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NEET CHEMISTRYSolutionsMedium

Question

The freezing point depression constant for water is 1.86C m11.86 ^\circ\text{C m}^{-1}. If 5.00 g5.00 \text{ g} Na2SO4Na_2SO_4 is dissolved in 45.0 g45.0 \text{ g} H2OH_2O, the freezing point is changed by 3.82C-3.82 ^\circ\text{C}. The Van’t Hoff factor for Na2SO4Na_2SO_4 is:

A

2.63

B

3.11

C

0.381

D

2.05

Step-by-Step Solution

The depression in freezing point (ΔTf\Delta T_f) for an electrolyte is given by the equation: ΔTf=i×Kf×m\Delta T_f = i \times K_f \times m.

  1. Calculate Molality (mm): Molar mass of Na2SO4=2(23)+32+4(16)=142 g mol1Na_2SO_4 = 2(23) + 32 + 4(16) = 142 \text{ g mol}^{-1}. Moles of solute (n2n_2) = 5.00 g142 g mol10.0352 mol\frac{5.00 \text{ g}}{142 \text{ g mol}^{-1}} \approx 0.0352 \text{ mol}. Mass of solvent (w1w_1) = 45.0 g=0.045 kg45.0 \text{ g} = 0.045 \text{ kg}. m=0.0352 mol0.045 kg0.782 mm = \frac{0.0352 \text{ mol}}{0.045 \text{ kg}} \approx 0.782 \text{ m}.

  2. Calculate Van't Hoff Factor (ii): Given ΔTf=3.82C\Delta T_f = 3.82 ^\circ\text{C} (magnitude of change) and Kf=1.86C m1K_f = 1.86 ^\circ\text{C m}^{-1}. Rearranging the formula: i=ΔTfKf×mi = \frac{\Delta T_f}{K_f \times m}.

  • i=3.821.86×0.782=3.821.45452.626i = \frac{3.82}{1.86 \times 0.782} = \frac{3.82}{1.4545} \approx 2.626.

Rounding to two decimal places, i2.63i \approx 2.63.

Exam Context & Concepts Covered

This question aligns with the NEET CHEMISTRY syllabus, specifically targeting concepts from Solutions. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

CHEMISTRYSolutionsfreezingdepressionconstantcirctextcdissolved

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