The osmotic pressure of a solution is given by the relation

$18 \ g$ glucose (Molar mass $= 180 \ g/mol$) is dissolved in $100 \ mL$ of water at $300 \ K$. If $R = 0.0821 \ L \ atm \ K^{-1} \ mol^{-1}$,what is the osmotic pressure of the solution (in $atm$)?

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 $g$)

Assuming complete ionisation,arrange the following solutions in order of increasing osmotic pressure.
$(a)$ $0.5 \ m \ Li_2SO_4$
$(b)$ $0.5 \ m \ KCl$
$(c)$ $0.5 \ m \ Al_2(SO_4)_3$
$(d)$ $0.1 \ m \ BaCl_2$

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: $(R = 0.083 \text{ L bar mol}^{-1} \text{ K}^{-1})$

At a constant temperature,the osmotic pressures of a $1.5 \ M \ NH_4NO_3$ aqueous solution and an $x \ M \ Al_2(SO_4)_3$ aqueous solution are approximately equal. Find the value of $x$. (Assume $100 \%$ dissociation of the solute in the solution.)