Chemistry A Level Question

100cm3 of H2SO4 solution was measured out into a 250cm3 conical flask using a measuring cylinder.
Excess Barium Chloride was added to the conical flask.
The mixture was filitered and the precipitate of BaSO4 collected and dried.
Mass: 2.57g

Calculate the number of moles of BaSO4 formed.
Calculate the number of moles of H2SO4 in the 100cm3 of solution and hence calculate the concentration of the sulfuric acid to the appropriate number of siginicant figures.


PLEASE I FOUND THIS IN THE TEXTBOOK BUT THERE ARE NO ANSWERS. I didn't start sixth form yet I'm self studying preparing for 11th of September.

The first equation is this:

BaCl2 + H2SO4 = BaSO4 + 2HCl

Moles= mass/mr
So moles of BaSO4 is 2.57/233.38 = 0.01101020833
= 0.01

Moles of H2SO4 is also 0.01 due to balanced equation 1:1 ratio

Moles = concentration x volume
Hence concentration is 0.11 mol dm^-3

Omg!!! I got the 0.01 but I crossed it out as I thought it's wrong! That's amazing thank you so much!

Learn the formulas.

Learn the importance of the balanced equation as it gives the ratios you need to do questions like this.

250/100 is 2.5

use number of moles is concentration x volume

The first equation is this:

BaCl2 + H2SO4 = BaSO4 + 2HCl

Moles= mass/mr
So moles of BaSO4 is 2.57/233.38 = 0.01101020833
= 0.01

Moles of H2SO4 is also 0.01 due to balanced equation 1:1 ratio

Moles = concentration x volume
Hence concentration is 0.11 mol dm^-3

How did you get 0.11?? What volume do you use 100 or 250?

The question says 100cm^3 which is 0.1dm^3 which is the volume you use for concentration

But 0.01/0.1 is 0.1

I used exact values. Never round early.

I used exact values. Never round early.

I'm guessing you have the same book as I, could you tell me what module 1 question 5CII is? I'm so lost as I can't see anything which answers it in Module 1. Are the questions from module 1 on all modules or something?

I have this book:

Oh alright nevermind then 😭😂

It's telling me to calculate a value for the enthalpy of combustion for methanol. Mass of spirit difference: 0.44g. Volume of water: 100 and temperature change if water is 18.

First step is q= mass x c x delta t
We take the mass to be the mass of the water. There is 100cm^3 so mass is 100g. Specific heat capacity of water is 4.18. Change in temperature is 18.
Hence q = 7524 J
Which is 7.524 kJ.

q is basically the energy released, here it is 7.524 kJ.

Next we work out the number of moles of methanol burnt.
0.44g was burnt so moles is 0.44/32.04 = 0.01248439451

You'll notice that the units for enthalpy change of combustion are kJ mol^-1

Hence we do 7.524/0.01248439451 = 603 kJ/mol

You haven't said if temperature increased or decreased. If it increased the answer would be -603 kJ/mol as reaction would be exothermic.

Sorry it increased! Thank you so much, makes sense!!!

First step is q= mass x c x delta t
We take the mass to be the mass of the water. There is 100cm^3 so mass is 100g. Specific heat capacity of water is 4.18. Change in temperature is 18.
Hence q = 7524 J
Which is 7.524 kJ.

q is basically the energy released, here it is 7.524 kJ.

Next we work out the number of moles of methanol burnt.
0.44g was burnt so moles is 0.44/32.04 = 0.01248439451

You'll notice that the units for enthalpy change of combustion are kJ mol^-1

Hence we do 7.524/0.01248439451 = 603 kJ/mol

You haven't said if temperature increased or decreased. If it increased the answer would be -603 kJ/mol as reaction would be exothermic.