Question

Amount of sodium hydroxide present in 500 ml of \[0.2\]M solutions is:
A.4 g
B.8 g
C.40 g
D.20 g

Answer 1

Hint:This question can be solved from the concept of molarity. Molarity is defined as the number of moles of a substance present per litre of the substance. It is also known as the molar concentration of the solution.
Formula used: ${\text{M = }}\dfrac{{\text{n}}}{{\text{v}}}$
where n is the number of moles of the substance and v is the volume of the solution in litres.

Complete step by step answer:
According to the question, the volume of the solution is: 500 ml
While the molarity of the solution is: \[0.2\;{\text{mol}}{{\text{L}}^{{\text{ - 1}}}}\]
Therefore it can be said from the equation of molarity that, in 1 L of sodium hydroxide, there are \[0.2\] moles of it.
So in 500 ml, the amount of sodium hydroxide present is:
$0.2 \times \dfrac{{500}}{{1000}} = 0.1$ mole.
Now, no. of moles, n is mathematically expressed by:
${\text{n = }}\dfrac{{\text{w}}}{{\text{M}}}$
$ \Rightarrow {\text{w = n}} \times {\text{M}}$
Where, w is the weight of the substance and M it the molecular weight of it.
Therefore, molecular weight of sodium hydroxide = $\left[ {23 + 16 + 1} \right]$= 40 ${\text{gmo}}{{\text{l}}^{ - 1}}$
$ \Rightarrow {\text{w}} = 0.1 \times 40 = 4{\text{ g}}$

So, the correct answer is option (A).

Note:
This problem can be solved alternatively as:
${\text{Molarity = }}\dfrac{{\text{w}}}{{\text{M}}} \times \dfrac{{1000}}{{\text{v}}}$
$ \Rightarrow 0.2 = \dfrac{{\text{w}}}{{40}} \times \dfrac{{1000}}{{500}}$
$ \Rightarrow {\text{w}} = 4{\text{g}}$
Molality is another unit besides Molarity to express the concentration of the solution.
Molality (m) is defined as the amount of a substance or solute dissolved per Kg of the solvent. The difference between molarity and molality is that, in molarity the amount of the substance is expressed in volumes while in molality it is expressed in the units of mass.