What weight of glucose must be dissolved in 100 g of water to lower the vapour pressure by 0.20 mm Hg? [2023]

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Solution

Q.
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 ${mol}^{- 1}$ . [2023]

1 3.69 g

2 2.59 g

3 3.59 g

4 4.69 g

Ans.
(1)

$For a solution \frac{P_{A}^{0} - P_{s}}{P_{s}} = \frac{n}{N} (for dilute solution)$

$\frac{54.2 - 54}{54} = \frac{w / 180}{100 / 18}$

$Amount of glucose (w) = \frac{200}{54} = 3.69 g$

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