Question

A 13% solution (by weight) of sulphuric acid with a density of 1.02g/ml. To what volume should 100$mL$ of this acid be diluted in order to prepare a 1.5N solution?

Answer 1

Molarity is number of moles of solute present in 1 liter of solution. Density is the ratio of mass to volume. Normality is a product of valency and molarity. For dilution following formula can be used, ${{N}_{1}}{{V}_{1}}={{N}_{2}}{{V}_{2}}$.

Complete Solution :
-molarity is the number of moles of solute present in one liter of solution.
-number of moles of solute can be calculated by dividing weight of solute to molecular mass of solute.
As $\%$ weight indicates weight of solute in grams present in 100 gm of solution. d is density.
-13$\%$w/v indicates 13 gm of solute is present in 100 gm of solution. As solution is composed of solvent and solute so 87gm of solvent is present in solution.
-Molecular mass of sulphuric acid is 98g/mol.

& M=\dfrac{\text{Weight of solute}}{molecular\text{ mass of solute}\times \text{Volume of solution in liters}} \\\ & M=\dfrac{\%W/W\times 10\times \text{d}}{molecular\text{ mass of solute}} \\\ \end{aligned}$$ $M=\dfrac{13\times 10\times 1.02}{98}$=1.35M -As we know normality is obtained by multiplying valency and molarity. -Valency of sulphuric acid is 2 as it is a strong acids and it can donate two protons so $N=M\times Valency=1.35\times 2$=2.7N A 13$\%$ solution (by weight) of sulphuric acid with a density of 1.02g/ml has normality as 2.7N. On dilution, resultant volume is calculated using following formula: $$\begin{aligned} & {{N}_{1}}{{V}_{1}}={{N}_{2}}{{V}_{2}} \\\ & 2.70\times 100=1.5\times {{\text{V}}_{2}} \\\ \end{aligned}$$ ${{V}_{2}}=180mL$ 100$$mL$$ of this acid be diluted to $$180mL$$ in order to prepare a 1.5N solution. **Note:** following formula can be used to relate molarity, % by weight and density. Normality is obtained by multiplying molarity and valency. $$\begin{aligned} & M=\dfrac{\text{Weight of solute}}{molecular\text{ mass of solute}\times \text{Volume of solution in liters}} \\\ & M=\dfrac{\%\W/W\times 10\times \text{d}}{molecular\text{ mass of solute}} \\\ \end{aligned}$$ Resultant normality and volume can be calculated using following formula: ${{N}_{1}}{{V}_{1}}={{N}_{2}}{{V}_{2}}$