Identifying a chloride of the unknown element from the masses of chlorine and compound as well as vapor density
I agree with the starting point of the Matthew's solution, namely the definition of vapor density, but I fail to understand the follow-up math, so I just post my approach.

$$M(\ce{ECl_x}) = D\cdot M(\ce{H2}) = 85\cdot\pu{2 g mol-1} = \pu{170 g mol-1}\tag{1}$$

By definition, molar mass $M(\ce{ECl_x})$ is also

$$M(\ce{ECl_x}) = M(\ce{E}) + x\cdot M(\ce{Cl})$$

$$\implies M(\ce{E}) = M(\ce{ECl_x}) - x\cdot M(\ce{Cl}) = (170 - x\cdot 35.5)~\pu{g mol-1}\label{eqn:2}\tag{2}$$

The unknown variable $x$ is defined by the molar ratio between the elements:

$$x = \frac{n(\ce{Cl})}{n(\ce{E})} = \frac{m(\ce{Cl})\cdot M(\ce{E})}{m(\ce{E})\cdot M(\ce{Cl})} = \frac{\pu{0.835 g}\cdot (170 - x\cdot 35.5)~\pu{g mol-1}}{(1 - 0.835)~\pu{g} \cdot \pu{35.5 g mol-1}} = 24.14 - 5.04\cdot x\label{eqn:3}\tag{3}$$

Solving \eqref{eqn:3}:

$$6.04\cdot x= 24.14 \quad\implies\quad x = 4.0 \tag{4}$$

Using $x$, we can now determine the element from \eqref{eqn:2}:

$$M(\ce{E}) = (170 - 4.0\cdot 35.5)~\pu{g mol-1} = \pu{28.0 g mol-1}\tag{5}$$

which appears to be silicon, valency 4.
The unknown compound is silicon tetrachloride $\ce{SiCl4}$, a fuming volatile liquid at NTP.