If $20 \ g$ of a solute was dissolved in $500 \ mL$ of water and the osmotic pressure of the solution was found to be $600 \ mm$ of $Hg$ at $15 \ ^oC$,then the molecular weight of the solute is:

Calculate the molar mass of solute in a solution prepared by dissolving $1 \ g$ in $0.3 \ dm^3$ solvent having osmotic pressure $0.2 \ atm$ at $300 \ K$. $\left[R=0.082 \ dm^3 \ atm \ K^{-1} \ mol^{-1}\right]$

$A$ solution has an osmotic pressure of $x \ kPa$ at $300 \ K$ having $1 \ mole$ of solute in $10.5 \ m^{3}$ of solution. If its osmotic pressure is reduced to $(\frac{1}{10})$th of its initial value,what is the new volume of solution (in $m^{3}$)?

The concentration in $g/L$ of a solution of cane sugar $(M = 342)$ which is isotonic with a solution containing $6 \ g$ of urea $(M = 60)$ per litre is :-

The concentration in $g$ per litre of a solution of cane sugar $(M = 342)$ which is isotonic with a solution containing $6 \ g$ of urea $(M = 60)$ per litre is

$A$ $5\% \text{ (w/v)}$ solution of cane sugar (molar mass = $342 \text{ g/mol}$) is isotonic with a $1\% \text{ (w/v)}$ solution of an unknown solute. What is the molar mass of the unknown solute in $\text{g/mol}$?