Read the passage carefully and answer the Questions.
A homogeneous mixture of two or more than two components has uniform composition and properties. The binary solutions have two components, which may be solid, liquid or gas. The resultant solution has a larger amount of solvent and a smaller amount of solute. The amount of solute in a given amount of solvent decides the concentration of the resulting solution. The concentration can be expressed in various ways, like molarity, molality, normality and mole fraction.
The molality of a 1 liter aqueous solution of 93% \( H_2SO_4 \) (mass/volume) is:
(Given: density of the solution = \( 1.84\, g\, mL^{-1} \), molar mass of \( H_2SO_4 = 98\, g\, mol^{-1} \))
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(10.43\, mol\, kg^{-1}\)
Molality (m) is the number of moles of solute per kilogram of solvent (not of solution):
\( \displaystyle m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \)
A 93% (mass/volume) solution has 93 g of \( \ce{H2SO4} \) in every 100 mL of solution. For 1 L (1000 mL) of solution:
Mass of \( \ce{H2SO4} \) = 930 g, so moles of \( \ce{H2SO4} \) = \( \frac{930}{98} = 9.49\,\mathrm{mol} \)
Mass of the solution = volume × density = \( 1000\,\mathrm{mL} \times 1.84\,\mathrm{g\,mL^{-1}} = 1840\,\mathrm{g} \)
Mass of water (solvent) = \( 1840 - 930 = 910\,\mathrm{g} = 0.910\,\mathrm{kg} \)
\( \displaystyle m = \frac{9.49\,\mathrm{mol}}{0.910\,\mathrm{kg}} = 10.43\,\mathrm{mol\,kg^{-1}} \)
So the molality is 10.43 mol kg−1.