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Q. A 1 % aqueous solution (M/V) of a certain substance is isotonic with a 3 % solution of glucose (M. Mass 180). The molar mass of the substance is

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Solution:

For isotonic solutions .
$\frac{n_1}{V_1} = \frac{n_2}{V_2}$
$\frac{m_1}{M_1 \times V_1} = \frac{m_2}{M_2 \times V_2}$
$\frac{1}{M_1 \times 100} = \frac{3}{180 \times 100}$
$M_1 = 60$