Question

If you need to make $2L$ of a $2M$ solution of NaOH, how many grams of NaOH would you need to make the solution?

Answer 1

Hint: We need to know about molarity before solving this question. A $2M$ solution of NaOH means $2$ moles of solute i.e., NaOH pellets are dissolved in $1\,L$ of solution. We should also be familiar with the mole concept in order to solve this question.

Formula used:$Molarity = \dfrac{{no.\,of\,moles\,of\,solute}}{{Volume\,of\,solution\,in\,L}}$ in units of $mol\,{L^{ - 1}}$ or M

Complete step-by-step answer:
We are given with the following values in the question
Volume of solution$ = 2L$
Molarity of NaOH solution $ = 2M$
By using the following formula,
$Molarity = \dfrac{{no.\,of\,moles\,of\,solute}}{{Volume\,of\,solution\,in\,L}}$
Substituting the values,
$2M = \dfrac{{no.\,of\,moles\,of\,NaOH}}{{2L}}$
Number of moles of NaOH $= 2\,mol\,{L^{ - 1}} \times 2L \\ = \,4\,mol \\
 $
We need to find the grams of NaOH required to make the given solution,
For this, we need to first find the molecular mass of NaOH
Molecular mass of NaOH $ = 23g\,mo{l^{ - 1}} + 16g\,mo{l^1} + 1g\,mo{l^1}$ $ = 40\,g\,mo{l^{ - 1}}$
Using the relation between the number of moles and molecular mass,
No. of moles $ = \dfrac{w}{M}$
Where w is the given weight in g.
M is the molecular mass in $g\,mo{l^{ - 1}}$
Substituting the values in the relation,
No. of moles $ = 4$
Molecular mass of NaOH $ = 40\,g\,mo{l^{ - 1}}$
$\Rightarrow w = n \times M$
$
\Rightarrow w = 4mol \times 40\,g\,mo{l^{ - 1}} \\
\Rightarrow 160\,g \\
 $
$160\,g$is the required mass of NaOH required to make $2L$ of a $2M$ solution.

Hence, $160\,g$ is the required answer.

Note: While doing these questions, we need to know which concentration term is used in the question as sometimes molality and normality are also used. We need to remember the molar weight of the elements as sometimes these are not provided in the question. Also, we need to see correctly what is asked in the question as sometimes numbers of moles are asked only. Hence, there is no need to calculate weight in such cases.