Question

What is the molarity of ${H_2}S{O_4}$ solution that has a density of $1.84g/cc$ at ${35^0}C$ and contains $98\% $ by weight?
A . $4.18M$
B . $8.14M$
C . $18.4M$
D . $18M$

Answer 1

Hint :In the mole concept we have studied about the molarity and we have studied that molar concentration of substance is also known as molarity. It is defined as the ratio of moles of substance to volume of solution in litre.

Complete answer:
> Mole is defined as weight in grams divided by the molecular weight of the substance. So we will find out the molarity by the formula; M=Number of moles of solute /volume of the solution in litre.
> As it is given in the problem that ${H_2}S{O_4}$ is $98\% $ by weight that means that $100g$ solution contains $98g$${H_2}S{O_4}$ by mass.
> Density of solution of sulphuric acid is given that is $1.84g/cc$; $Density = \dfrac{{mass}}{{Volume}}$ $ \Rightarrow Volume = \dfrac{{mass}}{{Density}}$
$ \Rightarrow Volume = \dfrac{{100}}{{1.84 \times 1000}}L$
$ \Rightarrow Volume = 0.0543$ Litre
> Molecular weight of sulphuric acid (${H_2}S{O_4}$) = 2×atomic weight of hydrogen atomic weight of sulphur +4×atomic weight of oxygen
Molecular weight of sulphuric acid (${H_2}S{O_4}$) $ = 2 \times 1 + 32 + 4 \times 16$
$ = 98$
- Now here weight of sulphuric acid in gram is $98g$ and molecular weight of sulphuric acid is $98g/mol$
- So first calculate the number of moles before calculating molarity. Number of moles of ${H_2}S{O_4}$=Weight in gram of ${H_2}S{O_4}$/ molecular weight of ${H_2}S{O_4}$
$ = 98/98$$ = 1$
- Hence the molarity of ${H_2}S{O_4}$=Number of moles /Volume in litre $=\dfrac{1}{{0.0543}}$
$ = 18.4M$
So the correct option is (c) that is $18.4M$.

Note : In mole problems we should always remember some important relation .Here in this problem we have used some relations to solve the problem as we have found the mass with the help of given density and then put the value of mass in the formula to find the number of moles. Once the number of moles is determined we calculate molarity of sulphuric acid easily with the help of molarity formula.