Question

A solution is created by dissolving 13.0 grams of ammonium chloride in enough water to make 295 mL of solution. How many moles of ammonium chloride are present in the resulting solution? =0.243 molesWhen 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? =0.824 MTo 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? _________ mL of solution?

Answer 1

Answer:

(a) Moles of ammonium chloride = 0.243 moles

(b) $Molarity_{ammonium\ chloride}=0.824\ M$

(c) 60.68 mL

Explanation:

(a) Mass of ammonium chloride = 13.0 g

Molar mass of ammonium chloride = 53.491 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{13.0\ g}{53.491\ g/mol}$

Moles of ammonium chloride = 0.243 moles

(b) Moles of ammonium chloride = 0.243 moles

Volume = 295 mL = 0.295 L ( 1 mL = 0.001 L)

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

$Molarity_{ammonium\ chloride}=\frac{0.243}{0.295}$

$Molarity_{ammonium\ chloride}=0.824\ M$

(c) Moles of ammonium chloride = 0.0500 moles

Volume = ?

Molarity = 0.824 M

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

$0.824\ M=\frac{0.0500}{Volume}$

Volume = 0.05 / 0.824 L = 0.06068 L = 60.68 mL