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The freezing point of aqueous solution t...

The freezing point of aqueous solution that contains `5%` by mass urea. `1.0%` by mass `KCl` and `10%` by mass of glucose is:
`(K_(f) H_(2)O = 1.86 K "molality"^(-1))`

A

290.2 k

B

285 .5 k

C

269 .9 k

D

250 k

Text Solution

Verified by Experts

The correct Answer is:
C
`n_("ghucose")=(10)/(180),n_("urea"=(5)/(60),n_(KCl)=2xx1/(74.5)`(electrolyte ,i=2)
`therefore DeltaT_f=K_fxxm=1.86 xx(((10)/(180)+5/(10)+2/(74.5)))/(84//1000)=3.7`
`therefore T_f =273 -3.7 = 269.3 K`
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