If a 5.25% , w/v solution of a substance is i | vedclass

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(D) For isotonic solutions,the osmotic pressures are equal: $\pi_1 = \pi_2$.Since the solvent and temperature are the same,the molar concentrations $(

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$210$

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(D) For isotonic solutions,the osmotic pressures are equal: $\pi_1 = \pi_2$.Since the solvent and temperature are the same,the molar concentrations $(C)$ must be equal: $C_1 = C_2$.The molar concentration is given by $\frac{w \, (g/L)}{Molar \, mass \, (g/mol)}$.For the substance: $C_1 = \frac{5.25 \, g/100 \, mL}{M_1} = \frac{52.5}{M_1} \, mol/L$.For urea: $C_2 = \frac{1.5 \, g/100 \, mL}{60 \, g/mol} = \frac{15}{60} \, mol/L = 0.25 \, mol/L$.Equating the two: $\frac{52.5}{M_1} = 0.25$.$M_1 = \frac{52.5}{0.25} = 210 \, g \, mol^{-1}$.

If a $5.25\% \, w/v$ solution of a substance is isotonic with a $1.5\% \, w/v$ solution of urea $(Molar \, mass = 60 \, g \, mol^{-1})$ in the same solvent,then the molar mass of the substance is ........ $g \, mol^{-1}$. (in $.0$)

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