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Genrish500 [490]
3 years ago
6

Determine the volume of water to be added to the nitric acid solution at a concentration of 8.61 mol / L to prepare 500 mL of th

e bulk concentration solution at 1.75 mol / L
Chemistry
1 answer:
Alex_Xolod [135]3 years ago
6 0

Answer:

398 mL

Explanation:

Using the equation for molarity,

C₁V₁ = C₂V₂ where C₁ = concentration before adding water = 8.61 mol/L and V₁ = volume before adding water, C₂ = concentration after adding water = 1.75 mol/L and V₂ = volume after adding water = 500 mL = 0.5 L

V₂ = V₁ + V' where V' = volume of water added.

So, From C₁V₁ = C₂V₂

V₁ =  C₂V₂/C₁

= 1.75 mol/L × 0.5 L ÷ 8.61 mol/L

= 0.875 mol/8.61 mol/L

= 0.102 L

So, V₂ = V₁ + V'

0.5 L = 0.102 L + V'

V' = 0.5 L - 0.102 L

= 0.398 L

= 398 mL

So, we need to add 398 mL of water to the nitric solution.

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3 years ago
An aqueous solution contains 32.7% KCl (wt/wt%). what is the mole fraction of KCl in the solution
inna [77]

The mole fraction of KCl in the solution is 0.1051

calculation

mole fraction of KCl in solution = moles of KCl / total number of moles(moles of KCl +moles of H2O)

moles=mass/molar mass

mass of KCl=32.7g

molar mass of KCl= 39 +35.5

moles of KCl is therefore= 32.7g/74.5 g/mol=0.439 moles

find the moles of H2O= mass of H2O/molar mass

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5 0
3 years ago
When 250 ml of water is added to 35 ml of 0.2 M HCl
Paladinen [302]
Assuming that you’re looking for the concentration of water in the solution, then it would be 0.028 M.

You would have to use the formula:
c1v1 = c2v2, where c =concentration and
v = volume

C1 = ?
V1 = 250 mL
C2 = 0.2 M
V2 = 35 mL

C1 x 250 mL = 0.2 M x 35 mL

C1 = (0.2 M x 35 mL) / 250 mL

C1 = 0.028 M of water added to 35mL of 0.2M HCl

Therefore, there is 0.028 M of water added to 35mL of 0.2M HCl
8 0
3 years ago
Read 2 more answers
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