Question

What is the molecular formula of the alkane, the $5.6L$ of which weighs $11g\text{at}STP$ ?

• A. $CH_{4}$
• B. $C_{4}H_{10}$
• C. $C_{3}H_{8}$
• D. $C_{2}H_{6}$

Answer 1

Answer: (C)

At $STP$ condition, the volume at $STP$ is $22.4L$ .
Now, the $5.6L$ of alkane has weigh $11g$ at $STP$
Then,
$1g$ of alkane has weight of $22.4LSTP$ is $=\frac{11 g \times 22 . 4 L}{5 . 6 L}$
$\Rightarrow 44gm$
The molecular mass of alkane at $STP$ is $44gm$ .
So, $C_{3}H_{8}$ has molecular mass if $M=\frac{w}{n}$ .
$C_{3}H_{8}=3\times 12+8\times 1$
$\Rightarrow 36+8$
$\Rightarrow 44u$