The difference between the boiling point and freezing point of an aqueous solution containing sucrose (molecular mass `= 342 g mol^(-1))` in 100 g of water is `105.04`. If `K_(f)` and `K_(b)` of water are 1.86 and `0.51 Kg mol^(-1)` respectively, the weight of sucrose in the solution is about

Options

A. 34.2g
B. 342g
C. 7.2g
D. 72g

Answer

D

Explanation

`DeltaT_(b) = K_(b) xx m` and `DeltaT_(f) = K_(f) xx m`
`:. DeltaT_(b) +DeltaT_(f) = (K_(b) +K_(f))m`
Now, `T_(b) - T_(f) = (T_(b)^(@) +DeltaT_(b)) - (T_(f)^(@)-DeltaT_(f))`
`105 = (DeltaT_(b) +DeltaT_(f)) + (T_(b)^(@) - T_(f)^(@))`
`105 = (DeltaT_(b) +DeltaT_(f)) +100`
`:. DeltaT_(b) + DeltaT_(f) = 5`
`:. m = (DeltaT_(b) + DeltaT_(f))/(K_(b) +K_(f)) = (5)/(1.86 +0.51) = (5)/(2.37) = 2.11`
Molalioty `= ("Moles of solute")/("Mass of solvent"(kg))`
`:.` Moles of solute `= 2.11 xx 0.1 = 0.211`
? `:.` Mass of solute `= 0.211 xx 342 = 72.16g`