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Bad White [126]
2 years ago
14

A student is performing a titration to determine the concentration of a 20.0 mL sample of hydrochloric acid. What did the studen

t determine the concentration of the HCl to be if the end point was reached after 50.0 mL of 0.1 M NaOH was added.
Chemistry
1 answer:
irakobra [83]2 years ago
8 0

Answer:

0.25M HCl

Explanation:

The reaction of HCl with NaOH is:

HCl + NaOH ⇄ H₂O + NaCl

<em>Where 1 mole of HCl reacts per mole of NaOH</em>

The end point was reached when the student added:

0.0500L × (0.1mol / L) = 0.00500 moles of NaOH

As 1 mole of HCl reacted per mole of NaOH, moles of HCl present are:

<em>0.00500 moles HCl</em>

The volume of the sample of hydrochloric acid was 20.0mL = 0.0200L, and concentration of the sample is:

0.00500 mol HCl / 0.0200L = <em>0.25M HCl</em>

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3 years ago
What is the boiling point of a solution of 76 g of water dissolved in 500 mL of acetic acid, CH3COOH?
tatuchka [14]

Answer:

127.3° C, (This is not a choice)

Explanation:

This is about the colligative property of boiling point.

ΔT = Kb . m . i

Where:

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i = Van't Hoff factor (number of particles dissolved in solution)

Water is not a ionic compound, but we assume that i = 2

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T° boling of solution - 118.1°C =  0.52°C . m . 2

Mass of solvent =  Solvent volume / Solvent density

Mass of solvent = 500 mL / 1.049g/mL → 476.6 g

Mol of water are mass / molar mass

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6 0
3 years ago
Read 2 more answers
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