$A$ $5\% \text{ (w/v)}$ 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 $\text{g/mol}$?

$x \% (w/v)$ solution of urea is isotonic with $4 \% (w/v)$ solution of a non-volatile solute of molar mass $120 \ g \ mol^{-1}$. The value of $x$ is

$A$ $25 \%$ solution of cane-sugar (molar mass $= 342 \ g \ mol^{-1}$) is isotonic with a $5 \%$ solution of a substance $A$. Find the molecular weight of $A$.

The process in which solvent molecules pass through a semipermeable membrane is called:

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.)

Calculate the molar mass of a solute when $4 \ g$ of it is dissolved in $1 \ dm^3$ of solvent,having an osmotic pressure of $2 \ atm$ at $300 \ K$. $[R = 0.082 \ dm^3 \ atm \ K^{-1} \ mol^{-1}]$