The molarity of a solution containing $15 \text{ g}$ of $\text{NaOH}$ in $900 \text{ ml}$ of solution will be . . . . . . . (in $\text{ M}$)

$A$ 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:

$A$ solution of two components containing $n_{1}$ moles of the $1^{st}$ component and $n_{2}$ moles of the $2^{nd}$ component is prepared. $M_{1}$ and $M_{2}$ are the molecular weights of component $1$ and $2$ respectively. If $d$ is the density of the solution in $g \ mL^{-1}$, $C_{2}$ is the molarity and $x_{2}$ is the mole fraction of the $2^{nd}$ component, then $C_{2}$ can be expressed as

By dissolving $0.35 \, \text{mole}$ of sodium chloride in water,$1.30 \, \text{L}$ of salt solution is obtained. The molarity of the resulting solution should be reported as $..... \, \text{M}$.

$A$ commercially sold concentrated $HCl$ is $35 \%$ $HCl$ by mass. If the density of this commercial acid is $1.46 \ g/mL$,the molarity of this solution is $....M$.
(Atomic mass : $Cl = 35.5 \ amu, H = 1 \ amu$)

Calculate the mass of Glucose $(C_6H_{12}O_6)$ required in making $2.5 \ kg$ of $0.25 \ molal$ aqueous solution. [Atomic wt : $H=1, O=16, C=12 \ amu$] (in $g$)