Question

When chemists work with solid materials, we simply weigh out amounts of solid reagents and calculate mole amounts when thinking about stoichiometry. However, when we dissolve a solid (also known as a solute) in a solvent to form a solution, the solute becomes evenly distributed throughout the solution and we need to know how many moles of solute are present in a particular volume of solution.

A solution is created by dissolving 10.0 grams of ammonium chloride in enough water to make 305 mL of solution. How many moles of ammonium chloride are present in the resulting solution?

When thinking about the amount of solute present in a solution, chemists report the concentration or molarity of the solution. Molarity is calculated as moles of solute per liter of solution. What is the molarity of the solution described above?

To carry out a particular reaction, you determine that you need 0.0500 moles of ammonium chloride. What volume of the solution described above will you need to complete the reaction without any leftover NH4Cl?

Answer 1

1) Moles of ammonium chloride in 305 mL water :

Moles = Mass (g) / Molar mass of ammonium chloride (g/mol)

= 10 g / 53.5 g/mol

= 0.186 moles

That is 0.186 moles of ammonium chlorides are present in 305 mL of solution.

2) Molarity : Molarity is calculated by,

Molarity ( M) = Moles / Volume (lit)

= 0.186 moles / 0.305 lit [ 305 mL = 305 mL / 1000 = 0.305 lit ]

= 0.60 M

Molarity of ammonium chloride is 0.60 m.

3) We can calculate the volume using equation,

Molarity = Moles / volume

Rearranging this equation for volume,

Volume ( lit ) = Moles / Molarity

= 0.0500 moles / 0.60 M

= 0.0833 lit

= 83.3 mL [ 0.0833 lit = 0.0833 x 1000 = 83.3 mL

For complete reaction, we require 83.3 mL of solution.