NVAMediumJEE 2023Colligative Properties

JEE Chemistry 2023 Question with Solution

Sea water contains 29.29%29.29\% NaCl and 19%19\% MgCl2_2 by weight of solution. The normal boiling point of the sea water is:

Answer

Correct answer:113

Step-by-step solution

Standard Method

Given: Sea water contains 29.29%29.29\% NaCl and 19%19\% MgCl2_2 by weight of solution.

Find: The normal boiling point of the sea water.

Use elevation in boiling point:

ΔTb=iKbm\Delta T_b = iK_bm

From the extracted working,

ΔTb=(23×2×1000(58.5)×57.5+3×19.5×100095×57.5)×0.52\Delta T_b = \left(\frac{23 \times 2 \times 1000}{(58.5) \times 57.5} + \frac{3 \times 19.5 \times 1000}{95 \times 57.5}\right) \times 0.52 =(7.86+6.16)×0.5257.5×10012.66= \frac{(7.86 + 6.16) \times 0.52}{57.5} \times 100 \simeq 12.66

Therefore, boiling point 113C\simeq 113^\circ \text{C}.

So, the required answer is 113113.

Common mistakes

  • Using the percentage composition directly as molality is incorrect because boiling point elevation depends on molality, not mass percent. First convert the given percentages into masses of solutes and solvent.

  • Ignoring the van't Hoff factor is wrong because NaCl and MgCl2_2 dissociate in solution. Their particle contributions must be counted through the factor ii while applying ΔTb=iKbm\Delta T_b = iK_bm.

  • Taking the whole solution mass as solvent mass is incorrect. The solvent mass must exclude the dissolved salts; otherwise the molality is underestimated or overestimated.

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