Atoms and Molecules
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 elements $$X$$ and $$Y$$ form two compounds, $$XY_2$$ and $$X_3Y_2$$. $$0.1$$ mol of former weighs $$10$$ g while $$0.05$$ mole of the latter weighs $$9$$ g. Find the atomic weight of $$X$$.
$$X+2Y\rightarrow XY_2$$
$$0.1$$ mol of $$XY_2=10 g$$
$$1$$ mol of $$XY_2=100 g$$
$$3X+2Y\rightarrow X_3Y_2$$
$$0.05$$ mol of $$X_3Y_2=9 g$$
$$1$$ mol of $$X_3Y_2=\dfrac {9}{0.05}=180 g$$
$$\therefore X+2Y=100; MW_{XY_2}=100$$
$$3X+2Y=180; MW_{X_2Y_3}=180$$
Solving for $$X$$ and $$Y$$, we get, $$X=40 g$$ and $$Y=30 g$$.
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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