Question

The volume of water added to $500{\text{ mL}}$ , ${\text{0}}{\text{.5 M NaOH}}$ solution so that its strength becomes ${\text{10 mg NaOH}}$ per ${\text{mL}}$ :

Answer 1

The strength of solution is the ratio of mass of solute and volume of solution. We will find the moles of ${\text{NaOH}}$ present in the solution. Then we can find the mass of ${\text{NaOH}}$ with the help of moles and molecular mass of ${\text{NaOH}}$ . Thus after addition of water we can find its strength by using the above relation.
$(i){\text{ number of moles = molarity }} \times {\text{ volume of solution}}$
$(ii){\text{ Strength = }}\dfrac{{mass{\text{ }}of{\text{ }}solute}}{{volume{\text{ }}of{\text{ }}solution}}$ .

Complete Step By Step Answer:
Since the volume of ${\text{NaOH}}$ solution is $500{\text{ mL}}$ and its molarity is given as $0.5{\text{ M}}$ . Therefore the number of moles can be find using relation as:
$\Rightarrow {\text{ number of moles = molarity }} \times {\text{ volume of solution}}$
$\Rightarrow {\text{ number of moles = 0}}{\text{.5 }} \times {\text{ 500 mL}}$
$\Rightarrow {\text{ number of moles = 0}}{\text{.5 }} \times {\text{ 500 }} \times {\text{ 1}}{{\text{0}}^{ - 3}}$
$\Rightarrow {\text{ number of moles = 0}}{\text{.25 mole}}$
Now we will find the mass of ${\text{0}}{\text{.25 mole NaOH}}$ as:
$\Rightarrow {\text{mass = moles }} \times {\text{ molecular mass}}$
The molecular mass of ${\text{NaOH}}$ is: $23 + 16 + 1 = 40$
Therefore the mass of ${\text{NaOH}}$ will be:
$\Rightarrow {\text{mass = 0}}{\text{.25 }} \times {\text{ 40}}$
$\Rightarrow {\text{mass = 10 g}}$
We know that strength of solution can be find as:
${\text{Strength = }}\dfrac{{mass{\text{ }}of{\text{ }}solute}}{{volume{\text{ }}of{\text{ }}solution}}$
Now let $x$ be volume of water is added to $500{\text{ mL}}$ of ${\text{NaOH}}$ , then its total volume will be $\left( {x + 500} \right){\text{ mL}}$ . Then its strength will be:
$\Rightarrow {\text{Strength = }}\dfrac{{mass{\text{ }}of{\text{ }}solute}}{{volume{\text{ }}of{\text{ }}solution}}$
$\Rightarrow {\text{Strength = }}\dfrac{{10}}{{\left( {x + 500} \right)}}$
It is given that strength of solution will be ${\text{10 mg NaOH}}$ per ${\text{mL}}$ , therefore:
$\Rightarrow 10{\text{ mg = }}\dfrac{{10}}{{\left( {x + 500} \right)}}$
$\Rightarrow 10{\text{ }} \times {\text{ 1}}{{\text{0}}^{ - 3}}{\text{ g = }}\dfrac{{10}}{{\left( {x + 500} \right)}}$ (Milligram is being converted into gram)
$\Rightarrow \left( {x + 500} \right){\text{ = }}\dfrac{{10}}{{10{\text{ }} \times {\text{ 1}}{{\text{0}}^{ - 3}}}}$
$\Rightarrow \left( {x + 500} \right){\text{ = 1}}{{\text{0}}^3}$
$\Rightarrow x + 500{\text{ = 1000}}$
$\Rightarrow x{\text{ = 500}}$
Therefore the volume of water will be $500{\text{ mL}}$ . Hence we can say that the volume of water added must be $500{\text{ mL}}$ .

Note:
We should convert milligram into gram and milliliter to litre while finding the number of moles of ${\text{NaOH}}$ in the solution otherwise the number of moles will be in millimoles. For all these conversions we have divided them by a factor of $1000$ . The volume of water can be found in litres by dividing it by a factor of $1000$ . The molecular mass of ${\text{NaOH}}$ can be found by algebraic sum of mass of each atom present in ${\text{NaOH}}$ .