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The freezing point of aqueous solution t...

The freezing point of aqueous solution that contains 5% urea by mass, 1.0% KCl by mass and 10% glucose by mass is : (K_(f) "for" H_(2)O = 1.86 Kkg//mol)` :

A

290.2 K

B

285.5 K

C

269.93 K

D

250 K

Text Solution

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The correct Answer is:
C
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The freezing point of aqueous solution that contains 5% by mass urea. 1.0% by mass KCl and 10% by mass of glucose is: (K_(f) H_(2)O = 1.86 K "molality"^(-1))

The freezing point of of aqueous solution that contains 3% urea. 7.45% KCl and 9% of glucose is (given K_(f) of water =1.86 and asume molarity = molality)

The freezing point of a solution that contains 10 g urea in 100 g water is ( K_1 for H_2O = 1.86°C m ^(-1) )

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 :

Calculate the freezing point of a solution containing 60 g glucose (Molar mass = 180 g mol^(-1) ) in 250 g of water . ( K_(f) of water = 1.86 K kg mol^(-1) )

Calculate the freezing point of an aqueous solution containing 10.50 g of MgBr_(2) in 200 g of water (Molar mass of MgBr_(2) = 184g ). ( K_(f) for water = 1.86" K kg mol"^(-1) )

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)] :

An aqueous solution contaning 5% by weight of urea and 10% weight of glucose. What will be its freezing point? (K'_(f) for H_(2)O "is" 1.86^(circ) mol^(-1) kg)

An aqueous solution contain 3% and 1.8% by mass urea and glucose respectively. What is the freezing point of solution ? ( 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) :

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