Midterm 1 Q!C

Natalie_Martinez_1I
Posts: 30
Joined: Mon Feb 26, 2018 3:00 am

Midterm 1 Q!C

Potassiumpermanganate,KMnO4,isaninorganicchemicalcompoundusedfor cleaning wounds. 5.00 g of KMnO4 is dissolved in a 150.00 mL flask of water.
If 20.00 mL of this solution is removed and placed in a new 2nd 250.00 mL flask and
filled with water, what is the concentration of the solution in the 2nd flask?
Molar Masses: K (39.10 g/mol) Mn (54.94 g/mol) O (16.00 g/mol)
16.00 x 4 = 64 64.00 + 54.94 + 39.10 = 158.04 g/mol of KMnO4
5.00 g/158.04 g/mol = 0.0316375 mol
0.0316375 mol/ 0.150 L = 0.210917 mol/L
20.00 ml/250.00 ml x 0.210917 mol/L = 0.016873 mol.L-1 or 1.69 x 10-2 mol.L-1

Can someone explain why you had to divide 20ml by 250ml to get the molarity in this question please?
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AnnaYan_1l
Posts: 96
Joined: Fri Apr 06, 2018 11:05 am
Been upvoted: 1 time

Re: Midterm 1 Q!C

What I did to solve this problem was:

$5.00 g KMnO_{4} \cdot \frac{1 mol KMnO_{4}}{158.04 g KMnO_{4}} = 0.0316 mol KMnO_{4}$
$M = \frac{n}{V} = \frac{0.0316 mol KMnO_{4}}{.150L} = 0.211 M KMnO_{4}$
$n = MV = (0.211 M KMnO_{4})(0.020 L) = 0.00422 mol KMnO_{4}$
$(0.00422 mol KMnO_{4}) = M(.250L)$
$M = \frac{0.00422 mol KMnO_{4}}{.250 L} = .0169 M KMnO4 = 1.69 \ast 10^{-2} M KMnO_{4}$

Hope that clarifies the process more!

Natalie_Martinez_1I
Posts: 30
Joined: Mon Feb 26, 2018 3:00 am

Re: Midterm 1 Q!C

It does thank you so much!