How many m. moles of sourose should be dissolved in 500 grams of water so as to get a solution which has a difference of `103.57^@C` between boiling point and freezing point ?
`(K_f=1.86 K kg "mol"^(-1), K_b=0.52 K Kg "mol"^(-1))`

Options

A. 500 m. moles
B. 900 m. moles
C. 750 m. moles
D. 650 m. moles

Answer

C

Explanation

Boiling point of solution =boiling point `+DeltaT_b=100+DeltaT_b`
Freezing point of solution =freezing point `-DeltaT_f=0-DeltaT_f`
Difference in temperature (given)=`100+DeltaT_b-(-DeltaT_f)`
`103.57=100+DeltaT_b+DeltaT_b+DeltaT_f=100+"molality"xxK_b+"molality"xxK_f`
`=100+"molality"(0.52+1.86)`
Molality=`(103.57-100)/238=3.57/2.38=1.5 m`
and molality =`("moles"xx1000)/W_(gm(solvent)), 1.5=("moles"xx1000)/500`
Moles of solute =`(1.5xx500)/1000=0.75` moles