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Varvara68 [4.7K]
3 years ago
10

You must make 1 L of 0.2 M acetic acid (CH3COOH). All you have available is concentrated glacial acetic acid (assay value, 98%;

specific gravity, 1.05 g/mL). It will take _________ milliliters of acetic acid to make this solution. Assume a gram molecular weight of 60.05 grams.
Chemistry
1 answer:
Sergeu [11.5K]3 years ago
7 0

Answer:

The correct answer is "11.44 ml".

Explanation:

Molarity,

= 0.2 M

Density,

= 1.05 g/ml

Volume,

= 1 L

As we know,

⇒  Molarity=\frac{No. \ of \ moles }{Volume \ of \ solution}

or,

⇒  No. \ of \ moles=Molarity\times Volume

On putting the values, we get

⇒                         =0.2\times 1

⇒                         =0.2 \ moles

Now,

⇒  No. \ of \ moles=\frac{Mass \ taken}{Molecular \ mass}

or,

⇒  Mass \ taken=No. \ of \ moles\times Molecular \ mass

⇒                      =0.2\times 60.05

⇒                      =12.01 \ gram

hence,

⇒  Density= \frac{Mass }{Volume}

or,

⇒  Volume=\frac{Mass}{Density}

⇒                =\frac{12.01}{1.05}

⇒                =11.44 \ ml

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