Answer : The correct option is, 11.77 L
Solution : Given,
Mass of bromine gas = 84 g
Molar mass of bromine gas = 159.8 g/mole
First we have to calculate the moles of bromine gas.
[tex]\text{Moles of }Br_2=\frac{\text{ given mass of }Br_2}{\text{ molar mass of }Br_2}=\frac{84g}{159.8g/mole}=0.5256moles[/tex]
Now we have to calculate the volume of bromine gas at STP.
As we know that at STP,
1 mole of bromine gas contains 22.4 L volume of bromine gas
So, 0.5256 mole of bromine gas contains [tex]22.4\times 0.5256=11.77L[/tex] volume of bromine gas
Therefore, the volume of bromine gas at STP is, 11.77 L
Option 4 - 11.77 L
Explanation
Given
Mass of Br2 gas: 84g
Molecular mass of Br2 gas: 159.808 g/mol
R= 0.082057 atm.dm mol-1 K-1
T= 0°C = 273 K
(Standard Temperature Pressure abbreviated as STP, defined as zero degrees celsius temperature 0°C or 273 K and pressure is equals to 1 atm or 760 torr or 760 mm of hg)
Required:
Volume at S.T.P=?
Formula:
V=nRT
since, n= m / M
Here, m= given mass
and M= molecular mass
Now,
V= m/M ×R × T
V= 84/ 159.808 × 0.082057 × 273
V= 11.77 L
Hence, volume of Br2 gas at STP will be 11.77 L.
