Question

4. A puddle of coastal seawater, caught in a depression formed by some coastal rocks at high tide, begins to evaporate as the tide goes out. If the volume of a puddle decreases to 23% of its original volume, what is the sodium chloride concentration if it was initially 0.449 M?

5. Iron (II) can be precipitated from a slightly basic aqueous solution by bubbling oxygen through the solution, which converts Fe (II) to Fe (III):

                  4 Fe(OH)⁺ (aq) + 4 OH^- (aq) + O₂(aq) + 2H₂O (L) ? 4 Fe(OH)₃ (s)

How many grams of oxygen are needed to precipitate all of the iron in 75 mL of 0.090 M Fe(II)?

Answer 1

4. A puddle of coastal seawater, caught in a depression formed by some coastal rocks at high tide, begins to evaporate as the tide goes out. If the volume of a puddle decreases to 23% of its original volume, what is the sodium chloride concentration if it was initially 0.449 M

Solution:

We can assume initial volume= 100 mL

According to the problem condition final volume will be 23% of initial volume

Therefore final volume = 100 mL * 23 / 100

= 23 mL

We have to find final concentration of NaCl, For that we use M1V1 = M2V2

Initial molarity of NaCl is 0.449 M = M1

M1V1 = M2 V2

M2 = M1V1/ V2

= 0.449 M * 100 mL / 23 mL

= 1.95 M

So final concentration of NaCl wold be 1.95 M

5. Iron (II) can be precipitated from a slightly basic aqueous solution by bubbling oxygen through the solution, which converts Fe (II) to Fe (III):

                  4 Fe(OH)⁺ (aq) + 4 OH^- (aq) + O₂(aq) + 2H₂O (L) -------> 4 Fe(OH)₃ (s)

How many grams of oxygen are needed to precipitate all of the iron in 75 mL of 0.090 M Fe(II)?

Solution:

This is stoichiometry related problem, Essential condition to balance the given reaction. Given reaction is already balanced.

We need to find mass of oxygen in gram. We get mass of oxygen once we get is moles required to react with 75 mL of 0.090 Fe(II).

For this we need to use mole ratio from the reaction stoichiometry. We know mole ratio between Fe(II) : O2 (g) is 4 : 1

To get moles of O2 we should calculate moles of Fe(II) from its volume and molarity.

We know Molarity = Number of moles / Volume in L

Number of moles = Molarity * volume in L

Number of moles of Fe(II)

= 0.090 M * 0.075 L

= 0.00675 mol Fe(II)

We get moles of O2 as follow

Moles of O2

= 0.00675 mol Fe(II) * 1 Mol O2 / 4 mol Fe(II)

=0.001688 mol O2

We know Mass in gram = Number of moles * Molar mass

Mass of O2 in gram = 0.001688 mol O2 * 31.998 g/mol

= 0.053997

=0.054 g O2