1. Glucose, CaHi20, is a sugar that is present in fruits. It is also known as \"
ID: 695597 • Letter: 1
Question
1. Glucose, CaHi20, is a sugar that is present in fruits. It is also known as "blood sugar" because it is found in the blood and it is the body's main source of energy. What is the molality of a solution containing 5.67 g of glucose dissolved in 25.2 g of H20? What are the mole fractions of glucose and water in a solution containing 5.67 g of glucose, CeH1206, dissolved in H20? 2. 3. Calculate the vapor pressure lowering (AP) of water when 5.67 g of glucose CeH1206, is dissolved in 25.2 g of water at 25°C. The vapor pressure of water at 25°C is 23.8 mm Hg. What is the final vapor pressure of the solution ?Explanation / Answer
1)
molality of solution = [moles of solute ]/mass of solvent in Kg
= [mass of solute /molar mass] / mass of solvent in Kg
= [5.67g/180g/mol] / 0.0252Kg
= 1.25m
2) moles of glucose = 5.67g/180g/mol
= 0.0315 mol
moles of H2O = 25.2g/18g/mol
= 1.4 mol
Mole fraction of glucose = moles of glucose/ total moles
= 0.0315/1.4315
= 0.022
mole fraction of H2O = 1.4/1.4315
= 0.977
3)
From Raoults law
delta P /P0 = mole fraction of solute(glucose)
Given P0 = 23.8mm
thus
delta P/23.8mm = 0.022
delt aP = 0.5236mm
thus the lowering of vapor pressure on adding glucose in water is 0.5236mm
and final pressure of the solution = P0 - delta P
= 23.8 - 0.5236
= 23.2764
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