5.0 g of solid al(no3)3 was added to enough water to make a 100.0 ml solution. al(no3)3 is very soluble in water.

1. write out the dissolution reaction that occurs when the solid al(no3)3 was added to water.

2. what is the concentration, in molarity, of al(no3)3 once it is dissolved in the water?

3. if 53 ml of water was added to the original al(no3)3 solution described above, what is the new concentration in molarity?

4. what is the concentration, in % w/v, of the solution created when 12.0 g of solid al(no3)3 was added to enough water to make a 150.0 ml solution?

1- Al (NO₃) ₃ ⇒ Al⁺³ + 3 NO₃⁻

2- 0.234 M

3- 0.153 M

4- 8% p / v.

Explanation:

1- When Al (NO₃) ₃ is added water, a dissociation reaction occurs, in which the compound is separated into ions that are soluble in water. It is:

Al (NO₃) ₃ ⇒ Al⁺³ + 3 NO₃⁻

2- Molarity (M) is defined as the amount of moles is solute, dissolved in 1000 mL (1L) of solution. To determine this concentration, the amount of moles of solute must be calculated using the periodic table, in this case, it would be Al (NO3) 3, and then establish the relationship according to the volume of solution we have.

a) Calculate moles of Al (NO₃)₃

Molar mass of Al (NO₃)₃ = mAl + 3 x (mN + 3 x mO) = 26.98g + 3 x (14g + 3 x 15.99g) = 212.89 g / mol

Knowing the mass of a mole, you can calculate the moles that represent the initial mass that the statement tells us:

212.89 g of Al (NO3) 3 1 mol

5.0 g of Al (NO3) 3 X = 0.0234 mol

Calculation: 5.0 g x 1 mol / 212.89 g = 0.0234 mol of Al (NO₃)₃

b) Calculate molarity of the solution

The amount of moles calculated is dissolved in 100 mL of solution; With this information, you can calculate the amount of moles that will be in 1 L of solution:

100 mL solution 0.0234 mol of Al (NO₃)₃

1000 mL solution X = 0.234 mol of Al (NO₃)₃ ⇒ 0.234 M

Calculation: 1000 mL x 0.0234 mol / 100 mL = 0.234 mol of Al (NO₃)₃

There is 0.234 mol of Al (NO₃)₃ in 1 L (1000mL) solution, therefore, the answer is that the molar concentration is 0.234 M

3 - Considering that the volumes are additive, the same amount of the initial Al (NO₃)₃ solute will be obtained, which is 0.0234 moles, but now the volume of the solution will be greater. The same calculation is made as in the previous point, but now using the new volume (100 mL + 53 mL = 153 mL).

153 mL solution 0.0234 mol of Al (NO₃)₃

1000 mL solution X = 0.153 mol of Al (NO₃)₃ ⇒ 0.153 M

Calculation: 1000 mL x 0.0234 mol / 153 mL = 0.153 mol of Al (NO3) 3

The new concentration will be 0.153 M

4- The concentration of % w / v indicates the mass of solute (in this case it would be the Al (NO3) 3) dissolved every 100 mL of solution. Knowing the mass of solute and the volume in which it is dissolved in the solution, this percentage can be calculated as follows:

150 mL of solution 12.0 g of Al (NO₃)₃

100 mL of solution X = 8 g of Al (NO₃)₃

Calculation: 100 mL x 12.0 g / 150 mL = 8 g of Al (NO₃)₃

There is 8 g of Al (NO₃)₃ dissolved per 100 mL of solution, therefore, the concentration is 8% w / v.