Turn on thread page Beta
    • Thread Starter
    Offline

    0
    ReputationRep:
    Hello TSR,

    I was wondering if someone could tell me where i have gone wrong here please:-

    Caclulate the molar mass of the acid H2(X)

    A student dissolved 1.571g of an acid, H2(X), in water and made the solution up to 250cm^3.

    She titrated 25.0cm^3 of this solution against 0.125mol dm^-3 sodium hydroxide, NaOH. 21.30cm^3 of NaOH were needed to reach the end point.

    The equation for this reaction is:-

    2NaOH + H2(X) - - - - Na2(X) + 2H2(O)

    My Calculation:-

    n = 0.125 * (21.30/1000) = 2.6625*10^-3 mol.

    (2.6625*10^-3)/2 = 1.33125*10^-3 mol. H2(X)


    M = (1.571g)/(1.33125*10^-3) = 1180.093897 g mol -1.

    The answer is 118 g mol -1, but i got 1180.093897... could anyone please tell me where i have gone wrong??

    Much Appreciated!! Thanks!
    Offline

    21
    ReputationRep:
    you have to divide the answer by 10 because remember - she made the solution up to 250cm^3 then used 25cm^3 of it.
    • Thread Starter
    Offline

    0
    ReputationRep:
    Oh right, thank you for that!!
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: January 19, 2010
The home of Results and Clearing

2,423

people online now

1,567,000

students helped last year

University open days

  1. Keele University
    General Open Day Undergraduate
    Sun, 19 Aug '18
  2. University of Melbourne
    Open Day Undergraduate
    Sun, 19 Aug '18
  3. Sheffield Hallam University
    City Campus Undergraduate
    Tue, 21 Aug '18
Poll
A-level students - how do you feel about your results?

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.