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Old May 12th 2017, 11:53 AM   #1
jaged82
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Question Cannot grasp how to find number of moles & mass of a solute

I'm in week 2 of college chemistry and I am completely lost on how to work out one of my chapter problems. It's:

Calculate the number of moles and the mass of the solute in each of the following solutions:
(a) 325 mL of 8.23 10^−5 M KI, a source of iodine in the diet
(b) 75.0 mL of 2.2 10^−5 M H2SO4, a sample of acid rain
(c) 0.2500 L of 0.1135 M K2CrO4, an analytical reagent used in iron assays
(d) 10.5 L of 3.716 M (NH4)2SO4, a liquid fertilizer

I only need help with one (preferably -a-), so I can learn how to do these on my own. It needs to be broken down step by step though, because everything I have come across just gives an answer with no or little explanation (I also could not seem to find it in my book). I don't know if my brain has just burnt out or what...

Thanks for any help you can give me.
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Old Jul 2nd 2017, 07:28 PM   #2
bkdem
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Calculate the number of moles and the mass of the solute in each of the following solutions:
(a) 325 mL of 8.23 10^−5 M KI
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You'll have to use the Molarity(M) formula to get the number of moles

Molarity = moles of solute/ liters of solution

And then find the mass, in grams(g), by converting moles to grams using molar mass of a compound.

Molar Mass = grams/moles
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Steps for (a):
1. convert mL to L: 325mL *(1L/1000mL) =0.325L
2. moles KI = (8.23 10^−5 mol/L) *(0.325L) = 2.67*10^-5 mol KI
3. Find molar mass of KI by adding up the atomic weights of potassium(K) and Iodine(I). Refer to periodic table-
molar mass KI = 39.0 + 126.9 = 165.9g/mol KI
4. grams KI = (2.67*10^-5 mol KI) * (165.9g/mol KI) = 0.00443g KI
or 4.43*10^-3 g KI
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