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(B) For isotonic solutions,the molar concentration is the same,so $C_1 = C_2$. Given that $0.05 \ M$ glucose solution has the same osmotic pressure as

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$C_2H_4O_2$

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(B) For isotonic solutions,the molar concentration is the same,so $C_1 = C_2$. Given that $0.05 \ M$ glucose solution has the same osmotic pressure as the compound solution,the molarity of the compound solution is $0.05 \ M$. Assuming the volume of the solution is $V \ L$,the number of moles of the compound is $n = M 	imes V = 0.05 	imes V$. Also,$n = \frac{	ext{mass}}{	ext{molar mass}} = \frac{3}{M_X}$. Equating these,$\frac{3}{M_X} = 0.05 	imes V$. Assuming the volume is $1 \ L$ for standard comparison,$M_X = \frac{3}{0.05} = 60 \ g/mol$. The empirical formula mass of $CH_2O$ is $12 + 2(1) + 16 = 30 \ g/mol$. The value of $n$ is $\frac{	ext{molecular mass}}{	ext{empirical formula mass}} = \frac{60}{30} = 2$. Therefore,the molecular formula is $2 	imes CH_2O = C_2H_4O_2$.

The empirical formula of a non-electrolyte is $CH_2O$. $A$ solution containing $3 \ g$ of the compound exerts the same osmotic pressure as that of $0.05 \ M$ glucose solution. The molecular formula of the compound is

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