\right)\times V{}_1=\left(3.43\right)\times 200$

8. $V_1=18.2mL$