Question

Mass of NaOH required for formation of 100ml of its decimolar solution is ???
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Answer 1

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Volume of NaOH = 500 mL

Molarity of the aqueous solution = 0.5 M

Molar mass of NaOH= (23+16+1)g = 40g

Let the mass of NaOH be x.

Now , according to formula number of moles of a given substance = given mass/molar mass

Therefore, no. Of moles of NaOH= x/40

Since, molarity = number of moles of solute/ volume of solution in lit.

Hence, 0.5 = (x/40)/(500/1000)

Solving this we get x= 10g

Hence, 10g of NaOH should be taken to obtain a 0.5 M solution of volume 500 mL.

I hope I could explain the numerical well and help you.

Answer 2

Given:

Volume of $NaOH$ solution $\[ = 100\,ml\]$

Molarity of $NaOH$ solution $\[ = 0.1\,M\]$

To Find: Mass of $NaOH$  solution

Solution:

The molar mass of $NaOH$ can be given as:

$\[\begin{gathered} m = 23 + 16 + 1 \hfill \\ m = 40 \hfill \\ \end{gathered} \]$

Suppose that the mass of $NaOH$ is $x$.

Since

Number of moles $\[ = \frac{{Mass}}{{Molar\,mass}}\]$

Therefore,

Number of moles of $NaOH$ $\[ = \frac{x}{{40}}\]$

Molarity of a solution is given as:

$\[M = \frac{{Moles\,of\,solute}}{{Volume\,of\,solution\,in\,litre}}\]$

$\[\begin{gathered} M = \frac{{x/40}}{{100/1000}} \hfill \\ 0.1 = \frac{{1000x}}{{40 \times 100}} \hfill \\ \end{gathered} \]$

Therefore,

$\[\begin{gathered} x = \frac{{0.1 \times 40 \times 100}}{{1000}} \hfill \\ x = 0.4\,g \hfill \\ \end{gathered} \]$

Hence, the mass of $NaOH$  solution required for formation of $\[100\,ml\]$ of its decimolar solution is $\[0.4\,g\]$.

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