[fatghost13]

A set of standard solutions are being prepared for the construction of a calibration curve for the determination of chromium by atomic absorption spectroscopy. A solution of potassium dichromate of concentration 4.810 x 10^-2 mol dm^-3 is available. Calculate the volume of this solution that must be diluted to 250 cm^3 to give a solution of chromium concentration equal to 20 mg dm^-3.

i know i need to change the dm^-3 to mol dm^-3, but how can i find the volume of the solution diluted.

[Hunter2]

1. Calculate the total  mass of chromium in the 250 cm³
2. Calculate how much mole it is
3. Calculate the corresponding amount of moles of the Dichromate.
4. Calculate the volume of the necessary solution.

[fatghost13]

1. Calculate the total  mass of chromium in the 250 cm³
2. Calculate how much mole it is
3. Calculate the corresponding amount of moles of the Dichromate.
4. Calculate the volume of the necessary solution.

1. 20 x (1000/250) = 80 mg dm ³
2. 0.08 x 52 =4.16 mol dm ³
3. the corresponding amount of moles of the Dichromate = 0.0481 x 2 = 0.0962 mol dm ³
    (As potassium dichromate (2 mole)  =  chromium (1 mole)
4  How to calculate the volume of the necessary solution.?

[magician4]

1. 20 x (1000/250) = 80 mg dm ³

what would that be? a mass, like proposed by Hunter2 ?

besides: 20 (mL) x (1000 mL / 250 mL) = 80 mL

  it is always useful to consider dimensions also, as this more often than not is meaningfull  (i.e. might show you where you've goofed, for example)

the correct calculation would have been:
m(Cr) _{total, 250 mL} = 20 mg dm^-3 x 250 dm³ = 5 g

2. 0.08 x 52 =4.16 mol dm ³

again, dimensions would have shown the error

with n = m/M ,the moles of Cr belonging to 5 g of Cr would be n = 5g / 52 g/mole [itex]\approx [/itex] 96.154 mmol

3. the corresponding amount of moles of the Dichromate = 0.0481 x 2 = 0.0962 mol dm 3
    (As potassium dichromate (2 mole)  =  chromium (1 mole)

these are not the corresponding moles of dichromate belonging to  96.154 mmol chromium, but would be the (virtual) concentration of named chromium in the original solution of dichromate, if it wasn't condensed
... and hence kind of irrelevant with respect to the path Hunter2 did propose / the question #3 he put

correct answer would have been , that you'd need to have 0.5 * 96.154 mmol = 48.077 mmol content in the volume of the starting solution containing named dichromate (as each dichromate will count for 2 chromium-atoms)

ref # 4:
x dm³ * 4.810 x 10^-2 mol dm^-3 = 48.077 * 10^-3 mol    x = ?


regards

Ingo

[DrCMS]

@magician4 As 1mL ≠ 1dm³ all your maths is wrong.

[magician4]

jeesss....

you're right DrCMS !

1 dm³ = 1000 ml would be soo much better...

...and I apologise for the confusion resulting thereof

hopefully, fatghost13 might work out the necessary corrections all by himself


regards

Ingo