(3)
how many liters are in 1.35 x 105 grams of oxygen at stp?

V = 8.97×10⁴ L

Explanation:

Given data:

Mass of oxygen = 1.35×10⁵ g

Litters of oxygen = ?

Solution:

First of all we will calculate the number of moles.

Number of moles = mass/ molar mass

Number of moles = = 1.35×10⁵ g / 32 g/mol

Number of moles = 0.04 ×10⁵ mol

Litters of oxygen:

PV = nRT

V = nRT/P

V = 0.04 ×10⁵ mol × 0.0821 atm. L/ mol. K ×273 K / 1atm

V = 0.897×10⁵ atm.L / 1atm

V = 8.97×10⁴ L

9.46 x 10⁴ L

Explanation:

Data Given:

mass of oxygen (O₂) = 1.35 x 10⁵

Liters of  oxygen O₂ (volume) = ?

Solution:

First find no. of moles of oxygen

Formula used

            no. of moles = mass in gram / molar mass

molar mass of oxygen (O₂) = 2(16)

molar mass of oxygen (O₂) = 32 g/mol

Put values in above formula

             no. of moles = 1.35 x 10⁵ / 32 g/mol

             no. of moles = 4219 mol

Now we will find volume of oxygen

Formula used

            no. of moles = Volume of gas / Molar volume

Rearrange the above equation to find volume

        Volume of gas = no. of moles x Molar Volume . . . . . . (1)

Where

Molar volume = 22.42 L /mol

Put values in above equation 1

             Volume of gas = 4219 mol x 22.42 L /mol

             Volume of gas = 94584 L

             Volume of gas = 9.46 x 10⁴ L

So,

Volume of oxygen (O₂) = 9.46 x 10⁴ L

explanation:

i think option no d is the correct answer

the answer for sodium is 0.186

the answer for rubidium is 0.251