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Standard IX

Chemistry

Mole Concept & Avogadro Number

1g CH4 and ...

Question

$1 g$ $C H_{4}$ and $4 g$ of compound $X$ have equal number of moles. Thus, molar mass of $X$ is :

16 g

m o l^{- 1}

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32 g

m o l^{- 1}

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4 g

m o l^{- 1}

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64 g

m o l^{- 1}

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Solution

The correct option is B $64 g$ $m o l^{- 1}$
No. of moles of $C H_{4}$ in $1$ g $= \frac{1}{16}$ .
No. of moles of X in $4$ g $= \frac{4}{m} = \frac{1}{16}$
$m = 64$ g.

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1g CH4 and 4g of compound X have equal number of moles. Thus, molar mass of X is :

The correct option is B 64g mol−1No. of moles of CH4 in 1 g =116.No. of moles of X in 4 g =4m=116 m=64 g.

$1 g$ $C H_{4}$ and $4 g$ of compound $X$ have equal number of moles. Thus, molar mass of $X$ is :

The correct option is B $64 g$ $m o l^{- 1}$
No. of moles of $C H_{4}$ in $1$ g $= \frac{1}{16}$ .
No. of moles of X in $4$ g $= \frac{4}{m} = \frac{1}{16}$
$m = 64$ g.