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What weight of sucrose will be dissolved in 250 g of water and obtain solution with boiling point at 100.2 degree Celsius |
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Answer» Answer: hope it will help you... Explanation: Let the molality of the solution is m. ΔT b = K f b× m ΔT f = K f × m ΔT f +ΔT b
= (K f + K b )× m= (1.86+ 0.51)× m
= 2.37× m Hence m= 5/2.37= 2.11 and 1000 2.11×100 = 0.211 gram of sucrose MUST be DISSOLVED in 100 gram of water. Required MASS of sucrose = 0.211mol× 342g/mole= 72.2g |
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