O level Chemistry Question: Mole Concept / Mole Calculations


Halo all students preparing for your O Level exams,

Do try out the following Chemistry Question from the Top 5 Sec Schools in Singapore during their Preliminary Examinations:

Question:
The number of atoms in 25g of Butane is:
A. 3 X 10^23
B. 3.6 x 10^24
C. 4.2 x 10^23
D. 4.2 x 10^24

Do try and send me your answers, probably by the email or leave a comment below the “COMMENT” section.

All the Best
Sean
Master Trainer =)

Related Articles:

Many people talked about teaching Chemistry, I simply LOVE it. I am a passionate Chemistry Coach based in Singapore, Southeast-Asia and aspire to be one of the most dynamic, powerful and humorous speakers in Asia. My 16+ years of coaching experience has equipped me to understand the true reasons why students are not able to perform well in Chemistry, and allow me to structure my teaching methodology to cater to different levels of learners. If you have found this post useful, please share it with your friends. I would really appreciate it! Sean Chua

17 Responses to O level Chemistry Question: Mole Concept / Mole Calculations

  1. Hi Brians,

    Do check out all the Mole Calculations questions in this website. We have alot of discussion on this topic and its related exam-based questions.

    Do a search on the right hand side with keywords: “Mole Calculations”.

    Regards,
    Sean Chua

  2. a good afternoon to you sir. Thank you very much for the advices that you have been giving my fellow students. They have been very useful to me in my chemistry lessons especially in mole concept.

  3. Sir I am from Pakistan & I am facing numerous problems in finding out concentrations.So plz help me …
    Thankyou
    Regards,
    Humza.

  4. 58 – mass
    riq – 25gm
    in 1 mole no. Of atom 6.02*10^23
    25*6.02*10^23
    = ————————–
    58
    answer B

  5. chemistry is a wonderful subject indeed but most of the students fails it in uganda ,why?????

  6. good evening,i have a problem solving complex equations and balancing them.i would like to be helped n this thank you.

  7. GMM of C4H10=4*12+1*10
    =48+10
    =58gm
    58gm 0f C4H10=1 mole
    hence, 25gm of C4H10= 25/58 moles
    1 mole=6.023*10^23 molecules
    hence, no.of atoms=(25/58*6.023*10^23)*14 atoms
    =3.634*10^24 atoms to be exact.
    and so according to the options present, the answer should be B.

  8. Hi,

    You can do a keyword search for “Mole Calculations” on the top right-hand corner of this website.
    From there you can find more discussions on mole calculations.

    Rgds,
    Sean Chua

  9. Hi Samuel,

    Thanks for your comment.

    For Mole Calculations, it requires you to be proficient with mathematical manipulations. You should have been taught the necessary mole calculations mathematics formulae required. All is required is to know how to apply them.

    I have posted quite a number of questions (and suggested solutions) previously regarding Mole Calculations.

    Go to the “SEARCH” field on the right hand side of this webpage and key in “Mole Calculations”. You will be directed to another page on the archive blogposts related to Mole Calculations.

    Hope this information is useful to you.

    Drop me a line or two. I would love to hear your success.

    To Your Academic Success in Chemistry,
    Sean Chua
    Master Trainer & Author
    WINNERS Education Group

    “Experience Learning with A Difference”

  10. GOODEVENING SIR, PLEASE I HAVE BEEN ENCANTERING DIFFICULTIES WHEN SOLVING MOLE CONCEPT QUESTIONS, PLEASE, ADVICE ME ON WHAT TO DO. THANKS, SAMUEL FROM NIGERIA.

  11. Well Done Lijun. This is a Cedar Girl O level Prelim MCQ and understand that couple of classes were tricked.

    For Anonymous, hope you understand what Lijun has done to get the answer. The tricky part is to multiple by 14. Avogadro’s number is formally defined as the number of PARTICLES in 1 mole of substance. Do note that PARTICLES can refer to molecules, ions or atoms.

    No. of molecules=(25/58)mole x(6×10^23)molecules/mole

    No. of atoms=(25/58)(6×10^23)molecules x number of atoms in C4H10(14atoms)=3.621×10^24 atoms

    Hope this helps. Let me know if you need more help in this.

    Rgds,
    Sean

  12. MR(C4H10)= 4(12) +8 =58
    no. of moles of C4H10=25/58
    no. of molecules=(25/58)x(6×10^23)
    no. of atoms=(25/58)(6×10^23)(14)=3.621×10^24

    hence answer should be B.

Leave a reply