The number of moles of BaCl2- 2H2O in a pure sample that contains 3-8 kilomoles of oxygen atoms is -

2 moles of O atoms are present in 1 mole of BaCl_2. 2H_2O. So, 3/8 kilomoles of O atoms will be present in nbsp; = 1/2×3/8× 1000 moles of BaCl_2.2H_2O = 187.5 ...

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Standard XII

Chemistry

Isobars and Isotopes

The number of...

Question

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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Solution

2 moles of O-atoms are present in 1 mole of $B a C l_{2} . 2 H_{2} O$ .
So, $\frac{3}{8}$ kilomoles of O-atoms will be present in
$= \frac{1}{2} \times \frac{3}{8} \times 1000$ moles of $B a C l_{2} .2 H_{2} O$
$= 187.5$ moles of $B a C l_{2} .2 H_{2} O$

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The number of moles of BaCl2.2H2O in a pure sample that contains 38 kilomoles of oxygen atoms is .

2 moles of O-atoms are present in 1 mole of BaCl2.2H2O. So, 38 kilomoles of O-atoms will be present in =12×38×1000 moles of BaCl2.2H2O =187.5 moles of BaCl2.2H2O

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 .

2 moles of O-atoms are present in 1 mole of $B a C l_{2} . 2 H_{2} O$ .
So, $\frac{3}{8}$ kilomoles of O-atoms will be present in
$= \frac{1}{2} \times \frac{3}{8} \times 1000$ moles of $B a C l_{2} .2 H_{2} O$
$= 187.5$ moles of $B a C l_{2} .2 H_{2} O$