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Anna11 [10]
4 years ago
12

A solution is created by dissolving 13.0 grams of ammonium chloride in enough water to make 295 mL of solution. How many moles o

f 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?
Chemistry
1 answer:
Masja [62]4 years ago
6 0

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}

<u>Moles of ammonium chloride = 0.243 moles</u>

(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}

<u>Volume = 0.05 / 0.824 L = 0.06068 L = 60.68 mL</u>

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