Question

Calculate van't Hoff-factor for 0.2 m aqueous solution of KCl which freezes at - 0.680°C. (K_f = 1.86 K kg mol^-1).

पर्याय

• 3.72

• 1.83

• 6.8

• 1.86

Answer 1

1.83

Explanation:

Given, Molality = 0.2 m

K_f = 1.86 K kg mol^-1

ΔT_f = 0.680°C = 0.680 K

As we know that

ΔT_f = `"i" xx "K"_"f" xx "m"`

`0.680 = "i" xx 1.86 xx 0.2`

`0.680/(1.86 xx 0.2)` = i

i = 1.83