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 315 mL of solution. How many moles of ammonium chloride are present in the resulting solution?

Answer 1

Answer:

0.187 mol

Explanation:

A solution is created by dissolving 10.0 grams of ammonium chloride in enough water to make 315 mL of solution.

The formula of ammonium chloride is NH₄Cl and its molar mass is 53.49 g/mol. We can use this information to calculate the moles of ammonium chloride associated with 10.0 grams.

10.0 g × (1 mol/53.49 g) = 0.187 mol

Answer 2

Answer:

0.593 mole

Explanation:

Step 1:

Data obtained from the question include:

Mass of ammonium chloride = 10g

Volume of solution = 315 mL = 315/1000 = 0.315L

Step 2:

Determination of the mass of ammonium chloride in 1L of the solution. This is illustrated below:

If 10g of ammonium chloride NH4Cl dissolves in 0.315L,

Therefore Xg of ammonium chloride NH4Cl will dissolve in 1L i.e

Xg of NH4Cl = 10/0.315

Xg of NH4Cl = 31.75g

Step 3:

Determination of the number of mole of NH4Cl in 31.75g of ammonium chloride NH4Cl.

This is illustrated below:

Molar Mass of NH4Cl = 14 + (4x1) + 35.5 = 14 + 4 + 35.5 = 53.5g/mol

Mass of NH4Cl = 31.75g

Number of mole of NH4Cl =?

Number of mole = Mass/Molar Mass

Number of mole of NH4Cl = 31.75/53.5

Number of mole of NH4Cl = 0.593 mole

Therefore, 0.593 mole of ammonium chloride NH4Cl is present in the solution.