January 23, 2021, 12:59:09 PM
Forum Rules: Read This Before Posting

### Topic: Problem in calculating substrate concentration  (Read 4726 times)

0 Members and 1 Guest are viewing this topic.

#### zerokun

• Regular Member
• Posts: 12
• Mole Snacks: +0/-0
##### Problem in calculating substrate concentration
« on: May 28, 2008, 12:19:39 AM »
In an experiment,
20 microLitre of 20 mili Mol / Litre of phenylphosphate was used.
What is the  concentration of phenylphosphate?

I try to calculate by first converting 20 microLitre to 2.0 X 10 ^-3   L
and then, i divide by 20 X 10^-6 L because i use  20  microlitre of phenylphosphate.
So i got 1000 mol L^-1

which is wrong, the correct answer is 0.02

does anyone comes out with any idea in how to calculate this I really need your help.Thxxxx in advance

#### cliverlong

• Full Member
• Posts: 611
• Mole Snacks: +60/-14
##### Re: Problem in calculating substrate concentration
« Reply #1 on: May 28, 2008, 03:09:59 AM »
In an experiment,
20 microLitre of 20 mili Mol / Litre of phenylphosphate was used.
What is the  concentration of phenylphosphate?

I try to calculate by first converting 20 microLitre to 2.0 X 10 ^-3   L
That is wrong

micro = 10-6
Quote
and then, i divide by 20 X 10^-6 L because i use  20  microlitre of phenylphosphate.
So i got 1000 mol L^-1

which is wrong, the correct answer is 0.02

But I don't understand why you use / are given the volume data - it is irrelevant in this context

You wrote above

20 mili Mol / Litre of phenylphosphate

This is already a concentration (amount in moles / volume)

So converting to

20 x 10-3 mol litre-1

This gives a concentration of 0.02 mol litre-1 (note the units) - which is the answer you are trying to understand

Quote
does anyone comes out with any idea in how to calculate this I really need your help.Thxxxx in advance
Does what I have written make sense to you?

Clive

#### zerokun

• Regular Member
• Posts: 12
• Mole Snacks: +0/-0
##### Re: Problem in calculating substrate concentration
« Reply #2 on: May 28, 2008, 08:41:49 AM »
Yes, it make sense

The volume is given because there is another question which require too calculate using the formula C1V1=C2V2
I get confuse with the unit but i manage to solve it during lab today

Thxxxx anyway ! ! ! !