Answer – We are given, mass of Na2CO3 = 2.06*10-4 g
Volume = 2.25 L
First we need to calculate the moles of Na2CO3
We know,
Moles of Na2CO3 = 2.06*10-4 g / 105.98 g.mol-1
= 1.94*10-6 moles
So, 1 moles of Na2CO3 =1 moles of CO32-
So, 1.94*10-6 moles of Na2CO3 = ? moles of CO32-
= 1.94*10-6 moles of CO32-
1 moles of Na2CO3 = 2 moles of Na+
So, 1.94*10-6 moles of Na2CO3 = ?
= 3.88*10-6 moles of Na+
So, molarity of Na2CO3 = 1.94*10-6 moles / 2.25 L
= 8.64*10-7 M
Molarity of Na+ = 3.88*10-6 moles / 2.25 L
= 1.73*10-6 M
molarity of CO32- = 1.94*10-6 moles / 2.25 L
= 8.64*10-7 M
We know, 1 ppm = 1 mg/ L
So, mass of Na+ = 3.88*10-6 moles * 22.990 g/mol
= 8.93*10-5 g
= 0.0893 mg
So, pppm of Na+ = 0.0893 mg / 2.25 L
= 0.0397 ppm
Mass of CO32- = 1.94*10-6 moles * 60.08 g/mol
= 0.000117 g
= 0.117 mg
So, ppm of CO32- = 0.117 mg / 2.25 L
= 0.0519 ppm
General chemistry 2. Help please. Determine the concentrators of Na_2CO_3 Na^+. and a solution prepared by...
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Stuck on how to calculate PPM. Can someone
please explain this to me!
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Please help with all the questions.
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