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Q.Calculate the mass of ascorbic acid (Molar mass = 176 g mol−1mol^{-1}) to be dissolved in 75 g of acetic acid, to lower its freezing point by 1⋅5 °C1·5\,°C. (Kf=3⋅9K_f = 3·9 K kg mol−1mol^{-1})

CBSECBSE Class XII Board 2020Subjective· 3mImportance★★★★★
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This is a classic colligative-properties problem where the freezing-point depression ΔTf\Delta T_f is directly proportional to the molality of the solute. Using ΔTf=Kf⋅m\Delta T_f = K_f \cdot m, we find the required molality, then convert it to mass of ascorbic acid. The answer is 5.08 g.

The key idea here is that freezing-point depression depends only on the number of solute particles, not on their identity. That’s why we can use the formula ΔTf=Kf⋅m\Delta T_f = K_f \cdot m, where mm is the molality (moles of solute per kg of solvent). Once we know the molality, we can find the moles of ascorbic acid needed, and then its mass using the given molar mass.

Let’s walk through it step by step.

  1. Write down the freezing-point depression formula The relationship is:

ΔTf=Kf⋅m\Delta T_f = K_f \cdot m

where ΔTf\Delta T_f is the depression in freezing point, KfK_f is the cryoscopic constant (given as 3.9 K kg mol−13.9\ \text{K kg mol}^{-1}), and mm is the molality of the solution.

  1. Plug in the known values We are told ΔTf=1.5 °C\Delta T_f = 1.5\ °\text{C} (which is numerically equal to 1.5 K1.5\ \text{K}, since a change of 1 °C1\ °\text{C} equals a change of 1 K1\ \text{K}). So:

1.5=3.9×m1.5 = 3.9 \times m

Solving for mm:

m=1.53.9=1539=513≈0.3846 mol kg−1m = \frac{1.5}{3.9} = \frac{15}{39} = \frac{5}{13} \approx 0.3846\ \text{mol kg}^{-1}

  1. Interpret molality in terms of the given solvent mass Molality mm is moles of solute per kilogram of solvent. Here the solvent is acetic acid, with mass 75 g=0.075 kg75\ \text{g} = 0.075\ \text{kg}. So the number of moles of ascorbic acid required is:

moles=m×mass of solvent (in kg)=513×0.075\text{moles} = m \times \text{mass of solvent (in kg)} = \frac{5}{13} \times 0.075

Calculate:

513×0.075=5×0.07513=0.37513=37513000=3104≈0.028846 mol\frac{5}{13} \times 0.075 = \frac{5 \times 0.075}{13} = \frac{0.375}{13} = \frac{375}{13000} = \frac{3}{104} \approx 0.028846\ \text{mol}

  1. Convert moles to mass using the molar mass Molar mass of ascorbic acid is 176 g mol−1176\ \text{g mol}^{-1}. So: mass=moles×molar mass=3104×176\text{mass} = \text{moles} \times \text{molar mass} = \frac{3}{104} \times 176 …

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