How many moles of sucrose should be dissolved in 500 g of water so as to get a solution which has a difference of 104°C between boiling point and freezing point. (K f = 1.86 K Kg mol –1 , K b = 0.52 K Kg mol –1 )
Text Solution
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Boiling point of solution = boiling point + Δ T b = 100 + Δ T b
Freezing point of solution = freezing point – Δ T f = 0 – Δ T f
Difference in temperature (given) = 100 + Δ T b – (– Δ T f )
104 = 100 + Δ T b + Δ T f = 100 + molality × K b + molality × K f = 100 + molality (0.52 + 1.86)
∴ Molality = 
= 
= 1.68 m
and molality = 
; 1.68 = 
∴ Moles of solute = 
= 0.84 moles.
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