$A$ solution of $8 \ g$ of a certain organic compound in $2 \ dm^3$ of water develops an osmotic pressure of $0.6 \ atm$ at $300 \ K$. Calculate the molar mass of the compound. $\left[R = 0.082 \ atm \ dm^3 \ K^{-1} \ mol^{-1}\right]$

The osmotic pressure of a dilute solution is given by

$A$ $0.6 \%$ solution of urea (molecular mass $= 60$) would be isotonic with

Calculate the osmotic pressure exerted by a solution containing $0.822 \ g$ of solute in $300 \ mL$ of water at $300 \ K$. (Molar mass of solute $= 340 \ g \ mol^{-1}, R = 0.0821 \ L \ atm \ mol^{-1} \ K^{-1}$) (in $atm$)

If osmotic pressure of a solution is $2 \ atm$ at $273 \ K$,then at $546 \ K$,the osmotic pressure is ........... $atm.$

$A$ solution is prepared by dissolving $0.3 \ g$ of a non-volatile non-electrolyte solute '$A$' of molar mass $60 \ g \ mol^{-1}$ and $0.9 \ g$ of a non-volatile non-electrolyte solute '$B$' of molar mass $180 \ g \ mol^{-1}$ in $100 \ mL$ $H_2O$ at $27^{\circ}C$. Osmotic pressure of the solution will be
[Given: $R=0.082 \ L \ atm \ K^{-1} \ mol^{-1}$] (in $atm$)