A hydrocarbon contains $82.8 %$ of carbon. Its molecular formula if its vapour density is $29$ is :
$[H = 1, C = 12]$

C_{3} H_{6}

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C_{4} H_{10}

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C_{4} H_{6}

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C_{4} H_{8}

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Solution

The correct option is B $C_{4} H_{10}$
As we know,
Molar mass $= 2 \times v a p o u r d e n s i t y$
Molar mass of $X$ $= 2 \times 29 = 58 g m$
Mass of $C$ $= 82.8 \times \frac{58}{100} = 48 g m$
Mass of $H$ $= 10 g m$
So, compound contains 4 C atoms and 10 H atoms and compound is $C_{4} H_{10}$ .

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A hydrocarbon contains 82.8% of carbon. Its molecular formula if its vapour density is 29 is :[H=1, C=12]

The correct option is B C4H10As we know,Molar mass =2×vapour densityMolar mass of X =2×29=58gmMass of C =82.8×58100=48gmMass of H =10gmSo, compound contains 4 C atoms and 10 H atoms and compound is C4H10.

A hydrocarbon contains $82.8 %$ of carbon. Its molecular formula if its vapour density is $29$ is :
$[H = 1, C = 12]$

The correct option is B $C_{4} H_{10}$
As we know,
Molar mass $= 2 \times v a p o u r d e n s i t y$
Molar mass of $X$ $= 2 \times 29 = 58 g m$
Mass of $C$ $= 82.8 \times \frac{58}{100} = 48 g m$
Mass of $H$ $= 10 g m$
So, compound contains 4 C atoms and 10 H atoms and compound is $C_{4} H_{10}$ .