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=‌
