Moles, Concentration & Volume

A 1.3 dm³ sample of sulfuric acid (H₂SO₄) was prepared with a concentration of 2.5 mol dm³.

What was the mass (in grams) which was used to prepare this sample?

(a)  

A 1.3 dm³ sample of sulfuric acid (H₂SO₄) was prepared with a concentration of 2.5 mol dm³.

How many moles were used to prepare this solution?

(a)  

A sample of sulfuric acid (H₂SO₄) was prepared with 3.25 moles to create a concentration of 2.5 mol dm³.

What was the volume of this solution in dm³ ?

(a)  

A sample of sulfuric acid (H₂SO₄) was prepared by adding 3.25 moles to a volume of 1300 mL.

What was the concentration of this solution in mol dm^-3 ?

(a)  

A sample of sulfuric acid (H₂SO₄) was prepared by adding 0.01 grams to a volume of 0.05 dm³.

What was the concentration of this solution in mol dm^-3 ?

(a)  

A sample of sulfuric acid (H₂SO₄) was prepared by adding 0.01 grams to make a solution with a concentration of 0.2 mol dm^-3.

What was the volume of this solution in dm³ ?

(a)  

A sample of hydrochloric acid (HCl) was prepared by adding 0.004 moles to make a solution with a concentration of 1.25 mol dm^-3.

What was the volume of this solution in dm³ ?

(a)  

A sample of hydrochloric acid (HCl) was prepared by adding 0.146 grams to make a solution with a concentration of 1.25 mol dm^-3.

What was the volume of this solution in dm³ ?

(a)  

A sample of hydrochloric acid (HCl) was prepared by adding 0.004 moles to make a solution with a volume of 3.2 cm³.

What was the concentration of this solution in mol dm^-3 ?

(a)  

A sample of hydrochloric acid (HCl) was prepared with a concentration of 1.25 mol dm^-3 and a volume of 3.2 cm³.

How many grams of HCl were added to the solution?

(a)  

Answer explanation

A sample of hydrochloric acid (HCl) was prepared with a concentration of 1.25 mol dm^-3 and a volume of 3.2 cm³.

Step 1) Convert volume to correct units by dividing 3.2 by 1000 to give 0.0032 dm³.

Step 2) n = c x V, n = 1.25 x 0.0032 = 0.004 moles

Step 3) Calculate molar mass of HCl: (1.0 x 1) + (35.5 x 1) = 36.5 g/mol

Step 4) mass = n x M, 0.004 x 36.5 = 0.146 grams