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What weight of glucose must be dissolved in 100 g of water to lower the vapour pressure by 0.20 mm Hg? (Assume dilute solution is being formed) Given : Vapour pressure of pure water is 54.2 mm Hg at room temperature. Molar mass of glucose is 180 g mol−1

1. 3.59 g
2. 3.69 g
3. 4.69 g
4. 2.59 g

Correct answer: 3.69 g

Solution

The correct option is right because it accurately reflects the calculation based on Raoult's Law, which states that the change in vapor pressure is proportional to the mole fraction of the solute. By determining the number of moles of glucose needed to achieve the specified decrease in vapor pressure, the weight of glucose can be calculated, leading to the correct answer of 3.69 g.

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