Question

The density of 3 M solution of NaCl is 1.25 g mL-1. Calculate molality of the solution.NCERT Class XIIChemistry - Main Course Book IChapter 1. Some Basic Conpepts of Chemistry

Answer 1

Method 1 :-
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3M , means 3 mole of solute present in 1 Litre of solution .
density of solution = 1.25 g/ml
so,
mass of solution = 1.25 × 1000 = 1250 g

mass of solution = mass of solute + mass of solvent.
here solvent is water .
we know density of water is 1 g/ml
so,
mass of solute = 3 × 58.5 = 175.5 g
mass of solvent = 1 g/ml × V

175.5 + 1g/ml × V = 1250
V = 1250 - 175.5 = 1074.5 g =1.0745 kg

now ,
molatity = mole of solute/mass of solvent in kg
= 3/1.0745 molal
= 2.79 molal ( answer)


method 2 :-
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use formula ,
1/m = d/M - Ms/1000
where ,
m = molatity
M = molarity
d = density of solution
Ms = mplar mass of solute

1/m = 1.25/3 - 58.5/1000
m = 2.79 molal ( answer )