Question

What is the volume of 84 g of bromine gas (Br2) at STP?

42.6L

0.085L

599L

11.77L

Answer 1

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.

`\text{Moles of }Br_2=\frac{\text{ given mass of }Br_2}{\text{ molar mass of }Br_2}=(84g)/(159.8g/mole)=0.5256moles`

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
`22.4* 0.5256=11.77L` volume of bromine gas

Therefore, the volume of bromine gas at STP is, 11.77 L

Answer 2

Option 4 - 11.77 L

Step-by-step 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.