Home
Class 11
CHEMISTRY
0.900g of a solute was dissolved in 100 ...

0.900g of a solute was dissolved in 100 ml of benzene at `25^(@)C` when its density is 0.879 g/ml. This solution boiled `0.250^(@)C` higher than the boiling point of benzene. Molal elevation constant for benzene is `"2.52 K.Kg.mol"^(-1)`. Calculate the molecular weight of the solute.

Text Solution

Verified by Experts

`"Weight of benzene "=100xx0.879=87.9g`
Molality of solution, m `=(0.900//M_(2))/(87.9)xx1000`
`=(900)/(87.9M_(2))`
`DeltaT_(b)=K_(b)m" (or) 0.52"xx(900)/(87.9M_(2))`
`therefore" "M_(2)=(900xx2.52)/(87.9xx0.250)`
`M_(2)="103.2g/mole"`
`therefore" "{:("Molecular weight"),("of the solute"):}}="103.2 g/mole"`
Promotional Banner

Topper's Solved these Questions

  • COLLIGATIVE PROPERTIES

    NCERT GUJARATI|Exercise Questions(Problems)|5 Videos

  • CHEMICAL KINETICS - I

    NCERT GUJARATI|Exercise Questions(E. Explain briefly on the following)|6 Videos

  • DETECTION AND ESTIMATION OF ELEMENTS

    NCERT GUJARATI|Exercise Problems (Estimation of halogens)|5 Videos

Similar Questions

Explore conceptually related problems

1.00 g of a non-electrolyte dissolved in 50.5g of benzene lowered its freezing point by 0.40K. The freezing point depression constant of benzene is "5.12K.kg mol"^(-1) . Find the molecular mass of the solute.

A solution containing 2.5 g of a non-volatile solute in 100 gm of benzene boiled at a temperature 0.42K higher than at the pure solvent boiled. What is the molecular weight of the solute? The molal elevation constant of benzene is "2.67 K kg mol"^(-1) .

0.15 g of a substance dissolved in 15 g of solvent boiled at a temperature higher at 0.216^(@) than that of the pure solvent. Calculate the molecular weight of the substance. Molal elevation constant for the solvent is 2.16^(@)C