Chemistry - Mole Concept Question with Solution | TestHub

Molecular weight$=2\times \mathrm{V}.\mathrm{D}.=2\times 95=190$
Approx atomic weight$=\frac{\mathrm{6.4 }}{\mathrm{s}\mathrm{p}\mathrm{e}\mathrm{c}\mathrm{i}\mathrm{f}\mathrm{i}\mathrm{c }\mathrm{h}\mathrm{e}\mathrm{a}\mathrm{t}}=\frac{6.4}{0.13}=49.23$
Suppose the formula of metal chloride$=\mathrm{M}\mathrm{C}{\mathrm{l}}_{\mathrm{X}}$
Molecular weight$=$$49.23+35.5\mathrm{X}=190$
So, X = 4(Nearest whole no.)
Formula is${\text{MCl}}_{\text{4}}$
$\text{mol}\text{.}\text{wt}\text{.= }\underset{\text{metal}\left(\text{exact}\right)}{\text{at}\text{. wt}\text{. of}}\text{ +}\left(35.5\times 4\right)$
$190-140=\text{at}\text{.wt}\text{.}$
$\text{at}\text{.}\text{wt}\text{.=}\text{50}$