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Question: A solution of \(2.5{\text{ molar}}\) sulphuric acid contains the following: A) \(29.5{\text{ g}}\)...

A solution of 2.5 molar2.5{\text{ molar}} sulphuric acid contains the following:
A) 29.5 g29.5{\text{ g}} of sulphuric acid per litre of solution.
B) 245.0 g245.0{\text{ g}} of sulphuric acid per 1000 ml1000{\text{ ml}} of solution.
C) 24.5 g24.5{\text{ g}} of sulphuric acid per 100 ml100{\text{ ml}} of solution.
D) 59.0 g59.0{\text{ g}} of sulphuric acid per litre of solution.

Explanation

Solution

To solve this we must know that the molarity of a solution is the number of moles of solute per litre of solution. We are given 2.5 molar2.5{\text{ molar}} sulphuric acid. Calculate the mass of sulphuric acid in 2.5 molar2.5{\text{ molar}} sulphuric acid to find the correct answer.

Complete solution:
We know that the molarity of a solution is the number of moles of solute per litre of solution. Thus,
Molarity(M)=Number of moles(mol)Volume of solution(litre){\text{Molarity}}\left( {\text{M}} \right) = \dfrac{{{\text{Number of moles}}\left( {{\text{mol}}} \right)}}{{{\text{Volume of solution}}\left( {{\text{litre}}} \right)}}
We are given 2.5 molar2.5{\text{ molar}} sulphuric acid. The 2.5 molar2.5{\text{ molar}} sulphuric acid indicates that there are 2.5 moles2.5{\text{ moles}} of sulphuric acid per litre or per 1000 ml1000{\text{ ml}} of solution.
We know that the number of moles of any substance is the ratio of its mass to molar mass. Thus,
Number of moles(mol)=Mass(g)Molar mass(g/mol){\text{Number of moles}}\left( {{\text{mol}}} \right) = \dfrac{{{\text{Mass}}\left( {\text{g}} \right)}}{{{\text{Molar mass}}\left( {{\text{g/mol}}} \right)}}
Rearrange the equation for the mass as follows:
Mass(g)=Number of moles(mol)×Molar mass(g/mol){\text{Mass}}\left( {\text{g}} \right) = {\text{Number of moles}}\left( {{\text{mol}}} \right) \times {\text{Molar mass}}\left( {{\text{g/mol}}} \right)
We know that the molar mass of sulphuric acid i.e. H2SO4{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}} is 98.079 g/mol98.079{\text{ g/mol}}. Now, calculate the mass of sulphuric acid in 2.5 moles2.5{\text{ moles}} of sulphuric acid using the equation as follows:
Mass of H2SO4=2.5 mol×98.079 g/mol{\text{Mass of }}{{\text{H}}_2}{\text{S}}{{\text{O}}_4} = {\text{2}}{\text{.5 mol}} \times 98.079{\text{ g/mol}}
Mass of H2SO4=245.0 g{\text{Mass of }}{{\text{H}}_2}{\text{S}}{{\text{O}}_4} = {\text{2}}45.0{\text{ g}}
Thus, 2.5 molar2.5{\text{ molar}} sulphuric acid contains 245.0 g{\text{2}}45.0{\text{ g}} of sulphuric acid per 1000 ml1000{\text{ ml}} of solution.
Now, if 2.5 molar2.5{\text{ molar}} sulphuric acid contains 245.0 g{\text{2}}45.0{\text{ g}} of sulphuric acid per 1000 ml1000{\text{ ml}} of solution then 100 ml100{\text{ ml}} solution contains 24.5 g24.5{\text{ g}} of sulphuric acid.
Thus, 2.5 molar2.5{\text{ molar}} sulphuric acid contains 24.5 g24.5{\text{ g}} of sulphuric acid per 100 ml100{\text{ ml}} of solution.
Thus, 2.5 molar2.5{\text{ molar}} sulphuric acid contains 245.0 g{\text{2}}45.0{\text{ g}} of sulphuric acid per 1000 ml1000{\text{ ml}} of solution and 24.5 g24.5{\text{ g}} of sulphuric acid per 100 ml100{\text{ ml}} of solution.

Thus, the correct options are (B) and (C).

Note: Remember that the molarity of a solution is the number of moles of solute per litre of solution. Molarity is a common term used to express concentration of the solution. The units of molarity are mol/L{\text{mol/L}} or M{\text{M}}.