The amount of water that should be added to $500 \, mL$ of $0.5 \, N$ solution of $NaOH$ to give a concentration of $10 \, mg$ per $mL$ is (in $, mL$)

$50 \ g$ of a substance is dissolved in $1 \ kg$ of water at $90^{\circ} C$. The temperature is reduced to $10^{\circ} C$. The density is increased from $1.1 \ g \ cm^{-3}$ to $1.15 \ g \ cm^{-3}$. What is the $\%$ change of molarity of the solution?

Find out the molarity of a solution when $0.01 \ mol$ of a substance is dissolved in $10 \ g$ of water. (in $M$)

$60 \ g$ of $CH_3COOH$ is dissolved in $1 \ dm^3$ of solvent. What is the molality of the solution (in $m$)? (Density $= 1.25 \ g / cm^3$)

The molality of $90\%$ ${H_2SO_4}$ solution is [density $= 1.8 \ g/mL$].

What is the molarity of a solution containing $0.8 \ g$ of $NaOH$ (molar mass $40 \ g \ mol^{-1}$) in $150 \ cm^{3}$ of water?