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When 25 g of a non-volatile solute is dissolved in 100 g of water, the vapour pressure is lowered by 2.25 × 10-1 mm of Hg. If the vapour pressure of water at 20oC is 17.5 mm of Hg, what is the molecular weight of the solute?
  • a)
    206
  • b)
    302
  • c)
    350
  • d)
    276
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
When 25 g of a non-volatile solute is dissolved in 100 g of water, th...
Given data:
Mass of solute (m1) = 25 g
Mass of solvent (m2) = 100 g
Vapour pressure of pure water (P1°) = 17.5 mm of Hg
Vapour pressure of solution (P2°) = 17.5 - 2.25 × 10-1 = 17.275 mm of Hg

To find: Molecular weight of the solute

Formula used:

Relative lowering of vapour pressure = P1° - P2°/ P1° = w2 / M2 / (w1 / M1 + w2 / M2)

Where w1 and w2 are the masses of solvent and solute respectively and M1 and M2 are their respective molecular weights.

Calculation:

Relative lowering of vapour pressure = P1° - P2°/ P1° = 2.25 × 10-1 / 17.5 = 0.01286

Let the molecular weight of the solute be M2.

Substituting the given values in the formula, we get:

0.01286 = 25 / M2 / (100 / 18 + 25 / M2)

Solving this equation, we get:

M2 = 350

Therefore, the molecular weight of the solute is 350.

Hence, option (c) is the correct answer.
Free Test
Community Answer
When 25 g of a non-volatile solute is dissolved in 100 g of water, th...
Given,
weight of non-volatile solute,
w = 25 g
Weight of solvent, W = 100 g
Lowering of vapour pressure,
− ps = 0⋅225mm of Hg
Vapour pressure of pure solvent,
= 17⋅5 mm of Hg
Molecular weight of solvent (H2 O) , M = 18 g
Molecular weight of solute, m= ?
According to Raoult's law
= 350 g
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When 25 g of a non-volatile solute is dissolved in 100 g of water, the vapour pressure is lowered by 2.25 × 10-1 mm of Hg. If the vapour pressure of water at 20oC is 17.5 mm of Hg, what is the molecular weight of the solute?a)206b)302c)350d)276Correct answer is option 'C'. Can you explain this answer?
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