What is the osmotic pressure of a solution prepared by dissolving $3 \ g$ of solute in $2 \ dm^3$ of water at $300 \ K$ (in $atm$)? (Molar mass of solute $= 60 \ g \ mol^{-1}$,$R = 0.0821 \ dm^3 \ atm \ K^{-1} \ mol^{-1}$)

The osmotic pressure (in $atm$) of an aqueous solution containing $0.01 \ mol$ of $NaCl$ (degree of dissociation $0.94$) and $0.03 \ mol$ of glucose in $500 \ mL$ at $27^{\circ} C$ is $\left(R=0.082 \ L \ atm \ K^{-1} \ mol^{-1}\right)$

$200 \ mL$ of an aqueous solution of a protein contains its $1.26 \ g$. The osmotic pressure of this solution at $300 \ K$ is found to be $2.57 \times 10^{-3} \ bar$. The molar mass of protein will be ......... $g \ mol^{-1}$.
$(R = 0.083 \ L \ bar \ mol^{-1} \ K^{-1})$

The osmotic pressure exerted by a solution prepared by dissolving $2.0 \ g$ of protein of molar mass $60 \ kg \ mol^{-1}$ in $200 \ mL$ of water at $27^{\circ} C$ is $.....Pa$. [integer value]
(use $R = 0.083 \ L \ bar \ mol^{-1} \ K^{-1}$)

Isotonic solutions have the same

Blood is isotonic with