Question

I.E1 and I.E2 of a metal M are 600 and 1600 kJ/mole respectively. If 1000kJ of energy supplied to one mole of vapour of M, then the resulting mixture mole percentage of M+2 is 5X.the value of X is​

Answer 1

Answer:

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Explanation:

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Answer 2

Answer:

The value of X is 5.

In resulting mixture, mole percentage of M⁺ ions in resulting mixture is 75% and percentage of M²⁺ ions is 25%.

Explanation:

Ionization of metal M:-

M $\longrightarrow$   M⁺   +   e⁻                  I.E.₁= 600kJmol⁻¹

M⁺ $\longrightarrow$   M²⁺    +  e⁻               I.E₂ = 1600KJmol⁻¹

Given, The first ionization energy of the one mole of M = 600KJ

Second ionization energy of  one mole of M = 1600KJ

The amount of energy supplied to 1 mole of vapour of M =1000KJ

Amount of energy left unused after 1st ionization =1000-600=400KJ

The percentage of M⁺ converted into  M²⁺:

$=\frac{400}{1600}*100$

$=25\%$

The percentage of M⁺ ions = 100-25 = 75%

Given, the resulting mixture mole percentage of M²⁺ = 5X

$5X=25$

$X=5$

Therefore, the value of X is 5.