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Question: \( 9.8grams\,of\,{H_2}S{O_4} \) is present in \( {\text{2 }}litres \) of a solution. The molarity of...

9.8grams of H2SO49.8grams\,of\,{H_2}S{O_4} is present in 2 litres{\text{2 }}litres of a solution. The molarity of the solution is:
A. 0.1M0.1M
B. 0.05M0.05M
C. 0.2M0.2M
D. 0.01M0.01M

Explanation

Solution

Hint : In order to solve the given question, first we will rewrite the given mass of sulphuric acid and then find the no. of moles of sulphuric acid. And with the help of no. of moles and the given mass of the sulphuric acid, we can find the molarity of the given solution.

Complete Step By Step Answer:
As we know that, the molar mass of sulphuric acid is 98g/mol98g/mol .
And, a mass of sulphuric acid is given, which is 9.8g9.8g .
Now, we will find the number of moles of Sulphuric acid:
No. of moles of H2SO4=Mass of H2SO4Molar mass of  H2SO4 =9.8g98g/mol=0.1molNo.\,of\,moles\,of\,{H_2}S{O_4} = \dfrac{{Mass\,of\,{H_2}S{O_4}}}{{Molar\,mass\,of\,\,{H_2}S{O_4}\,}} = \dfrac{{9.8g}}{{98g/mol}} = 0.1mol
Now, we can find the molarity of the given solution:
Molarity of the solution=No. of moles of  H2SO4Volume of solution=0.1mol2L=0.05MMolarity\,of\,the\,solution = \dfrac{{No.\,of\,moles\,of\,\,{H_2}S{O_4}}}{{Volume\,of\,solution}} = \dfrac{{0.1mol}}{{2L}} = 0.05M
Hence, the correct option is (B).

Note :
Molarity is represented by MM , which is termed as molar. One molar is the molarity of a solution where one gram of solute is dissolved in a litre of solution. As we know, in a solution, the solvent and solute blend to form a solution, hence, the total volume of the solution is taken.