Chapter 13, Problem 13.22QP

Interpretation Introduction

Interpretation:

The molality and the molarity of Sulphur acid solution is to be determined.

Concept introduction:

Number of moles from its given mass is,

$\text{Number of moles}\text{=}\frac{\text{Given}\text{mass}}{\text{Molecular}\text{mass}}$

Molality (m):

Molality is the number of moles of solute present in one kilogram of solvent.

$\text{Molality (m) =}\frac{\text{Number}\text{of}\text{moles}\text{of}\text{solute}}{\text{1kg}\text{of}\text{solvent}}$

Molarity (M):

Molarity is number of moles of the solute present in the one liter of the solution.

$\text{Molarity (M) =}\frac{\text{Number}\text{of}\text{moles}\text{of}\text{solute}}{\text{1}\text{liter}\text{of}\text{solution}}$

Answer to Problem 13.22QP

The molarity of Sulphur acid solution is $\text{18}\text{.28}M$.

The molality of Sulphur acid solution is $499.5\text{m}$.

Explanation of Solution

For molarity of sulfuric acid:

Given,

$\begin{array}{c}MasspercentofSulphuricacid=98\%\\ \text{Density of }Sulphuricacid\text{= }\text{1}\text{.83}\text{g/mL}\\ MolecularmassofSulphuricacid=98.079g\end{array}$

Amount of sulphuric acid divided into its molecular mass to gives mole. Moles are divided by liter to get molarity.

$MassofSulphuricacidin1Lofthesolution=\frac{1830g}{L}\times 0.98=1793.4g$

$\frac{\text{1793}\text{.4}\text{g/L}}{\text{98}\text{.079}\text{g}}\text{=}\text{18}\text{.28}M$

Therefore,

The molarity of sulphuric acid solution is $\text{18}\text{.28}M$

For molality of sulfuric acid:

Given,

$\begin{array}{c}MasspercentofSulphuricacid=98\%\\ \text{Density of }Sulphuricacid\text{= }\text{1}\text{.83}\text{g/mL}\\ MolecularmassofSulphuricacid=98.079g\end{array}$

Subtract the mass of sulphuric acid from mass of the solution to get mass of water. Moles of sulphuric acid divided by mass of water to get molality

$\begin{array}{c}Massofwater=1830gsolution-1793.4g\\ =36.6g\\ =0.0366kg\end{array}$

The molarity of sulphuric acid solution is $\text{18}\text{.28}M$ which means $1L$ of the solution contains $\text{18}\text{.28}\text{mol}$ of sulfuric acid.

Therefore,

The molality of sulphuric acid solution,

$\begin{array}{c}\text{Molality (m) =}\frac{\text{Number}\text{of}\text{moles}\text{of}\text{solute}}{\text{1kg}\text{of}\text{solvent}}\\ \text{=}\frac{\text{18}\text{.28}\text{mol}}{0.0366kg}\\ =499.5\text{m}\end{array}$

The molality of sulfuric acid solution is $499.5\text{m}$

Conclusion

The molarity of sulfuric acid solution was determined as $\text{18}\text{.28}M$.

The molality of sulfuric acid solution was determined as $499.5\text{m}$.