Anonymous
Anonymous asked in Science & MathematicsChemistry · 2 months ago

# Molarity Calculations HELP??

Its probably simple but im having trouble any help would be great :)) Relevance
• Anonymous
2 months ago

Molars are teeth.  They don't have anything to do with chemistry.

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• 2.

BaCl2 → Ba{2+} + 2 Cl{-}

(2.50 L) x (1.25 mol BaCl2/L) x (2 mol Cl{-} / 1 mol BaCl2) =

6.25 mol Cl{-} from BaCl2

NaCl → Na{+} + Cl{-}

(3.50 L) x (2.45 mol NaCl/L) x (1 mol Cl{-} / 1 mol NaCl) = 8.575 mol Cl{-} from NaCl

((6.25 mol Cl{-}) + (8.575 mol Cl{-})) / (6.00 L) = 2.47 mol/L Cl{-}

3.

H2SO4 + 2 NaOH → Na2SO4 + 2 H2O

(10.00 mL) x (0.300 M H2SO4) x (2 mol  NaOH / 1 mol H2SO4) / (40.00 mL NaOH) =

0.150 M NaOH

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• *** Q2 ***

assuming 100% dissociation

.... .... .... .... .... ... ... .... 2.50 L.. 1.25 mol BaCl2.. ... 2 mol Cl-

.. mol Cl- from BaCl2 = ---- ---- x ---- ----- ---- ----- x ---- --- ---- --- = 6.250 mol Cl-

... .... ... ... ... ... ... ... .. .... 1.. ...... ... . 1 L.. ... .. ... .1 mol BaCl2

.... .... .... .... .... ... ... .. 3.50 L.. 2.45 mol NaCl.. ... 1 mol Cl-

.. mol Cl- from NaCl = ---- ---- x ---- ----- ---- ---- x ---- --- ---- --- = 8.575 mol Cl-

... .... ... ... ... ... ... ... .. ... 1.. ...... ... . 1 L.. ... .. ... .1 mol NaCl

... .... ... .... ... ... .. mol Cl-.. .... 6.250 mol + 8.575 mol

.. molarity Cl- = ----- ----- --- = ----- ----- ----- ---- ---- ---- = 2.47M... (3 sig figs)

... .... ... ... .... ... L solution... .. ... 2.50L + 3.50L

*** Q3 ***

from the balanced equation

.. 2 NaOH + 1 H2SO4 --> 1 Na2SO4 + 2 H2O

we can see the ratio of H2SO4 to NaOH is 1:2

then.. since M = moles solute / L solution = mmol solute / mL solution

.. 10.00mL H2SO4... 0.300 mmol H2SO4... 2 mmol NaOH

------ ---- ----- ----- x ----- ----- ---- ------ ---- x ---- ----- ----- ---- = 0.150M NaOH (3 s.f.)

.. 40.00mL NaOH... ... . 1 mL H2SO4... .... . 1 mmol H2SO4

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• 2.

Moles of BaCl₂ = (1.25 mol/L) × (2.50 L) = 3.125 mol

Moles of NaCl = (2.45 mol/L) × (3.50 L) = 8.575 mol

Each mole of BaCl₂ contains 2 moles of Cl⁻ ion, while each moles of NaCl contains 1 mole of Cl⁻ ion.

Total moles of Cl⁻ ion = (3.125 mol) × 2 + (8.575 mol) = 14.825 mol

Molarity of Cl⁻ ion = (14.825 mol) / (6.00 L) = 2.47 M

====

3.

Equation for the reaction:

2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O

Mole ratio NaOH : H₂SO₄ = 2 : 1

Millimoles of H₂SO₄ = (0.300 mmol/mL) × (10.00 mL) = 3.00 mmol

Millimoles of NaOH = (3.00 mmol) × 2 = 6.00 mmol

Molarity of NaOH = (6.00 mmol) / (40.00 mL) = 0.150 M

OR:

[(0.300 mol H₂SO₄ / 1000 mL H₂SO₄ solution) × (10.00 mL NaOH solution) × (2 mol NaOH / 1 mol H₂SO₄)] / (40.00/1000 L)

= 0.150 M

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• Hi there!

