A student is supplied with 100.0 cm3 of a solution of 0.400 mol dm–3 magnesium iodide, MgI2.

A student plans to dilute this solution so that the iodide concentration is 0.250 mol dm–3. What volume of water, in cm3, does the student need to add?

i keep getting 160 but the mark scheme says 220 idk where im going wrong

A student plans to dilute this solution so that the iodide concentration is 0.250 mol dm–3. What volume of water, in cm3, does the student need to add?

i keep getting 160 but the mark scheme says 220 idk where im going wrong

Please forgive me for making my explanation long-winded and messy.

I’d start off the question by thinking about how many moles of magnesium iodide are present in the solution. This remains unchanged when the solution is diluted, so it can give us something to work with.

Remember that concentration = moles/volume

Therefore: moles = concentration x volume

moles of MgI2 = 0.400 mol dm^-3 x 100 cm^3 x 1/1000 dm^3 cm^-3 = 0.040 mol

The question asks for the iodide ion concentration, rather than the concentration of the magnesium iodide.

When 1 mole of magnesium iodide dissolves in water, it forms 2 moles of iodide ions (and 1 mole of magnesium ions, but that’s not particularly relevant). This tells us that the moles of iodide ions is 2 times the moles of magnesium iodide calculated earlier (meaning there are 0.080 mol of iodide ions).

Again, you use the formula concentration = moles/volume, but you rearrange it as volume = moles/concentration

The iodide ion concentration of the diluted solution is 0.250 mol dm^-3, so the total volume of the solution after diluting it is (0.080 mol)/(0.250 mol dm^-3) = 0.320 dm^3

To convert this to cm^3, you just multiply it by 1000. This tells us the total volume of the diluted solution is 320 cm^3.

Because there was initially 100 cm^3 of the solution, to get the total volume to 320 cm^3, 220 cm^3 of water must have been added to the original.

I think the two reasons you kept getting 160 cm^3 were: (A): not doubling the moles of MgI2 and (B): not subtracting the 100 cm^3 at the very end.

I’d start off the question by thinking about how many moles of magnesium iodide are present in the solution. This remains unchanged when the solution is diluted, so it can give us something to work with.

Remember that concentration = moles/volume

Therefore: moles = concentration x volume

moles of MgI2 = 0.400 mol dm^-3 x 100 cm^3 x 1/1000 dm^3 cm^-3 = 0.040 mol

The question asks for the iodide ion concentration, rather than the concentration of the magnesium iodide.

When 1 mole of magnesium iodide dissolves in water, it forms 2 moles of iodide ions (and 1 mole of magnesium ions, but that’s not particularly relevant). This tells us that the moles of iodide ions is 2 times the moles of magnesium iodide calculated earlier (meaning there are 0.080 mol of iodide ions).

Again, you use the formula concentration = moles/volume, but you rearrange it as volume = moles/concentration

The iodide ion concentration of the diluted solution is 0.250 mol dm^-3, so the total volume of the solution after diluting it is (0.080 mol)/(0.250 mol dm^-3) = 0.320 dm^3

To convert this to cm^3, you just multiply it by 1000. This tells us the total volume of the diluted solution is 320 cm^3.

Because there was initially 100 cm^3 of the solution, to get the total volume to 320 cm^3, 220 cm^3 of water must have been added to the original.

I think the two reasons you kept getting 160 cm^3 were: (A): not doubling the moles of MgI2 and (B): not subtracting the 100 cm^3 at the very end.

(edited 2 years ago)

Original post by TypicalNerd

Please forgive me for making my explanation long-winded and messy.

I’d start off the question by thinking about how many moles of magnesium iodide are present in the solution. This remains unchanged when the solution is diluted, so it can give us something to work with.

Remember that concentration = moles/volume

Therefore: moles = concentration x volume

moles of MgI2 = 0.400 mol dm^-3 x 100 cm^3 x 1/1000 dm^3 cm^-3 = 0.040 mol

The question asks for the iodide ion concentration, rather than the concentration of the magnesium iodide.

When 1 mole of magnesium iodide dissolves in water, it forms 2 moles of iodide ions (and 1 mole of magnesium ions, but that’s not particularly relevant). This tells us that the moles of iodide ions is 2 times the moles of magnesium iodide calculated earlier (meaning there are 0.080 mol of iodide ions).

Again, you use the formula concentration = moles/volume, but you rearrange it as volume = moles/concentration

The iodide ion concentration of the diluted solution is 0.250 mol dm^-3, so the total volume of the solution after diluting it is (0.080 mol)/(0.250 mol dm^-3) = 0.320 dm^3

To convert this to cm^3, you just multiply it by 1000. This tells us the total volume of the diluted solution is 320 cm^3.

Because there was initially 100 cm^3 of the solution, to get the total volume to 320 cm^3, 220 cm^3 of water must have been added to the original.

I think the two reasons you kept getting 160 cm^3 were: (A): not doubling the moles of MgI2 and (B): not subtracting the 100 cm^3 at the very end.

I’d start off the question by thinking about how many moles of magnesium iodide are present in the solution. This remains unchanged when the solution is diluted, so it can give us something to work with.

Remember that concentration = moles/volume

Therefore: moles = concentration x volume

moles of MgI2 = 0.400 mol dm^-3 x 100 cm^3 x 1/1000 dm^3 cm^-3 = 0.040 mol

The question asks for the iodide ion concentration, rather than the concentration of the magnesium iodide.

When 1 mole of magnesium iodide dissolves in water, it forms 2 moles of iodide ions (and 1 mole of magnesium ions, but that’s not particularly relevant). This tells us that the moles of iodide ions is 2 times the moles of magnesium iodide calculated earlier (meaning there are 0.080 mol of iodide ions).

Again, you use the formula concentration = moles/volume, but you rearrange it as volume = moles/concentration

The iodide ion concentration of the diluted solution is 0.250 mol dm^-3, so the total volume of the solution after diluting it is (0.080 mol)/(0.250 mol dm^-3) = 0.320 dm^3

To convert this to cm^3, you just multiply it by 1000. This tells us the total volume of the diluted solution is 320 cm^3.

Because there was initially 100 cm^3 of the solution, to get the total volume to 320 cm^3, 220 cm^3 of water must have been added to the original.

I think the two reasons you kept getting 160 cm^3 were: (A): not doubling the moles of MgI2 and (B): not subtracting the 100 cm^3 at the very end.

omg thank you soo much i was struggling with this a lot im glad for the help you gave!

Original post by lotus9765.

A student is supplied with 100.0 cm3 of a solution of 0.400 mol dm–3 magnesium iodide, MgI2.

A student plans to dilute this solution so that the iodide concentration is 0.250 mol dm–3. What volume of water, in cm3, does the student need to add?

i keep getting 160 but the mark scheme says 220 idk where im going wrong

A student plans to dilute this solution so that the iodide concentration is 0.250 mol dm–3. What volume of water, in cm3, does the student need to add?

i keep getting 160 but the mark scheme says 220 idk where im going wrong

omg same

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