titration calculation help

Er...How do you get 40cm³ of Cr(III)? Could you scan in the whole question? I'm a bit confused atm lol

Its above this post.
sorry its bit long.
Really, really appriciate your help.

Oh. You only used 40/250 of the volume of solution that you put the Cr(III) ions in, so the answer is only 40/250 of the total mass of the Cr ions in that 2.2g Should get 0.272g, which is 12.3% of 2.2g.

Those were the answers to the 2 questions
You used the right method for Q1:

$\frac{25.05}{1000} \times 0.1001 = 2.508 \times 10^{-3}\ \mathrm{moles\ of\ Fe}$

$= 8.36 \times 10^{-4}\ \mathrm{moles\ of\ Cr}$

$= 4.35 \times 10^{-2}\mathrm{g}\ \mathrm{in}\ 40\mathrm{cm^3}$

Because the salt's in 250cm³ of solution, you only have 40/250 of the salt, so divide by that to get the total mass of the Cr = 0.272g, which is 12.3% of 2.2g.

Those were the answers to the 2 questions
You used the right method for Q1:

$\frac{25.05}{1000} \times 0.1001 = 2.508 \times 10^{-3}\ \mathrm{moles\ of\ Fe}$

$= 8.36 \times 10^{-4}\ \mathrm{moles\ of\ Cr}$

$= 4.35 \times 10^{-2}\mathrm{g}\ \mathrm{in}\ 40\mathrm{cm^3}$

Because the salt's in 250cm³ of solution, you only have 40/250 of the salt, so divide by that to get the total mass of the Cr = 0.272g, which is 12.3% of 2.2g.

Redo your first calculation of 25.05/1000 x 0.1001 And the very last line should be x 250/40 rather than the other way round; you use 40cm³ out of 250, so you need to divide by 40 to get what's in 1 cm³ then multiply by 250.

What's this for btw? A level? Seems like you've been given an awful lot of time to do it

Yeah, except you still have the very first line wrong At least, my calculator says 25.05/1000 x 0.1001 = 2.508 x 10^-3. Using 52 as the atomic mass of chromium is fine btw

Haha, you wont be given anything close to that in an exam I think what they mean is use the calculator value rather than a rounded value when you're doing calculations with those numbers.

Oh cool i can put my heart at rest now. Thank you a billion.