The average atomic mass of a mixture containing 79 mole % of $^{24} M g$ and remaining 21 mole % of $^{25} M g$ and $^{26} M g$ , is 24.31. The % mole of $^{26} M g$ in the mixture is :

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Solution

The correct option is C 10
Assume mole $%$ of $^{26} M g$ is $x$ .

Then, mole % of $^{25} M g = 21 - x$
So, avg. atomic mass of mixture $= \frac{79 \times 24 + x \times 26 + (21 - x) \times 25}{100}$ $= 24.31$ (given)
So, $x = 10$ %

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Similar questions

Q. The average atomic mass of a mixture containing 79 mole percent of

^{24} M g

and remaining 21 mole percent of

^{25} M g

and

^{26} M g,

is 24.31. Find the mole percent of

^{26} M g

Q. What is the average atomic mass of Mg if the naturally occurring isotopes are

^{24} M g

^{25} M g

and

^{26} M g

. Their masses and abundances are as follows:

\begin{matrix} Isotope & Atomic Mass & Isotopic abundance^{24} M g & 23.98504 & 78.70 %^{25} M g & 24.98584 & 10.13 %^{26} M g & 25.98259 & 11.17 % \end{matrix}

Q. Assertion :The average mass of one Mg atom is

24.305

amu, which is not equal to the actual mass of one Mg atom. Reason: Three isotopes,

^{24} M g,^{25} M g

and

^{26} M g

of Mg can be found in nature.

Q. Calculate the reported average elemental atomic mass of

M g

, if the naturally occurring isomers are

^{24} M g,^{25} M g, a n d^{26} M g

. Their masses and abundance are as follows:

\begin{matrix} Isotope & Atomic Mass & Isotopic Abudance^{24} M g & 23.98504 & 78.70 %^{25} M g & 24.98584 & 10.13 %^{26} M g & 25.98259 & 11.17 % \end{matrix}

One mole of magnesium in the vapour state absorbed 1200 kJ ${m o l e}^{- 1}$ of energy. If the first and second ionization energies of Mg are 750 and 1450 kJ ${m o l e}^{- 1}$ respectively, the final composition of the mixture is

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The average atomic mass of a mixture containing 79 mole % of 24Mg and remaining 21 mole % of 25Mg and 26Mg, is 24.31. The % mole of 26Mg in the mixture is :

The correct option is C 10Assume mole % of 26Mg is x. Then, mole % of 25Mg=21−xSo, avg. atomic mass of mixture =79×24+x×26+(21−x)×25100 =24.31 (given)So, x=10 %

The average atomic mass of a mixture containing 79 mole % of $^{24} M g$ and remaining 21 mole % of $^{25} M g$ and $^{26} M g$ , is 24.31. The % mole of $^{26} M g$ in the mixture is :

The correct option is C 10
Assume mole $%$ of $^{26} M g$ is $x$ .

Then, mole % of $^{25} M g = 21 - x$
So, avg. atomic mass of mixture $= \frac{79 \times 24 + x \times 26 + (21 - x) \times 25}{100}$ $= 24.31$ (given)
So, $x = 10$ %