Calculation of equivalent weight


> 
> When a solid element is reacted with chlorine, a gaseous chloride of vapor density $68.75$ is formed. If this reaction is performed at constant temperature and pressure, the volume of the system reduces by $\frac{1}{3}$. What is the equivalent weight of the solid element?
> 
> 
> 


I found out the molar mass of the gas $68.75 \times 2$. But don't know how to proceed from here? 




I would express equivalent weight $M\_\mathrm{eq}(\ce{E})$ of unknown element $\ce{E}$ using the number of equivalents $x$ and corresponding molecular weights $M$:


\begin{align}
M(\ce{ECl\_x}) &= x \cdot M\_\mathrm{eq}(\ce{E}) + x \cdot M(\ce{Cl}) \tag{1} \\ \to \quad M\_\mathrm{eq}(\ce{E}) &= \frac{M(\ce{ECl\_x}) - x \cdot M(\ce{Cl})}{x} \tag{1a}
\end{align}


To find $x$, let's apply to the balanced reaction, assuming it is complete and there is no chlorine left:


$$\ce{E (s) + x/2 Cl2 (g) -> ECl\_x (g)}$$


The amounts $n$ of gaseous products are bound by stoichiometry:


$$n(\ce{Cl2}) = \frac{x}{2}n(\ce{ECl\_x}) \tag{2}$$


At constant temperature and pressure $n\_i V\_i = \mathrm{const}$. The problem says the volume of the system reduces by $1/3$, which means:


\begin{align}
V(\ce{Cl2}) &= \frac{3}{2}V(\ce{ECl\_x}) \tag{3} \\
\to \quad n(\ce{Cl2}) &= \frac{3}{2}n(\ce{ECl\_x}) \tag{3a}
\end{align}


Now, comparing/equating (2) and (3a) it's easy to see that $x = 3$:


$$\frac{x}{2} = \frac{3}{2} \quad \to \quad x = 3$$


You already correctly determined molecular weight from vapor density $\rho\_\mathrm{v}$:


$$M(\ce{ECl\_x}) = \rho\_\mathrm{v} \cdot M(\ce{H2}) = 68.75 \cdot \pu{2.016 g mol-1} = \pu{138.60 g mol-1} \tag{4}$$


All that's left now is to do the math with (1a):


$$M\_\mathrm{eq}(\ce{E}) = \frac{\pu{138.60 g mol-1} - 3 \cdot \pu{35.45 g mol-1}}{\pu{3 equiv}} = \pu{10.75 g mol-1 equiv-1}$$


I'm not sure this makes any sense from the chemistry prospective as this would suggest that element $\ce{E}$ is sulfur and so the gaseous product is $\ce{SCl3}$. I would've expected the compound to be something like $\ce{BCl3}$, but in this case the vapor density should've been $58.12$.


 1

I just made some rough calculations. And my answer (equivalent weight of the solid) comes out to be 10.33g.
As you have not mentioned the correct answer and since my answer has not matched with that of my previous answerer, so I am just providing an image of my drafting work and I am not going to type the full answer.
![enter image description here](https://i.stack.imgur.com/GqsBL.jpg)


 0