SOLUTION : Calculation of equivalent weight of arsenic
$\begin{array}{c}\frac{\text { wt. of silver }}{\text { wt. of arsenious chloride }}=\frac{\text { eq. wt. of silver }}{\text { eq. wt. of arsenious chloride }} \\\frac{100}{56}=\frac{107 \cdot 8}{E+35 \cdot 5} \quad \text { where } E=\text { eq. wt. of arsenic } \\\text { or, } \quad E+35 \cdot 5=\frac{107 \cdot 8 \times 56}{100}=60 \cdot 37, \quad \therefore \quad E=60 \cdot 37-35 \cdot 5=24 \cdot 87 .\end{array}$
Rough atomic wt. (by Dulong and Petit's law)
$=\frac{6 \cdot 4}{\text { sp. heat }}=\frac{6 \cdot 4}{c \cdot 082}=78 \cdot 05 \text { and valency }=\frac{78 \cdot 05}{24 \cdot 87}=3 \cdot 14 \approx 3$
The empirical formula of chloride of arsenic $= AsCl _{3}$
and thus molecular formula $=\left( AsCl _{3}\right) \times n$
and exact atomic wt. of arsenic $=3 \times 24 \cdot 87=74 \cdot 61$
Calculation of molecular weight of arsenious chloride :
Given that $V . D$. of arsenious chloride with respect to air $=6 \cdot 3$
The V.D. with respect to air
$\begin{aligned}&=\frac{\text { wt. of one litre of arsenious chloride at S.T.P. }}{\text { wt. of one litre of air at S.T.P. }} \\\therefore 6 \cdot 3 &=\frac{\text { wt. of one litre of arsenious chloride at S.T.P. }}{1 \cdot 293}\end{aligned}$
$\therefore \quad$ Wt. of one litre of arsenious chloride at ST.P. $=6 \cdot 3 \times 1 \cdot 293 g =8 \cdot 146 g$ Therefore, the wt. of $22 \cdot 4$ litres of arsenious chloride, that is, mol. wt.
$=8 \cdot 146 \times 22.4=182.47 g / \text { mole }$
Calculated mol. wt. of $\left( AsCl _{3}\right) \times n=($ at. wt. of $As +3 \times$ at. wt. of $Cl ) \times n$
$\begin{array}{c}=(74 \cdot 61+3 \times 35 \cdot 5) n=181 \cdot 11 \times n \\\therefore \quad 181 \cdot 11 \times n=182 \cdot 47 ; \quad \therefore \quad n=\frac{182 \cdot 47}{181 \cdot 11} \approx 1\end{array}$
The correct formula of arsenious chloride $= AsCl _{3}$.
