Question

A 20 % (W/W) solution of $NaOH$ is 5M. The density of the solution is:
(A) 1 g/ml
(B) 2 g/ml
(C) 0.5 g/ml
(D) 0.25 g/ml

Answer 1

Hint: The concentration of solution which describes the composition of the solution either qualitatively or quantitatively. To describe the concentration of solution quantitatively more methods are available. There are mass percentages, volume percentage, ppm, mole fraction, normality, molarity, and molality.

Complete step by step answer:
In the lower classes of chemistry, we have studied the basic physical chemistry parameters like molarity, molality, normality, mole fraction etc. and its calculation. Let us see these concepts in detail where the molarity calculation plays a role here and based on which we can approach the required answer.
- Molarity (M): It is defined as the number of moles of solute dissolved in one litre of solution. The value of molarity expressed in moles/litre which represents the concentration of a solution. The molarity of solution can be calculated by using the below formula, Molarity, $M=\dfrac{n}{V(ml)}\times 1000$
Where n = number of moles of solute
V = volume of solution in ml.
In the given problem, solute = $NaOH$
The gram molecular weight or molar mass of $NaOH$ = 40 g/mol
Weight percentage of $NaOH$, (W/W) = 20 %
The relation between weight percentage and molarity,
\[Molarity(M)=\dfrac{%(W/W)\times 10\times d}{GMW(NaOH)}\]
 D = density of the solution has to be found.
Given molarity of $NaOH$ = 5M
Then, $5 = \dfrac{20\times 10\times d}{40g/mol}$
5M = 5d
Therefore, d = 1g/ml
Hence, density of the $NaOH$ solution = 1g/ml
The correct answer is option “A” .

Note: The volume of the solution may change with increasing or decreasing the temperature of the solution when the moles of solute are constant. At lower concentration with temperature-independent then molarity and molality values will be equal.