Solve this.

Dear Student,

In given statement, CH₃COOH is 10% by weight. This means that CH₃COOH is 10 g in 100 g of solution. This tells that water is 90 g. To calculate molarity,

M o l a r i t y = \frac{m a s s o f s o l u t e}{m o l a r m a s s o f s o l u t e} \times \frac{1000}{v o l u m e o f s o l u t i o n}

Mass of solute = 10 g
molar mass of solute = molar mass of CH3COOH = 60 g/mol
volume of solution = mass of solution/density of solution = 100/1.2 = 83.33 ml

M o l a r i t y = \frac{10}{60} \times \frac{1000}{83.33}

= 2
Molarity = 2 M
To calculate molality,

M o l a l i t y = \frac{m a s s o f s o l u t e}{m o l a r m a s s o f s o l u t e} \times \frac{1000}{w e i g h t o f s o l v e n t}

Weight of solvent = 90 g

M o l a l i t y = \frac{10}{60} \times \frac{1000}{90}

= 1.85
Molality = 1.85 m
Answer is (2).

Regards

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