# A Simple AS Level QuestionWatch

Thread starter 9 years ago
#1
http://www.planet-master.com/Interna..._w08_qp_32.pdf

If someone could explain how to answer Q1. (c). I'd be awfully thankful.
0
9 years ago
#2
n = M/Mr
M = 28.44
Mr (KMnO4) = 158

28.44/158 = 0.18 mol
0
Thread starter 9 years ago
#3
(Original post by Woody.)
n = M/Mr
M = 28.44
Mr (KMnO4) = 158

28.44/158 = 0.18 mol
Sorry, that's incorrect. But thanks for your time.

I found out how to solve this.
0
9 years ago
#4
(Original post by aura1947)
Sorry, that's incorrect. But thanks for your time.

I found out how to solve this.
What was the solution?
0
Thread starter 9 years ago
#5
(Original post by Woody.)
What was the solution?
c = (g/dm^3) / Mr

n = c * v

i.e. n = ((g/dm^3) / Mr) * v

g/dm^3 = 28.44
Mr = 158
and assuming 0.042 dm^3 was transferred

n = ((28.44) / 158) * 0.042
n = 7.56*10^-3 mol

0
X

new posts
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

### University open days

• Cranfield University
Cranfield Forensic MSc Programme Open Day Postgraduate
Thu, 25 Apr '19
• University of the Arts London
Open day: MA Footwear and MA Fashion Artefact Postgraduate
Thu, 25 Apr '19
• Cardiff Metropolitan University
Undergraduate Open Day - Llandaff Campus Undergraduate
Sat, 27 Apr '19

### Poll

Join the discussion

#### Have you registered to vote?

Yes! (112)
38.75%
No - but I will (13)
4.5%
No - I don't want to (20)
6.92%
No - I can't vote (<18, not in UK, etc) (144)
49.83%