What is the normal boiling point of an aqueous solution whose freezing point is `-2.48^(circ)C`?
( Given `K_f `= 1.86 K / molal and `K_b` = 0.52 K / molal)

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)

Complete the following statements by selecting the correct alternative from the choices given : An aqueous solution of urea freezes at - 0.186^(@)C, K_(f) for water = 1.86 K kg. mol^(-1),K_(b) for water = 0.512 "K kg mol"^(-1) . The boiling point of urea solution will be :

What will be the freezing point of a 0.5 m KCl solution ? The molal freezing point constant of water is 1.86^@C m^(-1) .

The boiling point of an aqueous solution of a non - electrolyte is 100.52^@C . Then freezing point of this solution will be [ Given : k_f=1.86 " K kg mol"^(-1),k_b=0.52 "kg mol"^(-1) for water]

The boiling point of pure water is 373 K. calculate the boiling point of an aqueous solution containing 18 g of glucose (M.W. = 180) in 100g of water. Molal elevation constant of water is 0.52 K kg "mol"^(-1) .

The lowering in freezing point of 0.75 molal aqueous solution NaCI (80% dissociated) is [Given, K_f, for water 1.86 K kg mol^(-1)]

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 )

Calculate elevation in boiling point for 2 molal aqueous solution of glucose. (Given K_b(H_(2)O) = 0.5 kg mol^(-1) )

An aqueous glucose solution has boiling point of 374.19K . Another aqueous glucose solution has a freezing point of 272.22K . Find the molality ratio of two solution. Boiling point and freezing point of H_(2)O373.15K and 273.15K respectively. Kb and Kf for water is 0.512 and 1.86 respectively.

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.