Home
Class 12
CHEMISTRY
On dissolving 2.34g of non-electrolyte s...

On dissolving 2.34g of non-electrolyte solute in 40g of benzene, the boiling point of solution was higher than benzene by 0.81K. Kb value for benzene is 2.53 K `kgmol^(-1)` .Calculate the molar mass of solute. [Molar mass of benzene is 78 `gmol^(-1)`]

Text Solution

Verified by Experts

(a)Data: Mass of solute = `W_(2)` = 2.34 g,
`K_(b) = 2.53 K.kg mol^(-1)`,
Molar mass of solute = `M_(2)`= ?
Mass of solvent benzene = Wj = 40 g
Elevation in boiling point =0.81 K
Formula: `M2=W_(2)/W_(1) 1000`
Substituion: `M_(O_(2))=2.53xx2.34xx(1000)/(40)xx1/0.81`
Answer: `M_(2) = 182.72 g mol^(-1)`
(b) (ii) Solubility of gas decreases with increase in temperature and increases with increase in pressure.
Promotional Banner

Topper's Solved these Questions

  • SUPPLEMENTARY EXAMINATION QUESTION PAPER JUNE 2019

    SUBHASH PUBLICATION|Exercise PART-C|8 Videos
  • SUPPLEMENTARY EXAMINATION QUESTION PAPER JUNE - 2018

    SUBHASH PUBLICATION|Exercise PART -D|30 Videos
  • SUPPLEMENTARY EXAMINIATION QUESTION PAPER JUNE 2017

    SUBHASH PUBLICATION|Exercise PART E|22 Videos

Similar Questions

Explore conceptually related problems

On dissolving 3.24 g of sulphur in 40 g of benzene,the boiling point of the solution was higher than that of benzene by 0.81K .What is the molecular formula of sulphur? ( K_(b) for benzene = 2.53 K kg mol^(-1) , atomic mass of sulphur = 32 g mol^(-1) ).

Vapour pressure of benzene is 200 mm of Hg. When 2 gram of a non-volatile solute dissolved in 78 gram benzene, benzene has vapour pressure of 195 mm of Hg. Calculate the molar mass of the solute. [Molar mass of benzene is 78 g/ mol^(-1) ]

0.90g of a non-electrolyte was dissolved in 90 g of benzene. This raised the boiling point of benzene by 0.25^(@)C . If the molecular mass of the non-electrolyte is 100.0 g mol^(-1) , calculate the molar elevation constant for benzene.

1.0 g of non - electrolyte solute dissolved in 50 g of benzene lowered the freezing point of benzene by 0.4 K. Find the molar mass of the solute. [Given : Freezing point depression constant of benzene = 5.12 K. kg mol].

The boiling point of benzene is 353.23 K when 1.80 g of a non-volatile, non-ionising solute was dissolved in 90 g of benzene, the boiling point is raised to 354.11 K. Calculate the molar mass of solute. " " [Given K_(b) for benzene = 2.53 K kg mol^(-1) ]

The boiling point of benzene is 353.23 K . When 1.80 g of a non-volatile solute was dissolved in 90 g benzene, the boiling point is raised to 354.11 K . Calculate the molar mass of the solute. ( K_(b) for benzene is 2.53 K kg mol^(-1) )

A solution of solute 's' in benzene boils at 0.126^(@) higher than benzene. The molality of the solution is ( K_(b) for benzene = 2.52Km^(-1) ) :