Answer to Question #96500 in General Chemistry for TRICIA

Question

A solution is prepared by dissolving 104.1 grams of barium chloride, BaCl2 in 1000. grams of water? What is the freezing point of this solution? [The freezing point depression constant for water is 1.86°C/mole solute in 1000g of water

Answer 1

$BaCl_2 \to Ba^{+2}+2Cl^-$

t=0 1 0 0

t=T 0 1 2

Vant Hoff's factor for $BaCl_2(i)=\frac{2+1}{1}=3$

Mass of solute$=104.1 g$

Number of moles of solute$=\frac{104.1}{208.2}=0.5$ (where 208.2 gm is molar mass of solute)

Mass of solvent$=1000g$

Molality for solution$(m)=\frac{104.1}{1000}\times1000=0.5$

Depression in freezing point$=iK_fm=3\times 1.86 \times 0.5=3 \times 0.93\degree C=2.79\degree C$

So freezing point of water is$\ 0-2.79=-2.79\degree C$