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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?

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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 $58.12$.

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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

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  • $\begingroup$ Since you decided not to type the answer properly, I decided not to spend time trying to decipher the handwritten draft; but the equivalent weight of an element (from the header) is supposed to be given in $\pu{g mol-1 equiv-1}$, not just in grams. $\endgroup$ – andselisk Jan 2 '18 at 11:05
  • $\begingroup$ Actually I am a student and not quite used to the SE script writing.....so, it would have taken much time and by the way I am not sure of the answer. $\endgroup$ – DJ Koustav Jan 2 '18 at 11:08
  • $\begingroup$ It's not scripting language, it's two markup languages which are relatively easy to get used to, especially if you used sites like GitHub before. Please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown. Many people here are also students, some of them have jobs and families, and they don't use this as an excuse. $\endgroup$ – andselisk Jan 2 '18 at 11:13
  • $\begingroup$ Ok..I agree.... But can you please"decipher" my solution once... If I've done any thing wrong $\endgroup$ – DJ Koustav Jan 2 '18 at 11:15
  • $\begingroup$ I posted an answer below on how I'd approach this problem. Compare yours with that and you'll get the same result. If you are unsure/in doubt or have another question, post it as, well, another question. $\endgroup$ – andselisk Jan 2 '18 at 11:18

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