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IrinaVladis [17]
3 years ago
11

How many grams of ammonium chloride (gram formula mass= 53.5 g) are contained in .500 L of a 2.00 M solution?

Chemistry
1 answer:
Elza [17]3 years ago
8 0

Answer:

53.5g of NH4Cl

Explanation:

First, we need to obtain the number of mole of NH4Cl. This is illustrated below:

Volume = 0.5L

Molarity = 2M

Mole =?

Molarity = mole /Volume

Mole = Molarity x Volume

Mole = 2 x 0.5

Mole = 1mole

Now, let us convert 1mole of NH4Cl to gram. This is illustrated below:

Molar Mass of NH4Cl = 53.5g/mol

Number of mole = 1

Mass =?

Number of mole = Mass /Molar Mass

Mass = number of mole x molar Mass

Mass = 1 x 53.5

Mass = 53.5g

Therefore, 53.5g of NH4Cl is contained in the solution.

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If 200 ml of 0.15 M propionic acid (PA) is added to 300 ml of 0.02 M NaOH, what is the resulting pH of the solution? Round the a
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Answer:

pH = 4.543

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added 300 mL 0f 0.02 M NaOH:

⇒ <em>C</em> CH3CH2COOH = ((0.200 L)(0.15 M)) - ((0.300 L)(0.02 M))/(0.3 + 0.2)

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charge balance:

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(2) in (1):

⇒ [CH3CH2COOH] = 0.06 M - 0.02 M - [H3O+] = 0.04 M - [H3O+]

replacing in Ka:

⇒ 1.349 E-5 = [H3O+][([H3O+] + 0.02) / (0.04 - [H3O+])

⇒ (1.349 E-5)(0.04 - [H3O+]) = [H3O+]² + 0.02[H3O+]

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⇒ [H3O+]² + 0.02001[H3O+] - 5.396 E-7 = 0

⇒ [H3O+ ] = 2.867 E-5 M

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