Question

Assume that you were assigned BaCl2 in lab. The water in your test tube weighed 21.797 g. Following the procedure in the lab manual, you determined that freezing point of water is 0.02oC. You weighed out 1.168 g of salt and added it to the original test tube, then determined that the freezing point was -0.33oC. Based on these experimental parameters, calculate the van't Hoff factor for BaCl2.

Answer 1

Answer: Van't Hoff factor for $BaCl_2$ is 1.15

Explanation:

Depression in freezing point:

$T_f^0-T^f=i\times k_f\times \frac{w_2\times 1000}{M_2\times w_1}$

where,

$T_f$ = freezing point of solution = $-0.33^oC$

$T^o_f$ = freezing point of water =  $0.22^oC$

$k_f$ = freezing point constant of water = $1.86^0C/m$

m = molality

i = Van't Hoff factor = ?

$w_2$ = mass of solute = 1.168 g

$w_1$= mass of solvent (water) = 21.797 g

$M_2$ = molar mass of solute = 208 g/mol

Now put all the given values in the above formula, we get:

$(0.22-(-0.33))=i\times (1.86)\times \frac{(1.168)\times 1000}{208\times (21.797)}$

$i=1.15$

Thus van't Hoff factor for $BaCl_2$ is 1.15