The average atomic mass of magnesium is $24.31 amu$ this magnesium is composed of $79 mole % of {}_{24}{Mg}$ and remaining $21 mole % of {}_{25}{Mg} and {}_{26}{Mg}$ . Calculate mole $% of {}_{26}{Mg} ?$

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Solution

Step 1: Given data
The average atomic mass of magnesium $= 24.31 amu$
$Mole % of {}_{24}{Mg} = 79 Mole % of {}_{25}{Mg} and {}_{26}{Mg} = 21$
Step 2: Molecular mass calculation formula:
Now, let $x$ be the mole $% of {}_{25}{Mg}$ and $(21 - x) mole % of {}_{26}{Mg}$
$Molecular mass = \sum_{} x_{i} M_{i} x_{i} = Fraction of the isotope M_{i} = Mass of the isotope .$
Step 3: Finding the mole $% of {}_{26}{Mg}$ :
By solving the equation
$24.3 = (79 / 100) 24 + (x / 100) 25 + (21 - x) / 100 x 25 Thus, x = 11 Therefore, mole % of {}_{26}{Mg} is (21 - x) = (21 - 11) = 10 .$
Therefore, the mole $% of {}_{26}{Mg} is 10 %$

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Q. average atomic mass of magnesium is 24.31 amu. this magnesium is composed of 79 mole % of Mg-24 and remaining 21 mole % of Mg-25. calculate mole % of Mg-26.

Q. Average atomic mass of magnesium is 24.31. AMU this magnesium is composed of 79% mole of 24 mg and remaining 21 mole percent of 25 mg and 26 MG .calculate mole percent of 26 mg

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

^{24} Mg

and remaining 21 mole percent of

^{25} Mg

and

^{26} Mg

, is 24.31. Mole percent of

^{26} Mg

is :

Consider a $70 %$ efficient hydrogen-oxygen cell working under the standard condition at $1$ bar and $298$ K. Its cell reaction is

$H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l)$

The work derived from the cell on the consumption of $1.0 \times 10^{- 3}$ $mol of H_{2} (g)$ is used to compress $1.00 mol$ of a monoatomic ideal gas in a thermally insulated container. What is the change in the temperature (in $K$ ) of the ideal gas?

The standard reduction potentials for the two half-cells are given below

$O_{2} (g) + 4 H^{+} (aq) + 4 e^{-} \to 2 H_{2} O (l)$ , $E^{0}$ = $1.23 V$

$2 H^{+} (aq) + 2 e^{-} \to H_{2} (g)$ , $E^{0}$ = $0.00 V$

Use $F = 96500 {Cmol}^{- 1}$ , $R = 8.314 {Jmol}^{- 1} K^{- 1}$ .

Q. Question 23
Calculate the number of moles of magnesium present in a magnesium ribbon weighing 12 g. Molar atomic mass of magnesium is

24 g m o l^{- 1}

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The average atomic mass of magnesium is 24.31amu this magnesium is composed of 79mole%ofMg24 and remaining 21mole%ofMg25andMg26. Calculate mole %ofMg26?

The average atomic mass of magnesium is $24.31 amu$ this magnesium is composed of $79 mole % of {}_{24}{Mg}$ and remaining $21 mole % of {}_{25}{Mg} and {}_{26}{Mg}$ . Calculate mole $% of {}_{26}{Mg} ?$