XeF4 reacts with SbF5 to form |LIQUIDSOLUTIONJEEMain2023

JEE MAIN 2023

08-04-2023 S1

Question

XeF4 reacts with SbF5 to form $\mathrm{XeF}_4$ reacts with $\mathrm{SbF}_5$ to form $\left[\mathrm{XeF}_{\mathrm{m}}\right]^{n+}\left[\mathrm{SbF}_{\mathrm{y}}\right]^{z-}$

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Solution

Sol. $$ \begin{aligned} & \text { Percentage of Carbon }=\frac{12}{44} \times \frac{\text { mass of } \mathrm{CO}_2 \text { formed }}{\text { mass of compound taken }} \times 100 \\ & 60=\frac{12}{44} \times \frac{\text { mass of } \mathrm{CO}_2 \text { formed }}{0.5} \times 100 \\ & \text { Mass of } \mathrm{CO}_2 \text { formed }=\frac{60 \times 44 \times 0.5}{12 \times 100} \mathrm{~g} \\ & =1.1 \text { gram } \\ & =11 \times 10^{-1} \text { gram } \end{aligned} $$

Question Tags

JEE Main

Chemistry

Easy

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