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Offline Shea

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solutions
« on: August 10, 2006, 03:23:09 PM »
Could I have a little assistance with these?

How many mL of 0.5M Fe(NO3)2 solution are required so that we have 0.4 moles of Fe(NO3)2.?

.5M x .4 moles = .2L

.2L = 200mL

How many mL of 2.0M AgNO3 solution are required so that we have 20 g of AgNO3?

20g of AgNO3 = .117 moles of AgNO3

2M x .117 moles of AgNO3 = .234L

Where did I go wrong with them?

Offline Will

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Re: solutions
« Reply #1 on: August 10, 2006, 03:32:37 PM »
Where did I go wrong with them?

moles = concentration(M) x volume(dm3) (ie volume = moles / concentration)

(Also round up the moles of AgNO3 from 0.1177343207 to 0.1177 or 0.118) ;)

Offline Borek

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Re: solutions
« Reply #2 on: August 10, 2006, 03:36:36 PM »
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info

Offline sdekivit

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Re: solutions
« Reply #3 on: August 10, 2006, 05:27:06 PM »
i assume you have basic knowledge in maths. For me, the easiest way is to put the numbers in the original equation and solve the equation.

Thus for problem 1:

c = n/V --> 0,5 = 0,4 / V --> V = 0,4 / 0,5 = 0,8 L = 800 mL

Offline Shea

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Re: solutions
« Reply #4 on: August 12, 2006, 06:48:34 PM »
Volume = mole / concentration, right?

My teacher said I still had the wrong formula.

Offline Borek

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Re: solutions
« Reply #5 on: August 12, 2006, 06:53:55 PM »
Volume = mole / concentration, right?

My teacher said I still had the wrong formula.

The one you have listed is OK. Take the way similar to the one sdekivit proposed: start with the definition of molar concentration and solve for V.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info

Offline Shea

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Re: solutions
« Reply #6 on: August 12, 2006, 07:00:54 PM »
Ok, I think I got it this time.

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