Unfortunately, I only have time to answer number 2 tonight. Just a few notes before we begin! I realize the equations below are super hard to read as plain text.

I have the compounds/ions separated by parenthesis purely to make it easier to read, you don't usually do this. I will attach a picture so you can see the equations better. Also, I don't really know what you understand and what you don't, so I just put it all down, including the why behind it. Some of these steps are skip-able, just make sure you know what you're doing first. Good luck!

The first thing you're going to do is find the number of moles of each compound in the mixture. Remember that M = molarity = moles of substance/1 liter. Set it up like you would any other unit conversion problem. Do it for both compounds, like this:

(1.25 moles (BaCl2) / 1 L) * 2.50 L = 3.125 moles (BaCl2)

(2.45 moles (NaCl) / 1 L) * 3.50 L = 8.575 moles (NaCl)

Now, because all of these are ionic compounds, and molarity implies they are suspended in water, we know that the compounds split completely into ions, like this:

(BaCl2) → (Ba2+) + 2(Cl-)

(NaCl) → (Na+) + (Cl-)

Notice how two moles of Cl- (aka chloride) ions are formed for every mole of BaCl2 molecules that is dissolved, and one mole of Cl- (aka chloride) ions is formed for every mole of NaCl molecules that is dissolved. We can uses this to create ratios, like this:

1 mole (BaCl2) = 2 moles (Cl-) → 1 = 2 moles (Cl-) / 1 mole (BaCl2)

1 mole (NaCl) = 1 mole (Cl-) → 1 = 1 mole (Cl-) / 1 mole (NaCl)

Because each of these ratios is equal to one, we can use them to convert across compounds, like this:

(3.125 moles (BaCl2) / 1) * (2 moles (Cl-) / 1 mole (BaCl2)) = 6.250 moles (Cl-)(8.575 moles (NaCl) / 1) * (1 mole (Cl-) / 1 mole (NaCl)) = 8.575 moles (Cl-)

Now add the number of moles of Cl- together to get the total number of moles of Cl- in the whole solution, like this:

6.250 moles (Cl-) + 8.575 moles (Cl-) = 14.825 moles (Cl-) total

You're almost done! Now all you have to do is use the given number of liters total to find the molarity of Cl- in the final mixture, like this:

14.825 moles (Cl-) total / 6.00 L total = approx 2.471 M (Cl-)

As for number 3, you will use the same principles we just went over. Remember that (NaOH) will dissolve into (Na+) and (OH-), and (H2SO4) will dissolve into 2(H+) and (SO4-). (OH-) and (SO4-) are both polyatomic ions. The solution will be neutralized when the number of moles of (H+) ions is equal to the number of (OH-) ions. It might be helpful to convert all volumes to liters so you don't get confused, since molarity is always in liters. Good luck! • Login to reply the answers
• Totl volume=2.5+3.5=6L

Equations:

Bacl2--->Ba+2 + 2Cl-

NaCl --> Na+ + Cl-

moles of Cl in first equation: 2*(2.5)(1.25) = because C*v is moles of bacl2 and moles of Cl- in first equation is double that.

Moles of Cl in second equation: 3.5*2.45

add them together, that is the total moles of Cl and divide by total volume which gives you the concentration.

3) NaOH--> Na+  +  OH-

H2SO4 --> 2H+ + SO4-2

moles of H+ = 2*0.01*0.3

You need to add the same number of moles of OH- to neutralize it.

you have the volume it's 0.04 litres and you have the moles that you added which equals to the moles of H+, now do the C=n/v to find the concentration.

Skype: Live:damnitsjake21 (for any free chem help)

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