Question

# The concentrated sulfuric acid we use in the laboratory is 98% H_2SO_4 by mass. Calculate the molality and molarity of the acid solution. The density of the solution is 1.83 \frac{g}{mL} answers are 5\times10^{-2} m, 18.3M

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The concentrated sulfuric acid we use in the laboratory is 98% $$\displaystyle{H}_{{2}}{S}{O}_{{4}}$$ by mass. Calculate the molality and molarity of the acid solution. The density of the solution is 1.83 $$\displaystyle{\frac{{{g}}}{{{m}{L}}}}$$ answers are $$\displaystyle{5}\times{10}^{{-{2}}}$$ m, 18.3M

2020-11-03
Mass percent of $$\displaystyle{H}_{{2}}{S}{O}_{{4}}$$ solution=98%
98%(m/m) $$\displaystyle{H}_{{2}}{S}{O}_{{4}}$$ solution means 98 g of $$\displaystyle{H}_{{2}}{S}{O}_{{4}}$$ solute is present per 100 g of solution.
Mass of $$\displaystyle{H}_{{2}}{S}{O}_{{4}}$$ solute =98 g
Mass of water =(100-98)g
$$\displaystyle={2}{g}\times{\frac{{{1}\ {k}{g}}}{{{1000}\ {g}}}}$$
=0.002 kg
Mass of $$\displaystyle{H}_{{2}}{S}{O}_{{4}}$$ solute =98 g
Molar mass of $$\displaystyle{H}_{{2}}{S}{O}_{{4}}$$ solute =98.09 g/mol
No of moles of $$\displaystyle{H}_{{2}}{S}{O}_{{4}}$$ solute $$\displaystyle={98}{g}\times{\frac{{{1}{m}{o}{l}}}{{{98.09}{g}}}}$$
=0.999 mol
$$\displaystyle\text{Molality}={\frac{{\text{No of moles of }\ {H}_{{2}}{S}{O}_{{4}}\ \text{ solute}}}{{\text{mass of }\ {H}_{{2}}{O}\ \text{ solvent}}}}$$
$$\displaystyle={\frac{{{0.999}\ {m}{o}{l}}}{{{0.002}\ {k}{g}}}}$$
$$\displaystyle={5.0}\times{10}^{{2}}$$ mol/kg
$$\displaystyle={5.0}\times{10}^{{2}}$$ m
Therefore, the molality of the 98% $$\displaystyle{H}_{{2}}{S}{O}_{{4}}$$ solution is $$\displaystyle{5.0}\times{10}^{{2}}$$ m
Mass of $$\displaystyle{H}_{{2}}{S}{O}_{{4}}$$ solution=100g
Density of $$\displaystyle{H}_{{2}}{S}{O}_{{4}}$$ solution=1.83$$\displaystyle{\frac{{{g}}}{{{m}{L}}}}$$
Volume of solution $$\displaystyle={100}\ {g}\ {H}_{{2}}{S}{O}_{{4}}\times{\frac{{{1}\ {m}{L}}}{{{1.83}\ {g}\ {H}_{{2}}{S}{O}_{{4}}}}}\times{\frac{{{1}\ {L}}}{{{1000}\ {m}{L}}}}$$
=0.999 mol
$$\displaystyle\text{Molality}={\frac{{\text{No of moles of }\ {H}_{{2}}{S}{O}_{{4}}\ \text{ solute}}}{{\text{Volume of solution}}}}$$
$$\displaystyle={\frac{{{0.999}\ {m}{o}{l}}}{{{0.0546}\ {L}}}}$$
=18.3 mol/L
=18.3 M
Therefore, the molarity of $$\displaystyle{H}_{{2}}{S}{O}_{{4}}$$ solution is 18.3 M