1.

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))

Answer»

500 m. moles
900 m. moles
750 m. moles
650 m. moles

Solution :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


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