JEE Advanced levelNumerical answerDifficulty 3 of 3so049
Pure water has vapour pressure 23.8mmHg at 298K. A solution is prepared by dissolving 3.0g of urea (molar mass 60) and 3.6g of glucose (molar mass 180), both non-volatile non-electrolytes, in 90g of water (molar mass 18). Calculate the lowering of vapour pressure in mmHg to two decimal places.
Numerical answer type, answer in mmHg, tolerance 0.005.
Answer: 0.33 mmHg
Moles: urea =0.050, glucose =0.020, so n2=0.070; water =5.00mol. x2=0.070/5.070=0.01381. Δp=23.8×0.01381=0.3286mmHg, i.e. 0.33.