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

Hint: 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}} $ .