Question

Volume of 18 M ${{\text{H}}_{2}}\text{S}{{\text{O}}_{4}}$ required to prepare 1 L of a 0.9 M solution of ${{\text{H}}_{2}}\text{S}{{\text{O}}_{4}}$ is:
A. 50 mL
B. 10 mL
C. 500 mL
D. 5 mL

Answer 1

Hint: The molarities of both the solutions is given, both solutions are of sulphuric acid only. We need to equalize the moles of both the solutions of sulphuric acid and find the answer by putting the values in the formula $\text{moles =Molarity}\times \text{Volume}$.

Complete answer:
Let us solve this question step by step using the formula of molarity:
Step (1)- The moles of solution can be found by $\text{moles =Molarity}\times \text{Volume}$.
The moles of solution 1 are moles = $18\times \text{V}$;
Where molarity of solution is 18 M and the volume is what we have to find,
Assume it to be V.
So, moles are 18V.
Step (2)- The moles of solution can be found by $\text{moles =Molarity}\times \text{Volume}$.
The moles of solution 2 are moles = $0.9\times 1$ ;
Where molarity of solution is 0.9 M and the volume is 1 Litre.
So, the moles are 0.9.
Step (3)- We know that moles of both solutions should be the same,
So, moles of solution 1 are equal to moles of solution 2.
So, the result is $18\text{V=0}\text{.9}$ ,
The value of volume of solution 1 is 0.05 Litres.
1 litre is equal to 1000 mL.
So, the volume of ${{\text{H}}_{2}}\text{S}{{\text{O}}_{4}}$ solution is $0.05\times 1000$ which is equal to 50 mL.
Therefore, volume of 18 M ${{\text{H}}_{2}}\text{S}{{\text{O}}_{4}}$ required to prepare 1 L of a 0.9 M solution of ${{\text{H}}_{2}}\text{S}{{\text{O}}_{4}}$ is 50 mL.

Hence option A is the correct one.

Note:
This question can be solved directly using the formula ${{\text{M}}_{1}}{{\text{V}}_{1}}\text{ = }{{\text{M}}_{2}}{{\text{V}}_{2}}$. The value of ${{\text{M}}_{1}}$= 18 M, ${{\text{V}}_{1}}$= V (which we have to find), ${{\text{M}}_{2}}$= 0.9 M and ${{\text{V}}_{2}}$= 1 litre or 1000 mL. Put the values in the formula, we will get, $18\times \text{V=0}\text{.9}\times \text{1000}$ or V=50 mL.