Question

1.Determine the molarity pf a calcium nitrate solution that contains 40.0 grams in a total solution volume of 250 mL.


2. How many grams of lithium carbonate are required to prepare 250 mL of a 0.655 molar solution?

3. What volume of 2.00 molar H2SO4 is needed to prepare 750 mL of 0.0950 molar sulfuric acid?

Answer 1

1.Determine the molarity pf a calcium nitrate solution that contains 40.0 grams in a total solution volume of 250 mL.
solution :- molarity = moles / volume in liter

Moles = mass/ molar mass

Moles of Ca(NO3)2 = 40.0 g / 164.088 g per mol

                                  = 0.2438 mol

Molarity = 0.2438 mol / 0.250 L

                = 0.975 M

2. How many grams of lithium carbonate are required to prepare 250 mL of a 0.655 molar solution?
solution :-

Moles = molarity * volume in liter

          = 0.655 mol per L * 0.250 L

         = 0.16375 mol

Mass of Lithium carbonate = moles * molar mass

                                            = 0.16375 mol * 73.891 g per mol

                                            = 12.1 g

3. What volume of 2.00 molar H2SO4 is needed to prepare 750 mL of 0.0950 molar sulfuric acid?

Solution :- M1V1=M2V2

V1 = M2V2/M1

     = 0.0950 M * 750 ml / 2.00 M

     = 35.6 ml

So we need 35.6 ml of 2.00 M H2SO4 solution