AIEEE 2006 Solved Paper

2.05M solution of acetic acid in water means in 1 litre solution 2.05 moles of CH3‌COOH present.
Density of solution =1.02g∕ml (Given)
Assume the volume of solution =1 litre =1000‌ml
∴ Mass of solution =1000×1.02=1020‌gm
Molar mass of CH3‌COOH=60
So Mass of CH3‌COOH(m‌solute ‌)=2.05×60=123
∴ Mass of solvent (m‌solvent ‌)=1020−123=897‌gm=0.897‌kg
Formula of molality (m)=‌

‌ no of moles of solute ‌

‌ weight of solvent in ‌kg

∴m=‌

2.05

0.897

=2.28
Using Formula :
Molality (m)=‌

1000×M

1000×d−M×M‌solute ‌

Here M= molarity, M‌solute ‌= molecular mass of solute, d= density of solution
∴m=‌

1000×2.05

1000×1.02−2.05×60

=2.28