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:

An aqueous solution at freezes at -2.55^(@)C . What is its boiling point (K_(b)^(H_(2)O)=0.52 K m^(-1) , K_(f)^(H_(2)O)=1.86 K m^(-1) ?

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 :

An aqueous solution freezes at - 0.372^@C . If K_f and K_b for water are 1.86K kg "mol"^(-1) and 0.53 K kg "mol"^(-1) respectively, the elevation in boiling point of same solution in K is:

An aqueous solution freezes at - 0. 186^@C then elevation in boiling point is (K_b = 0.512, K_f = 1.86)

AN aqueous solution of a substacne freezes at -0.186^(@)C. The boiling point of the same solution (k _(f) =1.86 Km^(-1), K_(b)=0.512 Km^(-1)) is :