What weight of glucose must be dissolved in $100g$ of water to lower the vapour pressure by $0 . 20mmHg$ ?
(Assume dilute solution is being formed)
Given: Vapour pressure of pure water is $54 . 2mmHg$ at room temperature.Molar mass of glucose is $180g{mol}^{- 1}$

Question

What weight of glucose must be dissolved in

100g

of water to lower the vapour pressure by

0 . 20mmHg

?
(Assume dilute solution is being formed)
Given: Vapour pressure of pure water is

54 . 2mmHg

at room temperature.Molar mass of glucose is

180g{mol}^{- 1}

Quantrex Academy · Accepted Answer

Correct answer: B. 3 . 69g

Sol. Using the formula P ^ o - P _ s P ^ o = (Molesofsolute)/(Molesofsolvent)(for dilute solution) P ^ o = Vapour pressure of pure water P _ s = Vapour pressure of solution (54 . 2 - 54)/(54 . 2) = (w / 180)/((100)/(18))w = (0 . 2 × 1000)/(54 . 2) = (200)/(54 . 2) = 3 . 69

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