Home
Class 12
CHEMISTRY
1.355 g of a substance dissolved in 55 g...

`1.355 g` of a substance dissolved in `55 g`of `CH_(3)COOH` produced a depression in the freezing point of `0.618^(@)C`. Calculate the molecular weight of the substance `(K_(f)=3.85)`

Text Solution

Verified by Experts

`DeltaT = (K)_(f)(W_(B)xx1000)/(Mw_(B)xxW_(A))`
where`W_(B)`= mass of solute ,`Mw_(B)`= molar mass of solute,
`W_(A)`= mass of solvent
`0.618 = 3.85 xx (1.355xx1000)/(Mw_(B)xx55)`
`Mw_(B)=153.47`
Promotional Banner

Similar Questions

Explore conceptually related problems

10g of an organic substance when dissolved in two litres of water gave an osmotic pressure of 0.59 atm, at 7^(@)C . Calculate the molecular weight of the substance.

0.15 g of a substance dissolved in 15 g of a solvent boiled at a temp. higher by 0.216^@C then that of the pure solvent. The molecular weight of the substance is ( K_b for solvent is 2. 16^@C .)

By dissolving 13.6 g of a substance in 20 g of water, the freezing point decreased by 3.7^(@)C . Calculate the molecular mass of the substance. (Molal depression constant for water = 1.863K kg mol^(-1))

15 g of an unknown molecular substance was dissolved in 450 g of water. The resulting solution freezes at - 0.34^(@) C. what is molar mass of the substance ? (K_(f)" for water " = 1.86 k kg " mol"^(-1))

On dissolving 0.25 g of a non-volatile substance in 30 mL benzene (density 0.8 g mL^(-1)) , its freezing point decreases by 0.25^(@)C . Calculate the molecular mass of non-volatile substance (K_(f) = 5.1 K kg mol^(-1)) .

2.5 g of a substance is present in 200 mL of solution showing the osmotic pressure of 60 cm Hg at 15^(@)C . Calculate the molecular weight of substance.What will be the osmotic pressure if temperature is raised to 25^(@)C ?

The boiling point of a solution made by dissolving 12.0 g of glucose in 100 g of water is 100.34^(@)C . Calculate the molecular weight of glucose, K_(b) for water = 0.52^(@)C//m .

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) )

A solution containing 25.6 g of sulphur, dissolved in 1000 g of naphthalene whose melting point is 80.1^(@)C gave a freezing point lowering of 0.680^(@)C . Calculate the formula of sulphur ( K_(f) for napthalene = 6.8 K m^(-1) )

A solution containg 12 g of a non-electrolyte substance in 52 g of water gave boiling point elevation of 0.40 K . Calculate the molar mass of the substance. (K_(b) for water = 0.52 K kg mol^(-1))