Question

In a chemical determination of the atomic weight of vanadium , $ 2.8934g $ of pure $ VOC{l_3} $ was allowed to undergo a set of reactions as a result of which all the chlorine contained in this compound was converted to $ AgCl $ which weighed $ 7.1801g. $ Calculate the atomic weight of vanadium. $ \left( {Ag = 107.868,Cl = 35.453} \right)\; $

Answer 1

Hint :To calculate the atomic weight of vanadium, we need to consider the number of moles of chlorine in each compound. It is said that pure $ VOC{l_3}, $ which is vanadium oxytrichloride, undergoes several chemical reactions before its full chlorine content is converted to $ AgCl, $ which is silver chloride. The number of moles of chlorine in vanadium oxytrichloride is the same as the number of moles of chlorine in silver chloride. We can use the equation:
 $ {\text{Number of moles of a substance = }}\dfrac{{{\text{Mass of the substance in the compound}}}}{{{\text{Molar mass of the substance}}}} $ .

Complete Step By Step Answer:
First, let us calculate the number of moles of chlorine in $ VOC{l_3} $ and $ AgCl. $ We know that:
 $ {\text{Number of moles of chlorine in }}VOC{l_3} = {\text{Number of moles of chlorine in }}AgCl $
 $ \Rightarrow 3{\text{ moles of chlorine in }}VOC{l_3} = 1{\text{ mole of chlorine in }}AgCl $
In $ VOC{l_3}, $ $ 3{\text{ moles of chlorine in }}VOC{l_3} = \dfrac{{3 \times 2.8934}}{{x + 16 + \left( {3 \times 35.453} \right)}} $
 $ = \dfrac{{8.6802}}{{x + 122.359}} \to (1) $
Where 'x' is the atomic weight of vanadium. The atomic weight of oxygen is $ 16 $ . Unit has been ignored as the final expression is in moles.
In $ AgCl, $ $ 1{\text{ mole of chlorine in }}AgCl = \dfrac{{1 \times 7.1801}}{{107.868 + 35.453}} $
 $ = \dfrac{{7.1801}}{{143.321}} $
 $ = 0.05009\,moles \to (2) $
Equating $ {\text{(}}1{\text{) and (}}2{\text{),}} $ we get:
 $ \dfrac{{8.6802}}{{x + 122.359}} = 0.05009 $
Rearranging,
 $ x = \left( {\dfrac{{8.6802}}{{0.05009}}} \right) - 122.359 $
 $ x = 173.292 - 122.359 $
Hence,
 $ x = 50.933g $
The atomic weight of vanadium is $ 50.933g. $

Note :
The principle used in this question is POAC or the Principle of atom conservation. In terms of moles, this principle states that the number of moles of an element in a compound are conserved throughout the reaction undergone by the compound. According to the question, the principle has been applied to chlorine, as the number of moles of chlorine in vanadium oxytrichloride has been conserved as the compound undergoes change into silver chloride.