Question

5.0 g of sodium hydroxide (NaOH-molar mass 40 g mol^-1) is dissolved in little quantity of water and the solution is diluted up to 100 ml. what is the molarity of the resulting solution?

• A. $0.1\mathrm{moles}-{\mathrm{dm}}^{-3}$
• B. $1.0\mathrm{moles}-{\mathrm{dm}}^{-3}$
• C. $1.25\mathrm{moles}-{\mathrm{dm}}^{-3}$
• D. $0.125\mathrm{moles}-{\mathrm{dm}}^{-3}$

Answer 1

The correct option is C

$1.25\mathrm{moles}-{\mathrm{dm}}^{-3}$

Molarity(M): The number of moles of a solute dissolved in 1 L (1000 ml) of the solution is known as the molarity of the solution.
$\mathrm{Molarity}\mathrm{of}\mathrm{solution}\left(\mathrm{M}\right)=\frac{\mathrm{Moles}\mathrm{of}\mathrm{solute}}{\mathrm{volume}\mathrm{of}\mathrm{solution}\mathrm{in}\mathrm{litre}}$

As per the question,

$\mathrm{Moles}\mathrm{of}\mathrm{solute}=\frac{Givenmassofsolute}{Molecularmassofsolute}=\frac{5}{40}=0.125\mathrm{moles}$

$\mathrm{Volume}\mathrm{of}\mathrm{solution}(\mathrm{in}\mathrm{ml})=\frac{100}{1000}=0.1\mathrm{ml}$

$\mathrm{Molarity}\mathrm{of}\mathrm{solution}\left(\mathrm{M}\right)=\frac{\mathrm{Moles}\mathrm{of}\mathrm{solute}}{\mathrm{volume}\mathrm{of}\mathrm{solution}\mathrm{in}\mathrm{litre}}=\frac{0.125}{0.1}=1.25\mathrm{Mol}-{\mathrm{dm}}^{-3}$

Conclusion Statement:

Hence option(c) is the correct answer.