Glucose is added to water such that the mole fraction of water in solution becomes 0.8.the boiling point of this solution is
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We know πV = nRT
πV = wB/mB RT
mB = wB×RT/πV …(i)
where wB = mass of solute (21.6× 10−3 g)
mB = molar mass of solute
R = 0.0821 L atm K−1 mol−1
T = 298 K
V = 100/1000 = 0.1 L : π = 3.70/760 atm
From (i), mB = 21.6×10–3×0.0821×298/(3.70/760)×0.1 = 1085 g/mol
πV = wB/mB RT
mB = wB×RT/πV …(i)
where wB = mass of solute (21.6× 10−3 g)
mB = molar mass of solute
R = 0.0821 L atm K−1 mol−1
T = 298 K
V = 100/1000 = 0.1 L : π = 3.70/760 atm
From (i), mB = 21.6×10–3×0.0821×298/(3.70/760)×0.1 = 1085 g/mol
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