Question

calculate the amount of cacl2 which must be added to 500g water to lower it's freezing point by 2 K assuming that cacl2 is completely dissociated KF for water is 1.86 Kelvin kg per mole

Answer 1

first you 2+500*2-1.86*2 anser is come

Answer 2

Answer: 19.9g

Explanation:-

Formula used for lowering in freezing point is,

$\Delta T_f=i\times k_f\times m$

or

$\Delta T_f=i\times k_f\times \frac{\text {Given mass}}{\text{ Molar mass}\times \text{ Mass of solvent in kg}}$

where,

$CaCl_2\rightarrow Ca^{2+}+2Cl^{-}$

i = vant hoff factor = 3 (as it is completely dissociated)

$T_f$ = change in freezing point

$k_f$ = freezing point constant

m = molality

$2=3\times 1.86\times \frac{x}{111\times 0.5}$

$x=19.9g$