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Question
The vapour pressure of water at 20°C is 17 mm Hg. What is the vapour pressure of solution containing 2.8 g urea in 50 g of water?
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Solution
Given: Vapour pressure of pure water = `"P"_1^0` = 17 mm Hg
Mass of urea (W2) = 2.8 g
Mass of water (W1) = 50 g
To find: Vapour pressure of the solution (P1)
Formula: `(P_1^0 - P_1)/P_1^0 = (W_2M_1)/(M_2W_1)`
Calculation:
Molar mass of urea (NH2CONH2) = 14 + 2 + 12 + 16 + 14 + 2 = 60 g mol–1
Molar mass of water = 18 g mol–1
Now, using formula,
`(P_1^0 - P_1)/P_1^0 = (W_2M_1)/(M_2W_1)`
= `(17 "mm Hg – P"_1)/(17 "mm Hg") = (2.8 g xx 18 g mol^-1)/(50 g xx 60 g mol^-1)`
∴ `(17 "mm Hg – P"_1)/(17 "mm Hg") = 0.0168`
∴ 17 mm Hg – P1 = 0.0168 × 17 mm Hg
∴ 17 mm Hg – P1 = 0.2856 mm Hg
∴ P1 = 17 mm Hg – 0.2856 mm Hg = 16.71 mm Hg
Vapour pressure of the given solution is 16.71 mm Hg.
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