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2 years ago

Concentrated hydrochloric acid is 36% (Weight/Volume) hydrochloric acid and has a density of 1.18 g.cm -3.

Chemistry

1 answer:

dezoksy [38]2 years ago

8 0

The concentration refers to the amount of substance that is contained in solution.

<h3>What is the concentration?</h3>

The concentration refers to the amount of substance that is contained in solution. We can be able to obtain the concentration of the raw acid by the use of the relation;

Co = 10pd/M

M = molar mass of the acid

p = percentage of the acid

d = density of the acid

Co = 10 * 36 * 1.18/36.5

Co = 11.6 M

Using the dilution formula;

C1V1 = C2V2

10 * 11.6 = C2 * 1000

C2 = 10 * 11.6/1000

C2 = 0.116 M

Using again;

C1V1 = C2V2

0.116 * 5 = C2 * 20

C2 = 0.116 * 5 /20

C2 = 0.029 M

Learn more about concentration:brainly.com/question/10725862

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In a titration of 35.00 mL of 0.737 M H2SO4, __________ mL of a 0.827 M KOH solution is required for neutralization.

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In a titration of 35.00 mL of 0.737 M H₂SO₄, 62.4 mL of a 0.827 M KOH solution is required for neutralization.

Explanation:

The balanced reaction is

H₂SO₄ + 2 KOH ⇒ 2 H₂O + K₂SO₄

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction) 1 mole of H₂SO₄ is neutralized with 2 moles of KOH.

The molarity M being the number of moles of solute that are dissolved in a given volume, expressed as:

$Molarity=\frac{number of moles}{volume}$

in units of $\frac{moles}{liter}$

then the number of moles can be calculated as:

number of moles= molarity* volume

You have acid H₂SO₄

35.00 mL= 0.035 L (being 1,000 mL= 1 L)
Molarity= 0.737 M

Then:

number of moles= 0.737 M* 0.035 L

number of moles= 0.0258

So you must neutralize 0.0258 moles of H₂SO₄. Now you can apply the following rule of three: if by stoichiometry 1 mole of H₂SO₄ are neutralized with 2 moles of KOH, 0.0258 moles of H₂SO₄ are neutralized with how many moles of KOH?

$moles of KOH=\frac{0.0258moles of H_{2} SO_{4}*2 moles of KOH }{1mole of H_{2} SO_{4}}$