Question

What is the molarity of a solution containing 10gm of NaOH in 500mL of a solution?
(A) 0.25$mol{L^{ - 1}}$
(B) 0.75$mol{L^{ - 1}}$
(C) 0.5$mol{L^{ - 1}}$
(D) 1.25$mol{L^{ - 1}}$

Answer 1

Formula to find molarity is $M = \dfrac{{{\text{Weight of Solute(gm)}}}}{{{\text{Molecular weight of Solute(gmmo}}{{\text{l}}^{ - 1}}) \times {\text{Volume of Solution(L)}}}}$
Atomic weight of Sodium is 23${\text{gmmo}}{{\text{l}}^{ - 1}}$

Complete answer:
We know the formula of Molarity which is $M = \dfrac{{{\text{Weight of Solute(gm)}}}}{{{\text{Molecular weight of Solute(gmmo}}{{\text{l}}^{ - 1}}) \times {\text{Volume of Solution(L)}}}}$
We are given weight of solute(NaOH) which is equal to 10gm.
Let’s find the molecular weight of NaOH.
${\text{Molecular weight of NaOH = Atomic weight of Na + Atomic weight of O + Atomic weight of H}}$
${\text{Molecular weight of NaOH = 23 + 16 + 1}}$
{\text{Molecular weight of NaOH = 40}}$$$${\text{gmmo}}{{\text{l}}^{ - 1}}
Now, we are given a volume of solution in mL, let’s convert it into L.
$1000mL = 1L$
$500mL = \dfrac{{500 \times 1}}{{1000}}$L
$500mL = 0.5L$
Let’s put all these values in formula of molarity
$M = \dfrac{{10}}{{40 \times 0.5}}$
Molarity of given NaOH Solution, M= 0.5$mol{L^{ - 1}}$

So, correct answer of this question is (C) 0.5$mol{L^{ - 1}}$

Additional Information:
We have another formula to find the molarity of a solution where we can put the volume of solution directly in the mL unit. That formula is as given below.
$M = \dfrac{{{\text{Weight of Solute(gm)}} \times {\text{1000}}}}{{{\text{Molecular weight of Solute(gmmo}}{{\text{l}}^{ - 1}}) \times {\text{Volume of Solution(mL)}}}}$
Using this formula, we will need not to convert Volume of solution in mL to Litres.
Molarity is expressed in $mol{L^{ - 1}}$ unit.
Molarity expresses the concentration of the solute in the given volume of the solution.
With the change in temperatures, the molarity of solution also changes because volume of solution changes with change in temperatures.

Note:
Do not forget to put the weight of solution and volume of solution in proper units as described in the formula, not doing so will lead to errors. Also make sure that you find the true molecular weight of the compound. Remember atomic weight of simple atoms that are used in the regular compounds.