The concentration of $C = 85.45$ % and $H = 14.55$ % is not obeyed by the formula :

C_{4} H_{8}

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C_{2} H_{4}

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C_{2} H_{6}

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C H_{2}

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Solution

The correct option is C $C_{2} H_{6}$
As per the given data, the ratio of number of moles of $C$ and $H$ is given by,
$\begin{matrix} C & : & H \frac{85.45}{12} & : & \frac{14.55}{1} = 7.12 & : & 14.55 = 1 & : & 2 \end{matrix}$
Therefore, the empirical formula will be $C H_{2}$ .
We know,
$Molecular formula = (C H_{2})_{n}$
where, $n = \frac{Molecular weight}{Empirical weight}$
So, Molecular Formula can be $C_{2} H_{4}, C_{3} H_{6}$ or $C_{4} H_{8}$ , which are all integral multiple of $C H_{2}$ .
So, it cannot be $C_{2} H_{6}$ which is a multiple of $C H_{3}$ .

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Q. For the gas phase reaction,

C_{2} H_{4} + H_{2} ⇌ C_{2} H_{6}; [△ H = - 32.7 k c a l]

Carried out in a vessel, the equilibrium concentration of

C_{2} H_{4}

can be increased by:

For the gas phase reaction $C_{2} H_{4} + H_{2} ⇌ C_{2} H_{6},$ $Δ H$ = -32.7kcal carried out in a vessel, the equilibrium concentration of $C_{2} H_{4}$ can be increased by -

Q. For the gas phase reaction,

C_{2} H_{4} + H_{2}

⇌

C_{2} H_{6}

(Δ H = - 32.7 k c a l)

, carried out in a closed vessel. The equilibrium concentration of

C_{2} H_{4}

can be increased by?

Following gaseous reaction is undergoing in a vessel $C_{2} H_{4} + H_{2} ⇌ C_{2} H_{6}; Δ H = - 32.7 K c a l$ . Which will increase the equilibrium concentration of $C_{2} H_{6}$ ?

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The concentration of C=85.45% and H=14.55% is not obeyed by the formula :

The concentration of $C = 85.45$ % and $H = 14.55$ % is not obeyed by the formula :