The number of molecules of water in $333 g$ of $A l_{2} (S O_{4})_{3}$ . $18 H_{2} O$ is :

18.0 \times 6.02 \times 10^{23}

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9.0 \times 6.02 \times 10^{23}

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333.0 \times 6.02 \times 10^{23}

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2.0 \times 6.02 \times 10^{23}

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Solution

The correct option is B $9.0 \times 6.02 \times 10^{23}$
Molar mass of $A l_{2} (S O_{4})_{3} .18 H_{2} O$ :
$M o l a r m a s s = 2 \times a t o m i c m a s s o f A l + 3 \times a t o m i c m a s s o f s u l p h u r + 12 \times a t o m i c m a s s o f O$ $+ 18 \times (2 \times a t o m i c m a s s o f H + 1 \times a t o m i c m a s s o f O)$

$M o l a r m a s s = 2 \times 27 + 3 \times 32 + 12 \times 16$ $+ 18 \times (21 + 1 \times 16)$

$M = 666 g / m o l$
$M = 666 g / m o l$
Number of moles of $A l_{2} (S O_{4})_{3} .18 H_{2} O = \frac{333}{666} = 0.5$

Number of molecular of water $= 18 \times 0.5 \times 6.02 \times 10^{23}$
$= 9 \times 6.02 \times 10^{23}$

Hence the correct option is $B$ .

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