The molarity of the solution containing 7.1g of ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}$  in 100 mL of aqueous solution

Gram molar mass ,GMW of ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}$ $\text{=142 g}$

Volume of solution, $\text{V =100 mL}$

The formula of molarity is

             $\text{M = }\frac{\text{w}}{\text{GMW}}\times \frac{1000}{\text{V}\text{in mL}}$

Here, M represents the molarity of solution.

Substitute the known data in the above equation to get molarity of solution.

 After that ,

                     $\text{M = }\frac{7.1}{142}\times \frac{1000}{100}=0.5$

Therefore, molarity of solution containing 7.1g of ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}$  in 100 mL of aqueous solution is 0.5 M.