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JEE MainChemistryModeratePhysical Chemistry

A 1.5 m solution of acetic acid (I) is mixed with 3 m solution of
the acetic acid (II) to prepare 2m solution (m is molality of solution).
Select the correct statement(s)

Show answer and solution
Let mass of solvent in 1.5 m solution = m_1 Kg and let mass of
solvent in 3 m solution = m_2 Kg.
Therefore molality of resulting solution
=1.5m1+3m2m1+m2=2= \frac{1.5m_{1} + 3m_{2}}{m_{1} + m_{2}} = 2
∴m1m2=21\therefore\frac{m_{1}}{m_{2}} = \frac{2}{1}
Therefore required ratio =10959= \frac{109}{59}

Source: Grass-G / CH-05 Chemistry Paper 10 October.docx

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