Question

How many grams of sucrose (M.wt $=342$) should be dissolved in $100$g water in order to produce a solution with a ${105}^o$C difference between freezing point and boiling point temperature? (Kb$=0.51$ and Kf$=1.86$)

• A. $34.2$g
• B. $72$g
• C. $342$g
• D. $460$g

Answer 1

The correct option is B $72$g

Let the molality of the solution is m.

$\Delta T_f=K_f\times m$ and$\Delta T_b=K_{f}b\times m$

$\Delta T_f$ $+\Delta T_b$ $=(K_f+K_b)\times m=(1.86+0.51)\times m$

$=2.37\times m$

Hence$m=5/2.37=2.11$

and $\frac{2.11\times 100}{1000}=0.211$ gram of sucrose must be dissolved in 100 gram of water.

Required mass of sucrose $=0.211mol\times 342g/mole=72.2g$

So option B is correct.