How many grams of a 0.300m solution of hydrochloric acid would contain 6.20g of pure hydrochloric acid?

680.4 mL

Explanation:

Determine the no. of moles of HCl in 6.20 g of pure HCl as follows:

No. of mole = 6.20/36/46

                   = 0.170 mol

Now, determine the volume of solution that contains 0.170 mol of HCl

Molarity of the solution is 0.300 M

That means 0.300 moles of HCl is present in 1 L of the solution.

Therefore, 0.170 mol will be present in  

                                                 =

Density of HCl solution is 1.20 g/mL

Mass = density × volume

         = 1.20 g/mL × 567 mL

          = 680.4 mL

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