An aqueous solution boils at 100.50^(@)C .The freezing point of the solution would be (K_(b) for water =0.51^(@)C//m),(K_(f) for water =1.86^(@)C//m ) [no association or dissociation]
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)
The measured freezing point depression of a non-volatile solute in aqueous solution is 0.20^(@) C . The elevation in boiling point of the same solution will be [ K_(f) = 1.86 K/m, K_(b) = 0.52 K/m]
The boiling point of an aqueous solution is 100.18 .^(@)C . Find the freezing point of the solution . (Given : K_(b) = 0.52 K kg mol^(-1) ,K_(f) = 1.86 K kg mol^(-1))
An aqueous solution freezes at 272.814 K . The boiling point of the same solution is __________. (K_(r)=1.86 K m^(-1), K_(b) = 0.512 K m^(-1) )
A 0.2 molal aqueous solution of weak acid (HX) is 20% ionised. The freezing point of this solution is (Given, K_(f) = 1.86^(@) C m^(-1) for water)
NARENDRA AWASTHI-DILUTE SOLUTION-Level 3 - Match The Column
