$10 \ g$ of $NaOH$ is dissolved in $500 \ mL$ of aqueous solution. Calculate the molarity of this solution? (Given,formula weight of $NaOH = 40$)

If $1 \ M$ and $2.5 \ L$ $NaOH$ solution is mixed with another $0.5 \ M$ and $3 \ L$ $NaOH$ solution,then the molarity of the resultant solution will be $........ \ M$.

$A$ 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

$5.0 \ g$ of sodium hydroxide (Molar mass $40 \ g \ mol^{-1}$) is dissolved in a small quantity of water and the solution is diluted up to $100 \ mL$. What is the molarity of the resulting solution?

$3.0$ molal $NaOH$ solution has a density of $1.110 \ g/mL$. The molarity of the solution is :-

Normality $N$ of a solution is equal to