Home
Class 12
CHEMISTRY
A solution prepared by dissolving a 2.50...

A solution prepared by dissolving a `2.50g` sample of a unknown compound dissolved in `34.0g` of benzene `C_(6)H_(6)` boils at `1.38^(@)C` higher than pure benzene. Which expression gives the molar mass of the unknown compound?
`{:("Compound",K_(b)),(C_(6)H_(6),2.53^(@)Cm^(-1)):}`

A

`2.53 xx (2.50)/(1.38)`

B

`1.38 xx (34.0)/(2.53)xx2.50`

C

`2.50xx10^(3)xx (2.53)/(34.0)xx(1)/(1.38)`

D

`2.50 xx 10^(3) xx (1.38)/(34.0)xx2.53`

Text Solution

Verified by Experts

The correct Answer is:
C
`1.38 = (2.5)/(34xxM) xx 10^(3) xx 2.53`
`M = (2.5 xx 10^(3) xx 2.53)/(34 xx 1.38)`
Promotional Banner

Similar Questions

Explore conceptually related problems

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) .

Addition of 0.643g of a compound to 50mL of benzene (density: 0.879g mL^(-1)) lower the freezing point from 5.51^(@)C to 5.03^(@)C . If K_(f) for benzene is 5.12 K kg mol^(-1) , calculate the molar mass of the compound.

When 20g of naphtholic acid (C_(11)H_(8)O_(2)) is dissolved in 50g of benzene (K_(f) = 1.72 K kg mol^(-1)) a freezing point depression of 2K is observed. The van'f Hoff factor (i) is

When 20g of naphtholic acid (C_(11)H_(8)O_(2)) is dissolved in 50g of benzene (K_(f) = 1.72 K kg mol^(-1)) a freezing point depression of 2K is observed. The van'f Hoff factor (i) is

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.