Question

Density of a ${H_2}S{O_4}$ solution is $1.2g/ml$ and it is $40\% $ ${H_2}S{O_4}$ by weight . Determine molarity of this solution .
A.2.9 M
B.3.9 M
C.4.9 M
D.5.9 M

Answer 1

Hint: To determine the molarity of a solution we should know the number of moles of solute present in the solution and the volume of solution . Since in the above question volume and number of moles are not mentioned we will first find the value of number of moles and volume of the solution .

Complete step by step answer:
Let us first find the volume of the solution . Density of ${H_2}S{O_4}$ is given and we know density is defined as mass per unit volume . Therefore to find volume the formula will be
$Volume = \dfrac{{Mass}}{{Density}}$
The given solution is $40\% $ ${H_2}S{O_4}$ by weight which means 40g of ${H_2}S{O_4}$ is dissolved in 100g of solution .
On substituting the value of density as $1.2g/ml$ mass as 100 , we get
Volume of 100g of the solution = $\dfrac{{100g}}{{1.2g/ml}}$ = $83.33ml$
Therefore the volume of solution is $0.0833litres$ .
Now , to find out the number of moles of sulphuric acid let us first find its molar mass
molar mass of ${H_2}S{O_4}$ = $(2 \times 1) + 32 + (4 \times 16) = 98gmo{l^{ - 1}}$
Number of moles of ${H_2}S{O_4}$ = $\dfrac{w}{M}$ , where w = mass of solute and M= molar mass of solute
Therefore number of moles = $\dfrac{{40}}{{98}} = 0.408moles$
Now we have to find the molarity of this solution
Molarity is defined as the number of moles of the solute present per litre of the solution . It is represented by M .
$M = \dfrac{n}{V}$ , where n = number of moles and V = volume of the solution .
On substituting the values of n and V we get ,
$M = \dfrac{{0.408}}{{0.0833}}$ = $4.9M$
The molarity of given solution is $4.9M$

So, the correct answer is Option C .

Note:
While calculating molarity we should always keep in mind to take the volume of solution in litres .
The value of molarity is different at different temperatures. The molarity of a solution changes with change in temperature .