Question

1. A student prepares a 250.0 mL solution using 41.7 grams of potassium nitrite. They then take 29.9 mL of this solution and dilute it to a final volume of 250.0 mL. How many grams of potassium nitrite are in a 36.2 mL sample of this final diluted solution?

2.

If you wanted to make a 4.50 M solution of aluminum nitrate using a 400.00 mL volumetric flask, how many grams of this salt would you need to use? 3. H2SO4 + 2KOH  →  2H2O + K2SO4 How many mL of a 1.75 M solution of potassium hydroxide are required to titrate 15.72 mL of a 2.13 M solution of sulfuric acid?

Answer 1

Answer –

1) We are given, mass of KNO₂ = 41.7 g , volume = 250.0 mL

First we need to calculate the moles of KNO₂

Moles of KNO₂ = 41.7 g / 85.104 g.mol^-1

                      = 0.490 moles

So initial concentration of KNO₂ = 0.490 moles / 0.250 L

                                              = 1.96 M

So, M1= 1.96 M, V1 = 29.9 mL, V2 = 250.0 mL , M2 = ?

By dilution law

M1V1 = M2V2

So, M2 = M1V1/V2

            = 1.96 M * 29.9 mL / 250.0 mL

            = 0.234 M

So the final solution with 0.234 M and from this solution when we take 36.2 mL

So, moles of KNO₂ = 0.234 M * 0.0362 L

                            = 0.008486 moles

So, mass of KNO₂ = 0.008486 moles * 85.104 g.mol^-1

                           = 0.722 g of KNO₂

So, 0.722 grams of potassium nitrite are in a 36.2 mL sample of this final diluted solution.

2) Given, molarity of Al(NO₃)₃ = 4.50 M, volume = 400.0 mL

First we need to calculate the moles of Al(NO₃)₃

Moles of Al(NO₃)₃ = 4.50 M * 0.400 L

                           = 1.8 moles

So, mass of Al(NO₃)₃ = 1.8 mole * 212.996 g/mol

                                = 383.4 g of Al(NO₃)₃

3) Given, 15.72 mL of a 2.13 M solution of sulfuric acid, molarity of KOH = 1.75 M

Reaction - H₂SO₄ + 2KOH  →  2H₂O + K₂SO₄

First we need to calculate the moles of H₂SO₄

Moles of H₂SO₄ = 2.13 M * 0.01572 L

                         = 0.0335 moles of H₂SO₄

From the above balanced equation

1 moles of H₂SO₄ = 2 moles of KOH

So, 0.0335 moles of H₂SO₄ = ?

= 0.0670 moles of KOH

So volume of KOH = 0.0670 moles / 1.75 M

                           = 0.0383 L

                           = 38.3 mL

So, 38.3 mL of a 1.75 M solution of potassium hydroxide are required to titrate 15.72 mL of a 2.13 M solution of sulfuric acid