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AnnyKZ [126]
3 years ago
10

Molarity problem:

Chemistry
2 answers:
S_A_V [24]3 years ago
8 0
Density = 1.0 g/mL

5.3 (w/w)% => 5.3 / 100 => 0.053

molar mass acetic acid = 60.05 g/mol

M = 1000.  w . D / molar mass

M = 1000 . 0.053 . 1.0 / 60.05

M = 53 / 60.05

M = 0.882 mol/L

hope this helps!
Degger [83]3 years ago
4 0
<span>There are a number of ways to express concentration of a solution. This includes molarity. Molarity is expressed as the number of moles of solute per volume of the solution. 

5.3 % = 5.3 g acetic acid / 100 g solution

Molarity = </span>5.3 g acetic acid / 100 g solution ( 1.0 g solution / <span>.001 L solution</span> ) ( 1 mol acetic acid/ 60.05 g acetic acid ) = 0.8826 mol / L solution
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The figure above shows the Earth at two different positions in its orbit around the Sun. Which position corresponds to summer in
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Determine how many gmol, kmol, and lbmols there are in 50 kilograms of n-hexane.
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Answer: 581 gmol

0.581 kmol

1.28\times 10^{-3}lbmol

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According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number 6.023\times 10^{23} of particles.

To calculate the moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}=\frac{50\times 1000g}{86g/mol}=581mol

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581 mol= \frac{1}{1}\times 581=581gmol

2. The conversion for mol to kmol

1 mol = 0.001 kmol

581 mol= \frac{0.001}{1}\times 581=0.581kmol

3. The conversion for mol to lbmol

1 mol = 2.2\times 10^{-3}lbmol

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How many liters of 15.0 molar NaOH stock solution will be needed to make 17.5 liters of a 1.4 molar NaOH solution? Show the work
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2.0 L
The key to any dilution calculation is the dilution factor

The dilution factor essentially tells you how concentrated the stock solution was compared with the diluted solution.

In your case, the dilution must take you from a concentrated hydrochloric acid solution of 18.5 M to a diluted solution of 1.5 M, so the dilution factor must be equal to

DF=18.5M1.5M=12.333

So, in order to decrease the concentration of the stock solution by a factor of 12.333, you must increase its volume by a factor of 12.333by adding water.

The volume of the stock solution needed for this dilution will be

DF=VdilutedVstock⇒Vstock=VdilutedDF

Plug in your values to find

Vstock=25.0 L12.333=2.0 L−−−−−

The answer is rounded to two sig figs, the number of significant figures you have for the concentration od the diluted solution.

So, to make 25.0 L of 1.5 M hydrochloric acid solution, take 2.0 L of 18.5 M hydrochloric acid solution and dilute it to a final volume of 25.0 L.

IMPORTANT NOTE! Do not forget that you must always add concentrated acid to water and not the other way around!

In this case, you're working with very concentrated hydrochloric acid, so it would be best to keep the stock solution and the water needed for the dilution in an ice bath before the dilution.

Also, it would be best to perform the dilution in several steps using smaller doses of stock solution. Don't forget to stir as you're adding the acid!

So, to dilute your solution, take several steps to add the concentrated acid solution to enough water to ensure that the final is as close to 25.0 L as possible. If you're still a couple of milliliters short of the target volume, finish the dilution by adding water.

Always remember

Water to concentrated acid →.NO!

Concentrated acid to water →.YES!
8 0
3 years ago
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