Turn on thread page Beta
    • Thread Starter
    Offline

    0
    ReputationRep:
    The question is 'How many moles of MnO4- are present in 1000cm^3 of the standard 0.01 M KMnO4 solution'

    So i recently did a titration experiment in class and I got 12.1cm^3 of 0.01 M KMnO4.

    So i had a go at the question and did moles = c*v/1000

    1000cm^3=L

    moles MnO4-: 1L x 0.01M= 0.01

    moles MnO4-: 0.0121 L x 0.01 M= 0.000121 moles.

    Is this right?? I'm not too sure if it is

    since it's talking about releasing anions would it be

    1000cm^3 = 1 L

    0.01 M = 0.01 mole/1000

    0.01 mole/1L * 1 L = 0.01 M KMnO4

    1 mole of KMnO4 releases 1 mole of MnO4- anions


    instead?
    Offline

    11
    ReputationRep:
    A 0.01M solution has 0.01 Mole of the solute in each litre. This means that there is 0.01 Mole of KMnO4, K, Mn, or MnO4.

    If you have 12cm^3 then you will have \frac {12}{1000}\times 0.01=0.00012M \quad\text{of}\quad KMnO_4 which suggests you are correct.

    Whilst this is true "1 mole of KMnO4 releases 1 mole of MnO4- anions". I don't see what you are driving at?
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: March 2, 2016
Poll
Do you think parents should charge rent?

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.