Question

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

• A. $C{H}_4$
• 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 $=\frac{11g\times 22.4L}{5.6L}$
$\Rightarrow 44gm$
The molecular mass of alkane at $STP$ is $44gm$ .
So, $C_3{H}_8$ has molecular mass if $M=\frac{w}{n}$ .
$C_3{H}_8=3\times 12+8\times 1$
$\Rightarrow 36+8$
$\Rightarrow 44u$