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Density of 1 M solution of a non-electro...

Density of 1 M solution of a non-electrolyte `(C_(6)H_(12)O_(6))` is 1.18g/ml. If `K_(f) (H_(2)O)` is 1.86 K `mol^(-1)` kg, the solution freezes at :

A

`1.58^(@)C`

B

`-1.86^(@)C`

C

`-3.16^(@)C`

D

`1.86^(@)C`

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The correct Answer is:
B
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Density of 1M solution of a non-electrolyte C_(6)H_(12)O_(6) is 1.18 g/mL . If K_(f) (H_(2)O) is 1.86^(@) mol^(-1) kg , solution freezes at :

Mole fraction of a non-electrolyte in aqueous solution is 0.07 . If K_(f) is 1.86^(@) "mol"^(-1)kg , depression if f.p.,DeltaT_(f), is:

An aqueous solution of NaCI freezes at -0.186^(@)C . Given that K_(b(H_(2)O)) = 0.512K kg mol^(-1) and K_(f(H_(2)O)) = 1.86K kg mol^(-1) , the elevation in boiling point of this solution is:

The freezing poing of an aqueous solution of a non-electrolyte is -0.14^(@)C . The molality of this solution is [K_(f) (H_(2)O) = 1.86 kg mol^(-1)] :

What is the freezing point of a solution contains 10.0g of glucose C_(6)H_(12)O_(6) , in 100g of H_(2)O ? K_(f)=1.86^(@)C//m

What would be the freezing point of aqueous solution containing 17 g of C_(2)H_(5)OH in 100 g of water (K_(f) H_(2)O = 1.86 K mol^(-1)kg) :

The freezing point of an aqueous solution of non-electrolyte having an osmotic pressure of 2.0 atm at 300 K will be, (K_(f) "for " H_(2)O = 1.86K mol^(-1)g) and R = 0.0821 litre atm K^(-1) mol^(-1) . Assume molarity and molarity to be same :

The freezing point of a 0.05 molal solution of a non-electrolyte in water is: ( K_(f) = 1.86 "molality"^(-1) )

A solution is prepared by dissolving 1.5g of a monoacidic base into 1.5 kg of water at 300K , which showed a depression in freezing point by 0.165^(@)C . When 0.496g of the same base titrated, after dissolution, required 40 mL of semimolar H_(2)SO_(4) solution. If K_(f) of water is 1.86 K kg mol^(-1) , then select the correct statements (s) out of the following( assuming molarity = molarity):

Calculate the temperature at which a solution containing 54g of glucose, (C_(6)H_(12)O_(6)) in 250g of water will freeze. ( K_(f) for water = 1.86 K mol^(-1) kg)

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