When $1.04 \ g$ of $BaCl_2$ is dissolved in $10^5 \ g$ of water,what will be the concentration of the solution in $ppm$?

$0.1 \ mol$ of $NaCl$ is dissolved in $100 \ g$ of water. The mole fraction of $NaCl$ is

$A$ solution of sodium sulphate contains $92 \ g$ of $Na^{+}$ ions per kilogram of water. The molality of $Na^{+}$ ions in that solution in $mol \ kg^{-1}$ is:

What is the relation between molality $(m)$ of the solution and molar mass $(M_2)$ of the solute,where $W_2$ is the mass of solute in grams and $W_1$ is the mass of solvent in grams?

What is the mole fraction of the solute in a $1.00 \, m$ aqueous solution?

Molality $(m)$ of $3 \text{ M}$ aqueous solution of $NaCl$ is:
(Given: Density of solution = $1.25 \text{ g mL}^{-1}$, Molar mass in $\text{g mol}^{-1}$: $Na = 23$, $Cl = 35.5$) (in $\text{m}$)