Determine the molarity of these solutions:
(assume that all are aqueous)
a) 3.67 moles of na₂so₃ dissolved to make 3.04 liters of solution.
b) 0.629 moles of al₂o₃ to make 1.500 liters of solution.
c) 4.783 grams of na₂co₃ to make 10.00 liters of solution.
d) 0.897 grams of (nh₄)₂co₃ to make 250 ml of solution.
e) 0.0348 grams of pbcl₂ to form 45.0 ml of solution.

a. 1.20M

b. 0.42M

c. 4.51x10⁻³ M

d. 0.037M

e. 2.78x10⁻³M

Explanation:

Molarity shows the moles of solute that occupy 1L of solution:

a. 3.04 L of solution have 3.67 moles of solute

1 L of solution has  (3.67  .1) / 3.04 = 1.20M

b. 1.5 L of solution have 0.629 moles of solute

1L of solution has (0.629 .1) /1.5 = 0.42M

c. Molar mass Na₂CO₃ = 105.98

Mass / Molar mass = 4.783 g / 105.98 g/m = 0.045 moles

10 L of solution have 0.045 moles

1L of solution has (1 . 0.045)/10 = 4.51x10⁻³ M

d. Molar Mass (NH₄)₂CO₃ = 96.09 g/m

Mass / Molar mass = 0.897 g / 96.09 g/m = 9.33x10⁻³ moles

250mL have 9.33x10⁻³ moles of solute

1000 mL have ( 1000 . 9.33x10⁻³) /250 = 0.037M

e. Molar Mass  PbCl₂ = 278.1 g/m

Mass / Molar mass = 0.0348 g / 278.1 g/m = 1.25x10⁻⁴ moles

45 mL of solution have  1.25x10⁻⁴ moles of salt

1000 mL of solution have ( 1000 . 1.25x10⁻⁴) / 45 = 2.78x10⁻³M

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%

1.solid

2.carbon dioxide

3.helium