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One mole of magnesium in the vapor state...

One mole of magnesium in the vapor state absorbed `1200 kJ mol^(-1)` of enegry. If the first and second ionization energies of `Mg` are `750` and `1450 kJ mol^(-1)`, respectively, the final composition of the mixture is

A

0.69 mol `Mg^(+)` and 0.31 mol `Mg^(2+)`

B

0.59 mol `Mg^(+)` and 0.41 mol `Mg^(2+)`

C

0.49 mol `Mg^(+)` and 0.51 mol `Mg^(2+)`

D

0.29 mol `Mg^(+)` and 0.71 mol `Mg^(2+)`

Text Solution

Verified by Experts

Molar mass of Mg =24 g `mol^(-1)`
`:.` Amount of Mg=1 mol
Thus, 750 kJ of energy is used up to convert 1 mol of Mg to `Mg^(+)`
Energy used to convert `Mg^(+)` to `Mg^(2+)`
`=1200-750 =450 kJ`
`:.` Amount of magnesium converted to `Mg^(2+)`
`=(450 kJ)/(1450) =0.31 mol`
`:. ` Amount of magnesium converted to `Mg^(+)`
=1-0.31 =0.69 mol
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