Question

when 20g of naphthoic acid is dissolved in 50 g of benzene.

(KΦ =1.72k.mori) a freezing point depression of 2k is observed, the vant hoff factor (i) is -----

Answer 1

hey here is your answer

we know that

napthoic acid(C11H8O2)

then molecular weight of napthoic acid is

M = (11x12) + 8 + (2 x 16) = 132 + 8 + 32 = 172

mass = 20 g

and we also know that.

ΔTb = i. Kb . m

2 = (i x 1.72 x 20 x 1000) / (172 x 50)

i = 0.5

I hope it will help you..

Answer 2

Answer: vant hoff factor is 0.5.

Explanation: Depresson in freezing point :

Formula used for lowering in freezing point is,

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

$\Delta T_f=i\times K_f\times \frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}$

Weight of solvent= 50 g = 0.05 kg (1 kg=1000 g)

Molar mass of napthoic acid $(C_{11}H_8O_2)$= 172 g/mol

Mass of napthoic acid added = 20 g

where,

$T_f$ = change in freezing point  = 2K

$k_f$ = freezing point constant = 1.72 K/mol

m = molality

i= vant hoff factor= ?

$2K=i\times 1.72\times \frac{20}{172\times 0.05}$

$i= 0.5$