Home
Class 12
CHEMISTRY
0.90g of a non-electrolyte was dissolved...

`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.

Text Solution

Verified by Experts

The given values are:
`W_("solute")= 0.90 g`
`W_("solvent")= 90.00 g`
`DeltaT_(b) = 0.25 ^(@)C`
`Mw_("solute")= 100.0 g mol^(-1)`
`K_(b)` = ?
Using the formula
`K_(b) = (DeltaT_(b) xx W_("solvent") xx Mw_("solvent")) / (1000 xx Mw_("solute")) = 2.5 K m^(-1)`
`:. K_(b) = (0.25 xx 100 xx 90.0)/(1000 xx 0.90) = 2.5 K m^(-1)`
Thus, `K_(b)` is `2.5 K m^(-1)`.
Promotional Banner

Similar Questions

Explore conceptually related problems

If 10.0 g of a non-electrolyte dissolved in 100 g of water lowers the freezing point of water by 1.86^(@)C , the molar mass of the non-electrolyte is ( K_(f)=1.86 Km^(-1) )

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].

(a) 10 g of non-electrolyte solute dissolved in 50 g of benzene lowered the freexing 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 mor^(-1)] (b) How solubility of a gas in liquid varies with (i) Temperature and (ii) pressure?

1.5 g of a non-volatile, non-electrolyte is dissolved in 50 g benzene ( K_b = 2.5 kg mol^(-1) ). The elevation of the boiling point of the solution is 0.75 K. The molecular weight of the solute in g mol^(-1) is :

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) ]