Question

How many mili moles of sucrose should be dissolved in $500\, g$ of water so as to get a solution which has a difference of $103.57^{\circ} C$ between boiling point and freezing point of solution :-
$\left( K _{ f }=1.86\, K\, kg\, mol ^{-1}, \,K _{ b }=0.52\, K\, kg\, mol ^{-1}\right)$

• A. 500 mili moles
• B. 900 milli moles
• C. 750 milli moles
• D. 650 milli moles

Answer 1

Answer: (C)

molality $=\frac{ n }{ w _{ g }} \times 1000$
molality $=\frac{ n }{500} \times 1000=2 n$
$\Delta T _{ b }= T _{ b }- T _{ b }^{0}$
$\Delta T _{ f }= T _{ f }^{0}- T _{ f }$
$\Delta T _{ b }+\Delta T _{ f }=\left( T _{ b }- T _{ f }\right)+\left( T _{ f }^{0}- T _{ b }^{0}\right)$
$\left( K _{ b } \times m + K _{ f } \times m \right)=103.57+(0-100)$
$(0.52+1.86) \times 2 n =3.57$
$n =\frac{3.57}{2 \times 2.38}=\frac{357}{2 \times 238} mol$
no. of millimol $=\frac{357 \times 1000}{2 \times 238}=750$ millimol