Question

How many milligrams of MgI2 must be added to 257.7 mL of 0.087 M KI to produce a solution with [I−] = 0.1000 M?

Answer 1

Answer:

The answer is 465.6 mg of MgI₂ to be added.

Explanation:

We find the mole of ion I⁻ in the final solution

C = n/V -> n = C x V = 0.2577 (L) x 0.1 (mol/L) = 0.02577 mol

But in the initial solution, there was 0.087 M KI, which can be converted into mole same as above calculation, equal to 0.02242 mol.

So we need to add an addition amount of 0.02577 - 0.02242 = 0.00335 mol of I⁻. But each molecule of MgI₂ yields two ions of I⁻, so we need to divide 0.00335 by 2 to find the mole of MgI₂, which then is 0.001675 mol.

Hence, the weight of MgI₂ must be added is

Weight of MgI₂ = 0.001675 mol x 278 g/mol = 0.4656 g = 465.6 mg

Answer 2

We have the milligrams of MgI2 must be added to 257.4 mL of 0.078 M KI to produce a solution with [I−] = 0.1000 M is

$m_{MgI2 }=465.65mg$

From the Question we are told that

volume of solution =257.7 $ml =0.2577L$  

Initial conc of $[I^-]$  $C_1 = 0.087M$

Final conc of $[I^-]$ $C_2= 0.1M$

Generally

Net Concentration of  [I^-]

$C'=C_2-C_1\\\\C'= 0.1-0.087\\\\C'=0.013mole/L$

no of moles of I^- required

m = molarity* volume of solution in L

$m = 0.013*0.2577\\\\m= 0.00335 moles of I^-$

With ratio of MgI_2 to I^- as 1:2

Therefore

The no of moles of MgI2 needed = 0.00335/2      

The no of moles of MgI2 needed =0.001675

Hence

mass of MgI2

$m_{MgI2 }=$ no of moles * gram molar mass

$m_{MgI2 }=0.001675*278$

$m_{MgI2 }=0.46565g$

$m_{MgI2 }=465.65mg$

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