Calculate the freezing point of an aqueous solution containing 10% by mass of glucose, 5% by mass of urea and 1% by mass of `KCl. K_f (H_2O) = 1.86 K kg mol^(-1)`

Options

A. `-2.25^@`
B. `0^@C`
C. `-3.67^@C`
D. `-1.63^@C`

Answer

C

Explanation

Mass of glucose = 10
No. of moles of glucose = 10/180
Mass of urea = 5, No. of moles of urea = 5/60
Mass of KCl = 1, Moles of `KCI= 1/74.5`
Number of ions furnished by one molecule of KCl is 2.
Thus, actual number of particles `=2/(74.5)`
Total number of particles `= 10/180+ 5/60+2/(74.5)`
`=0.056 +0.083 +0.027 = 0.166`
Mass of water `= 100-10-5-1 = 84`
Molality `=(0.166)/84 xx1000= 1.97`
`DeltaT_f = 1.97 xx 1.86 = 3.67`
`T_f=T_f^@ - DeltaT_f= 0 – 3.67 = -3.67^@C`