Question

If a 1L aq. solution of urea (d = 1.02 g/ml.) has
conc. 3M, then the molality of the solution will be
(1) 3.57 m
(2) 3 m
(3) 2.7 m
(4) 6 m​

Answer 1

we have to find the molality of the solution, if 1 L aqueous solution of urea has concentration 3M.

solution : density of solution, d = 1.02 g/ml

volume of solution = 1000 ml

mass of solution = 1.02 g/ml × 1000 ml = 1020 g

concentration of solution = 3M

so no of moles of solute = 3 moles

mass of solute = no of moles × molar mass of solute (urea)

= 3 × 60 = 180g

so mass of solvent = mass of solution - mass of solute

= 1020g - 180g

= 840g

now molality = no of moles of solute/mass of solvent in kg

= 3/(840/1000)

= 300/84

= 3.57 m

Therefore molality of the solution will be 3.57 m.

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