SolutionHard
Question
How many grams of sucrose (molecular weight = 342) should be dissolved in 100 g water in order to produce a solution with a 104.76°C difference between the freezing point and the boiling point temperature? (Kf = 1.86, Kb = 0.52)
Options
A.34.2 g
B.68.4 g
C.684 g
D.313.06 g
Solution
$\Delta T_{f} + \Delta T_{b} = \left( K_{f} + K_{b} \right).m$
Or, $4.76 = (1.86 + 0.52) \times \frac{w/342}{100/1000} \Rightarrow w = 68.4\text{ gm}$
Create a free account to view solution
View Solution FreeMore Solution Questions
For each of the following dilute solutions, van’t Hoff factor is equal of 3, except...At 35oC, the vapour pressure of CS2 is 512 mmHg, and of acetone is 344 mmHg. A solution of CS2 and acetone in, which the...The molal boiling point elevation constant of water is 0.513°C kg mol−1. When 0.1 mole of sugar is dissolved in 200 g of...x mole of KCI and y mole of BaCl2 are both dissolved in 1 kg of water. Given that x + y = 0.1 and Kf for water is 1.85 K...A 0.2 molal aqueous solution of a weak acid (HX) is 20 per cent ionised. The freezing point of this solution is (Given k...