Find molality of aqueous solution of `CH_(3)COOH` whose molarity is `2M` and density `d = 1.2g//mL`
`"Hint":m=(M)/(d-MM_(s))xx1000`
Where d = density in `gL^(-1)`, ,M = Molartity, m = molality, `M_(S) =` molar mass of solute .

Explanation

To find the molality of an aqueous solution of acetic acid (CH₃COOH) with a molarity of 2M and a density of 1.2 g/mL, we can use the formula provided in the hint:

\[ m = \frac{M}{d - M \cdot M_s} \times 1000 \]

Where:
- \( m \) = molality
- \( M \) = molarity
- \( d \) = density in g/mL
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