(a) Define the following terms :
(i) Molarity
(ii) Molal elevation constant` (K_(b))`
(b) A solution containing 15 g urea (molar mass = 60 g `mol^(-1)`) per litre of solution in water has same osmotic pressure (isotonic) as a solution of glucose (molar mass = 180 g `mol^(-1)`) in water. Calculate the mass of glucose present in one litte of its solution.

Explanation

(a) (i) Molarity is defined as number of moles of solute dissolved in one litre of solution.
(ii) lt is equal to elevation in boiling point of 1 molal solution.
(b) For isotonic solutions :
`pi_("urea") = pi_("glucose")` `(W_("urea"))/(M_("urea" xx V_(S))) = (W_("Glucose"))/(M_("Glucose" xx V_(s)))`
(As volume of solution is same
`(W_("urea"))/(M_("urea")) = ("glocose")/("glocose")`
or `(15 g)/( 60 g" mol"^(-1)) = (w_("Glucose"))/(180 g "mol"^(-1))`
`W_("glucose") = (15 g xx 180 g "mol"^(-1))/(60 g "mol"^(-1))` = 45 g