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NEET CHEMISTRYSolutionsMedium

Question

1% (w/w) solution of a compound is isotonic with 5% (w/w) sucrose (sugar) solution. The molecular weight of the compound will be:

A

32.4

B

68.4

C

129.6

D

34.2

Step-by-Step Solution

Two solutions are isotonic if they have the same osmotic pressure at the same temperature . Osmotic pressure (Π\Pi) is given by Π=CRT\Pi = CRT, where CC is the molar concentration (nV\frac{n}{V}). For isotonic solutions: C1=C2C_1 = C_2 (assuming effectively equal densities for dilute solutions, molarity ≈\approx molality for calculation purposes or volumes cancel out for equal mass percentages).

Step 1: Identify given values. Solution 1: Mass of compound (w1w_1) = 1 g; Molecular weight (M1M_1) = ? Solution 2: Mass of sucrose (w2w_2) = 5 g; Molecular weight of sucrose (C12H22O11C_{12}H_{22}O_{11}) = 342 g/mol .

Step 2: Apply the isotonic condition. Since n1V=n2V\frac{n_1}{V} = \frac{n_2}{V} (for same volume of solution containing these percentages), w1M1=w2M2\frac{w_1}{M_1} = \frac{w_2}{M_2}

Step 3: Calculate M1M_1. 1M1=5342\frac{1}{M_1} = \frac{5}{342} M1=3425=68.4 g/molM_1 = \frac{342}{5} = 68.4 \text{ g/mol}

The molecular weight of the compound is 68.4.

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.

CHEMISTRYSolutionssolutioncompoundisotonicsucrosesolution

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