Number of moles of water in 488 g of $B a C l_{2} . 2 H_{2} O$ are _____________ .

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Solution

The correct option is C 4
$B a C l_{2} .2 H_{2} O$ , have molar mass of $244 g / m o l$

moles of $B a C l_{2} .2 H_{2} O$ in 488g = $\frac{given weight}{molecular weight} = \frac{488}{244} = 2$

from the molecular formula we know that, 1 mole of $B a C l_{2} .2 H_{2} O$ contains 2 moles of water,

so, 2 moles of $B a C l_{2} .2 H_{2} O$ have water = $2 \times 2 = 4 m o l e s$

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Similar questions

No. of moles of water in $488.6 g$ of $B a C l_{2} .2 H_{2} O$ are (molecular weight of $B a C l_{2} .2 H_{2} O = 244.33$ )

488.6

B a C l_{2} .2 H_{2} O

B a C l_{2} .2 H_{2} O

=

244.33

No. of moles of water in $488.6 g$ of $B a C l_{2} .2 H_{2} O$ are (molecular weight of $B a C l_{2} .2 H_{2} O = 244.33$ )

The number of moles of $B a C l_{2} . 2 H_{2} O$ in a pure sample that contains $\frac{3}{8}$ kilomoles of oxygen atoms is .

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