Question

Just answer part 2 and 3 please

(0.3 point) Part 1 ii See Periodic TableSee Hint A solution is created by dissolving 12.5 grams of ammonium chloride in enough water to make 355 mL of solution. How many moles of ammonium chloride are present in the resulting solution? moles of NH4CI 0.233 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. Q See Hint (0.3 point) Part 2 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? 0.0086265 M See Hint (0.3 point) Part 3 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 NH4CI? mL of solution 0.0043132

Answer 1

part-2

Molarity of a solution= number of moles of solute/volume of solution in Litre

for given solution

Number of moles of ammonium chloride= 0.233 moles

Volume of ammonium chloride solution= 355 ML= 0.355 L (1L=1000mL)

Molarity of a ammonium chloride solution= 0.233 moles/0.355 L=0.6563 M

part-3

Number of moles of a solute in solution= Molarity X volume in L

Number of moles of ammonium chloride= Molarity X volume in L

0.05 moles= 0.6563 M X volume in L

volume in L = 0.05/0.6563 = 0.07618 L= 76.2 mL