Question

One mole of magnesium in the vapour state absorbed 1200 kJ of energy. If the first and second ionization potentials of magnesium are 750 and 1450 kJ/mole respectively, the final composition of the mixture is:

• A. $69$,$31$
• B. $59$,$41$
• C. $49$,$51$
• D. $29$,$71$

Answer 1

Answer: (A)

$69$,$31$

Let mass of $M{g}^{+}$ ion=$x$ $g$

$M{g}^{2+}$ ion=$y$ $g$

and total$=1$

$x+y=1$

Molar mass of $Mg=24.g/mol$

No. of moles of $M{g}^{+}=\frac{x}{24}g/mol$

No. of mole of $M{g}^{2+}=\frac{y}{24}g/mol$

Energy absorbed to convert to $M{g}^{+}=\frac{x}{24}\times 750$

Energy absorbed to convert to $M{g}^{2+}=\frac{y}{24}\times (1450+750)$

Total energy absorbed= $\frac{x}{24}\times 740+\frac{y}{24}\times 2200=50$

$24\times 50=740x+2200y$

$=740x+2200-2200x$

$24\times 50=2200-1460x$

$1200-2200=-1460x$

$x=\frac{-1000}{1460}$

$=0.6849$

$y=1-0.6849=0.3151$

Mass % of $M{g}^{2+}$ ion=$\times 0.3151\times 100$

$=31.51$

Mass% of $M{g}^{+}$ ion=$0.6849\times 100$

$=68.49$

Mass of $M{g}^{+}$ ion=$68.49g$

Mass of $M{g}^{2+}$ ion=$31.51g$