Question

5 g of kcl was dissolved in 100 g of water and the solution originally at 200c froze at

Answer 1

Answer:

Change in Tf (freezing temperature) = solvent's freezing point - solution's freezing point. (coz on addition of solute, freezing point tends to decrease)

change in Tf = (0+273) - (-0.240+273) = 0.240 Kelvin

change in Tf = (i)(Kf)(m)

i is van't hoff factor

Kf is freezing point depression constant.

m is the molality.

moles of solute= 0.5/(39 + 35.5)

mass of solvent in kg = 100/1000

molality = moles of solute/ mass of solvent in kg

Kf is given.

solve for i.

next, (alpha) = (i-1)/(n-1)

n is 2, since no. of moles formed after dissociation of kcl in water is 2 (k+ and cl-)

alpha/100 is percentage dissociation.....

Hope this helps you.....

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