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

Hint: Molarity is the 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 step by step:
-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.
\[\begin{align}
  & M=\frac{\text{Weight of solute}}{molecular\text{ mass of solute}\times \text{Volume of solution in liters}} \\
 & M=\frac{%W/W\times 10\times \text{d}}{molecular\text{ mass of solute}} \\
\end{align}\]
$M=\frac{13\times 10\times 1.02}{98}$=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:
\[\begin{align}
  & {{N}_{1}}{{V}_{1}}={{N}_{2}}{{V}_{2}} \\
 & 2.70\times 100=1.5\times {{\text{V}}_{2}} \\
\end{align}\]
${{V}_{2}}=180mL$
100mL 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.
\[\begin{align}
  & M=\frac{\text{Weight of solute}}{molecular\text{ mass of solute}\times \text{Volume of solution in liters}} \\
 & M=\frac{%W/W\times 10\times \text{d}}{molecular\text{ mass of solute}} \\
\end{align}\]
Resultant normality and volume can be calculated using following formula: ${{N}_{1}}{{V}_{1}}={{N}_{2}}{{V}_{2}}$