Aqueous solution of urea has density 1.5 g/ml and it is 5 M. Calculate (a) Its molality (b) w/w % (c) w/v % (d) mole fraction of solute
Answers
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Given info : Aqueous solution of urea has density 1.5 g/ml and its concentration is 5 M.
To find :
- (a) its molality
- (b) w/w %
- (c) w/v %
- (d) mole fraction
solution : (a) concentration of solution is 5 M, it means 5 mol of urea is present in 1000 ml of solution.
mass of solution = volume of solution × density of solution
= 1000 ml × 1.5 g/ml = 1500 g
mass of solute = no of moles of solute (urea) × molar mass of solute
= 5 mol × 60 g/mol = 300 g [ ∵ molar mass of urea is 60 g/mol ]
now mass of solvent = mass of solution - mass of solute = 1500 g - 300 g = 1200 g
now molality = no of moles of solute/mass of solvent in kg
= 5/(1200/1000) = 5/1.2 = 4.167 molal
Therefore the molality of the solution is 4.67 molal
(b) w/w % = mass of solute/mass of solution × 100
= 300g/1500g × 100
= 20 %
(c) w/v % = mass of solute/volume of solution × 100
= 300g/1000 ml × 100
= 30 %
(d) no of moles of solvent = mass of solvent/molar mass of solvent
here solvent is water because solution is aqueous.
so, molar mass of solvent ( water) = 18g/mol
∴ no of moles of solvent = 1200g/18g/mol = 66.67 mol
now mole fraction solute = no of moles of solute/(no of moles of solute + no of moles of solvent)
= 5/(5 + 66.67)
= 5/71.67
≈ 0.07