What is the normal B.P.of an aquoues solution whose freezing point is `-2.48^(@)`C ? ( `K_(f)`=`1.86^(@)`C . kg/mol `K_(b)`=`0.512^(@)`C .kg /mol )
A) `102.5^@`C
B) `109.0^@`C
C) `100.7^@`C
D) `99.3^@`C

An aqueous solution freezes at -0.186^(@)C (K_(f)=1.86^(@) , K_(b)=0.512^(@) . What is the elevation in boiling point?

For an aqueous solution freezing point is -0.186^(@)C . The boiling point of the same solution is (K_(f) = 1.86^(@)mol^(-1)kg) and (K_(b) = 0.512 mol^(-1) kg)

An aqueous solution boils at 101^(@)C . What is the freezing point of the same solution? (Gives : K_(f) = 1.86^(@)C// m "and" K_(b) = 0.51^(@)C//m )

An aqueous solution freezes at -0.2^(@)C . What is the molality of the solution ? Determine also (i) elevation in the boiling point (ii) lowering in vapour pressure at 25^(@)C , given that K_(f)=1.86^(@)C" kg mol"^(-1), K_(b)=0.512^(@)C" kg mol"^(-1) , and vapour pressure of water at 25^(@)C is 23.756 mm.

Which of the following is/are true for a solution containing 30g urea in 100 g water? (K_(f) for water =1.9.^(@)C.kg//mol and K_(b) for water =0.5.^(@)Ckg//mol

Find the freezing point of a glucose solution whose osmotic pressure at 25^(@)C is found to be 30 atm. K_(f) (water) = 1.86 kg mol^(-1)K .

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

In a 0.5 molal solution KCl, KCl is 50% dissociated. The freezing point of solution will be ( K_(f) = 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)] :

A 0.025 m solution of monobasic acids has a freezing point of -0.060^(@)C . What are K_(a) and pK_(a) of the acid? (K_(f)=1.86^(@)C)