Home
Class 12
CHEMISTRY
The freezing point of a 0.05 molal solut...

The freezing point of a 0.05 molal solution of a non-eletrolyte in water is ( `K_(f)=1.86Km^(-1)`)

A

`-1.86^(@)C`

B

`-0.93^(@)C`

C

`-0.093^(@)C`

D

`0.093^(@)C`

Text Solution

Verified by Experts

The correct Answer is:
C

`DeltaT_(f)=1.86 xx 0.05 = 0.093`,
`:.T_(f)=0-0.093 =-0.093^(@)C`
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) )

The freezing point of a 0.05 m BaCl_(2) in water ( 100% ionisation) is about (K_(f)=1.86 Km^(-1)) :

The freezing point of a solution prepared from 1.25 g of non-electrolyte and 20 g of water is 271.9 K . If the molar depression constant is 1.86 K mol^(-1) , then molar mass of the solute will be

Calculate the freezing point of an aqueous solution of a non-electrolyte having an osmotic pressure 2.0 atm at "300 K. Kf = 1.86 k.kg.mol"^(-1.)" R = 0.0821 lit.atm.k"^(-1)" mol"^(-1)

The freezing point of 1 molal NaCl solution assuming NaCl to be 100% dissociated in water is (Molal depression constant of water=1.86Kkg/mol)

The freezing point of a 0.08 molal solution of NaHSO^(4) is -0.372^(@)C . Calculate the dissociation constant for the reaction. K_(f) for water = 1.86 K m^(-1)

What is the freezing point of solution containing 3g of a non-volatile solute in 20g of water. Freezing point of pure water is 273K, K_(f) of water = 1.86 Kkg/mol. Molar mass of solute is 300 g/mol. T^(@)-T=K_(f)m

A solution of urea in water has boiling point of 100.15^(@)C . Calculate the freezing point of the same solution if K_(f) and K_(b) for water are 1.87 K kg mol^(-1) and 0.52 K kg mol^(-1) , respectively.