At a certain temperature,a 1.6 % solution of | vedclass

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(B) Two solutions are isotonic if they have the same osmotic pressure $(\pi_1 = \pi_2)$.
Since $\pi = CRT$,for the same temperature,$C_1 = C_2$ (mola

Vedclass · Accepted Answer

$40$

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(B) Two solutions are isotonic if they have the same osmotic pressure $(\pi_1 = \pi_2)$.
Since $\pi = CRT$,for the same temperature,$C_1 = C_2$ (molar concentrations).
Concentration $C = \frac{n}{V} = \frac{w}{M 	imes V}$.
Given $1.6 \%$ solution means $1.6 \ g$ of solute in $100 \ mL$ of solution.
For the unknown substance: $w_1 = 1.6 \ g$,$V_1 = 100 \ mL$,$M_1 = ?$.
For Urea: $w_2 = 2.4 \ g$,$V_2 = 100 \ mL$,$M_2 = 60 \ g/mol$.
Equating the molar concentrations: $\frac{1.6}{M_1 	imes 100} = \frac{2.4}{60 	imes 100}$.
$\frac{1.6}{M_1} = \frac{2.4}{60} = 0.04$.
$M_1 = \frac{1.6}{0.04} = 40 \ g/mol$.

At a certain temperature,a $1.6 \%$ solution of an unknown substance is isotonic with a $2.4 \%$ solution of Urea. If both solutions have the same solvent and the same density of $1 \ g/cm^3$,what is the molecular mass of the unknown substance in $g/mol$? [Molecular mass of urea $= 60 \ g/mol$]

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