Question

Calculate the weight of $NaOH$ present in $500ml\,of\,0.2M\,NaOH\,$ solution.

Answer 1

Molarity of a solution can be used to calculate the weight of the solution as it is an intensive property and depends on the mass of the solute.Use the molarity formula to find out the value of mass of sodium hydroxide present.
Formula used:
$Molarity = \dfrac{{number\,of\,moles\,of\,solute}}{{Volume\,of\,solution}}$
$Number\,of\,moles = \dfrac{{Given\,mass}}{{Molecular\,mass}}$

Complete step by step answer:
Let us first understand what Molarity means and how it is affected when mixing takes place.
Molarity is a unit which is used to represent concentration.
$Molarity = \dfrac{{number\,of\,moles\,of\,solute}}{{Volume\,of\,solution}}$
This formula can be rearranged to determine the value of Molarity for the number of moles in the solute. $number\,of\,moles = Molarity(M) \times Volume(V)$
From the information given to us in the question:
$M = 0.2M,V = 500mL$
Substituting this value in the above equation.
$number\,of\,moles = 500mL \times 0.2M$
Solving this equation, we get:
$number\,of\,moles = 0.1$
Now, the relation between the number of moles and given mass can be given by:
$Number\,of\,moles = \dfrac{{Given\,mass}}{{Molecular\,mass}}$
To calculate the molecular mass of $NaOH$
$Mass\,of\,Na + Mass\,of\,O + Mass\,of\,H$
$Mass\,of\,Na = 23g,Mass\,of\,O = 16g,Mass\,of\,H = 1g$
From this we get that the mass of $NaOH$= $Mass\,of\,Na + Mass\,of\,O + Mass\,of\,H$
$Mass\,of\,NaOH = 23 + 16 + 1 \Rightarrow 40g$
Substituting this value in the above formula , we get:
$0.1moles = \dfrac{{given\,mass}}{{40g}}$
Solving for the value of given mass, we get:
$given\,mass = 4g$

Note: Molarity helps you to identify exactly how much of an element is present in the solution as a whole in comparison to other measurements of concentration such as molality which only tell you the amount of solute with respect to the solvent. Volume in the molarity formula refers to the volume of the solution, not the solvent. This volume had an addition of volume of solute as well as the volume of the solvent.