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Standard IX

Chemistry

Atomic Mass and Atomic Mass Unit

1.00 g of E...

Question

$1.00 g$ of $E u C l_{2}$ is treated with excess of aqueius $A g N O_{3}$ and all the chlorine is recovred as $1.29 g$ of $A G c L$ . Calculate the atomic weight of $E u (C l = 35.5, A g = 108)$ .

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Solution

% of chlorine in $A g C l = \frac{35.5}{35.5 + 107.87} = 24.7$ %
$24.7$ % of 1.29 kg $= \frac{24.7 \times 1.29}{100} = 0.319$ g Cl
All the chlorine come from $E u C l_{2}$
In 1 g of $E u C l_{2}$ , 0.319 g is Chlorine
$31.9$ % of $E u C l_{2}$ is Cl
$68.1$ % of $E u C l_{2}$ is Eu by weight
and of that each of the two Cl atoms in $E u C l_{2}$ contribute $\frac{31.9}{2} = 16.0$ %
Set up the proportion
$\frac{35.5}{0.160} = \frac{x}{0.681}$
x is the atomic weight of Eu
$x = \frac{35.5 \times 0.681}{0.160} = 150.9 g m o l^{-}$

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1.00 g of EuCl2 is treated with excess of aqueius AgNO3 and all the chlorine is recovred as 1.29 g of AGcL. Calculate the atomic weight of Eu(Cl=35.5,Ag=108).

$1.00 g$ of $E u C l_{2}$ is treated with excess of aqueius $A g N O_{3}$ and all the chlorine is recovred as $1.29 g$ of $A G c L$ . Calculate the atomic weight of $E u (C l = 35.5, A g = 108)$ .