Find the amount of solute in a solution obtained by dissolving it in $2.5 \ dm^3$ of water that generates an osmotic pressure of $0.245 \ atm$ at $300 \ K$. (Molar mass of solute $= 58 \ g \ mol^{-1}$,$R = 0.0821 \ dm^3 \ atm \ K^{-1} \ mol^{-1}$) (in $g$)

Solution $A$ is prepared by dissolving $1 \text{ g}$ of a protein (molar mass = $50000 \text{ g mol}^{-1}$) in $0.5 \text{ L}$ of water at $300 \text{ K}$. Its osmotic pressure is $x \text{ bar}$. Solution $B$ is made by dissolving $2 \text{ g}$ of the same protein in $1 \text{ L}$ of water at $300 \text{ K}$. The osmotic pressure of solution $B$ is $y \text{ bar}$. The entire solution of $A$ is mixed with the entire solution of $B$ at the same temperature. The osmotic pressure of the resultant solution is $z \text{ bar}$. $x, y$ and $z$ respectively are: $(R = 0.083 \text{ L bar mol}^{-1} \text{ K}^{-1})$

$A$ 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 $m^{3}$)?

In which case does osmosis not occur from solution $A$ to solution $B$?

Which one of the following is an isotonic pair of solutions?

The solution containing $4.0 \ g$ of a polyvinyl chloride polymer in $1 \ L$ of dioxane was found to have an osmotic pressure of $6.0 \times 10^{-4} \ atm$ at $300 \ K$. Given $R = 0.082 \ L \ atm \ mol^{-1} \ K^{-1}$,the molecular mass of the polymer is: