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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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Given:

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The reaction of sulphur with oxygen is

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Change                                -x            +x

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Volume = 50 L

Let us calculate moles of SO2 formed using ideal gas equation as

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putting other values

n = PV / RT = 2.86 X 50 / 1223.15 X 0.0821 = 1.42 moles

Moles of Sulphur required = 1.42 moles

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3 years ago
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In the problem, you need to prepare a 1.2m solution of ethanol (Solute) in t-butanol (solvent).

14.0g of butanol are <em>0.014kg </em>and as you want to prepare the 1.2m solution, you need to add:

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To convert moles of ethanol to mass you require molar mass (Molar mass ethanol, C₂H₅OH = 46.07g/mol). Thus, mass of 0.0168 moles are:

0.0168moles Ethanol ₓ (46.07g / mol) =

<h3>0.774g of ethanol</h3>

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Answer:

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5 0
3 years ago
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