One litre of a gas at S.T.P. weighs 1.16 g. It can possibly be

Question

One litre of a gas at S.T.P. weighs 1.16 g. It can possibly be (A) C2H2 (B) CO (C) O2 (D) CH4

Tardigrade Admin · Accepted Answer

At STP, ( V text olume of a gas /22.4)=( text W eight / text Molecular weight ) (1/22.4)=(1.16/ text Molecular weight ) Molecular weight =1.16 × 22.4 =26 gram / mole Therefore, the answer is C 2 H 2.

Q. One litre of a gas at S.T.P. weighs 1.16 g. It can possibly be

$C_{2} H_{2}$

CO

$O_{2}$

$C H_{4}$

At STP,
$\frac{V olume of a gas}{22.4} = \frac{W eight}{Molecular weight}$
$< b r / > \frac{1}{22.4} = \frac{1.16}{Molecular weight} < b r / >$
Molecular weight $= 1.16 \times 22.4$
$= 26 g r am / m o l e$
Therefore, the answer is $C_{2} H_{2}$ .

Q. One litre of a gas at S.T.P. weighs 1.16 g. It can possibly be

C2​H2​

CO

O2​

CH4​

At STP, 22.4V olume of a gas ​= Molecular weight W eight ​ <br/>22.41​= Molecular weight 1.16​<br/> Molecular weight =1.16×22.4 =26gram/mole Therefore, the answer is C2​H2​.

$C_{2} H_{2}$

$O_{2}$

$C H_{4}$

At STP,
$\frac{V olume of a gas}{22.4} = \frac{W eight}{Molecular weight}$
$< b r / > \frac{1}{22.4} = \frac{1.16}{Molecular weight} < b r / >$
Molecular weight $= 1.16 \times 22.4$
$= 26 g r am / m o l e$
Therefore, the answer is $C_{2} H_{2}$ .