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  4. How many grams of urea (M.W. 60 g mol -1) should be dissolved in 100 g of water in order to form a solution with the difference of 101° C between its boiling point and freezing point? [ K f ( H 2 O )=1.86 K kg mol -1. and .K b ( H 2 O )=0.52 K kg mol -1]

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Q. How many grams of urea $(M.W. 60\, g\, mol ^{-1})$ should be dissolved in $100\, g$ of water in order to form a solution with the difference of $101^{\circ} C$ between its boiling point and freezing point?
$\left[ K _{ f \left( H _{2} O \right)}=1.86\, K\, kg\, mol ^{-1}\right.$ and $\left.K _{ b \left( H _{2} O \right)}=0.52\, K\, kg\, mol ^{-1}\right]$

Solutions

Solution:

$\Delta T _{ b }= K _{ b } \times m$
$B.P$. of solution, $T _{ b }=100+\Delta T _{ b }=100+ K _{ b } m$
$F.P$. of solution, $T _{ f }=0-\Delta T _{ f }=0- K _{ f } m$
$T _{ b }- T _{ f }=\left(100+ K _{ b } m \right)-\left(- K _{ f } m \right)$
$101=100+0.52 \times m +1.86 \times m$
$101-100=2.38\, m$
$m =\frac{1}{2.38}=0.42$
$m =0.42\, kg\, mol ^{-1}$
$\therefore $ Mass of urea to be dissolved in $100\, g$ of water
$=\frac{0.42 \times 60}{1000} \times 100=2.52\, g$