The number of molecules of water in 333 g of Al2SO43⋅18H2O is :

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The number of molecules of water in 333 g of ${Al}_{2} {({SO}_{4})}_{3} \cdot 18 H_{2} O$ is :

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$18 \times 6.022 \times 10^{23}$

$9 \times 6.022 \times 10^{23}$

answer is B.

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Detailed Solution

Molar mass of ${Al}_{2} {({SO}_{4})}_{3} \cdot 18 H_{2} O$

$M = 2 \times 27 + 3 \times (32 + 3 \times 16) + 18 \times 18$

$M = 666 g / mol$

Number of moles of ${Al}_{2} {({SO}_{4})}_{3} \cdot 18 H_{2} O = \frac{333}{666} = 0.5$

Number of molecular of water $= 18 \times 0.5 \times 0.05 \times 10^{23}$

$= 9 \times 6.09 \times 10^{23}$

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