How many atoms of hydrogen are present in $7.8 g$ of $A l {(O H)}_{3}$ ?

6.0 \times 10^{22}

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1.8 \times 10^{23}

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1.7 \times 10^{23}

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5.1 \times 10^{22}

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Solution

The correct option is A $6.0 \times 10^{22}$
Molar mass of $A l (O H)_{3} = 27 + 3 \times 16 + 3 \times 1 = 78 g$
So, no. of moles of $A l (O H)_{3} = \frac{7.8}{78} = 0.1 m o l e s$
$1 m o l e$ of $A l (O H)_{3}$ has $6.023 \times 10^{23}$ atoms.
So, $0.1 m o l e s$ has $6.023 \times 10^{22}$ atoms.

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How many atoms of hydrogen are present in 7.8 g of Al(OH)3?

The correct option is A 6.0×1022Molar mass of Al(OH)3=27+3×16+3×1=78 gSo, no. of moles of Al(OH)3=7.878=0.1 moles1 mole of Al(OH)3 has 6.023×1023 atoms.So, 0.1 moles has 6.023×1022 atoms.

How many atoms of hydrogen are present in $7.8 g$ of $A l {(O H)}_{3}$ ?