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1. I have chemistry in my access course and i'm really struggling with the last 4 questions, Ive put some answers in but my lecturer told me they are wrong and told me i'd need the answer for D to answer the rest? If anyone who's good at chemistry could explain that would be great.

I used 25cm3 of sulphuric acid and made the solution up to 250cm3 using distilled water.

D) Calculate the amount in moles of sulphuric acid in your mean titre. (mean titre 25cm3)

E) Calculate the number of moles of sulphuric acid present in the 250cm3 solution that you prepared

F) Calculate the concentration, in mol dm-3 of sulphuric acid present in the original undiluted solution of sulphuric acid.

G) Calculate the concentration, in g dm-3 of sulphuric acid present in the original undiluted solution of sulphuric acid.

Any help would be appreciated alot as its due in Thursday.

Thanks.
2. These steps will really help you:1) Ensure you have a fully ballaced equation 2) Create a division between the two things you are working with like shown belowulfuric acid Distilled water V= 25cm3. V=250 cm3n= C x V n= x (25 divide by 1000)

Is there any equation or concentration given?
3. H2SO4 is the equation for sulphuric acid. just from that if you know how to work out moles you'll understand.
4. (Original post by onceandnever)
H2SO4 is the equation for sulphuric acid. just from that if you know how to work out moles you'll understand.
this is not what I meant, in order to carry out this equation there needs to be at least one concentration given also the reaction can't just be H2SO4 + H2O there needs to be more such as:

NaOH+ HCl-> NaCl + H2O
5. u need concentration for first one mate
6. (Original post by Schoolstudent01)
this is not what I meant, in order to carry out this equation there needs to be at least one concentration given also the reaction can't just be H2SO4 + H2O there needs to be more such as:

NaOH+ HCl-> NaCl + H2O
The balanced equation I have is H2SO4 + 2NaOH ---> Na2SO4 + 2H2O

I prepared 250cm3 standard solution of 0.106 mol dm-3 for my other solution. I worked out that i used 0.00265 moles of sodium hydroxide in my titration. Not sure if this is needed.

For the sulphuric acid solution I mixed 25cm3 up to 250cm3 with distilled water. Not sure how to get the concentration from this though.
7. (Original post by ehruhure)
u need concentration for first one mate
I prepared 250cm3 standard solution of 0.106 mol dm-3 for my other solution. I worked out that i used 0.00265 moles of sodium hydroxide in my titration. Not sure if this is needed.

For the sulphuric acid solution I mixed 25cm3 up to 250cm3 with distilled water. Not sure how to get the concentration from this though.
8. (Original post by luck2luke)
I prepared 250cm3 standard solution of 0.106 mol dm-3 for my other solution. I worked out that i used 0.00265 moles of sodium hydroxide in my titration. Not sure if this is needed.

For the sulphuric acid solution I mixed 25cm3 up to 250cm3 with distilled water. Not sure how to get the concentration from this though.
M/Mr=n=25/98=0.255mol
9. (Original post by Cheesus69)
M/Mr=n=25/98=0.255mol
Is that the answer to D? I had the same idea but I thought because sulphuric acid density is 1.84g per cm3 that id have to do 25x1.84 first?

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