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Atomic Number and Mass Number

1.00 g of B...

Question

$1.00$ g of $B a C l_{2}$ is treated with excess of aqueous $A g N O_{3}$ and all chlorine is recovered as $1.38$ g of AgCl. What is the atomic weight of $B a$ ? ( $C l$ $= 35.5$ , $A g$ $= 108$ )

137

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172.5

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33

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68.5

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Solution

The correct option is C $137$
$B a C l_{2} + 2 A g N O_{3} ==> 2 A g C l + B a (N O_{3})_{2}$

$1.38 g A g C l \times \frac{1 m o l e A g C l}{143.3 g A g C l} \times \frac{1 m o l e B a C l_{2}}{2 m o l e s A g C l} = 0.00482 m o l e s B a C l 2$

$m o l e s B a C l_{2} = \frac{g r a m s B a C l_{2}}{f o r m u l a w t . B a C l_{2}}$
$f o r m u l a w t . B a C l_{2} = \frac{g r a m s X C l_{2}}{m o l e s B a C l_{2}} = \frac{1.00}{0.00482} = 208$

So $B a C l_{2} = 208 a m u$
$B a = 208 - 2 C l = 208 - 2 (35.5) = 137 a m u$
So atomic weight of Ba is 137

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