Question

16 g of an ideal gas SOx occupied same volume as 20g SO3 at STP. how many atoms of oxygen are contained in given amount of SOx?

Answer 1

Answer: Two atoms of oxygen are contained in given amount of $SO_2$

Explanation:

Given that at STP, 16 g of$SO_x$ gas occupies same volume occupied by 20 gram of $SO_3$ gas at STP.

1) For $SO_3$ gas

Volume occupied by the 20 grams of $SO_3$ gas.

Number of moles of $SO_3$ gas $=\frac{20g}{80g/mol}=0.25mol$

Pressure at STP,P = 1 atm

Temperature at STP ,T= 273 Kelvins

R = universal gas constant = 0.0821 atm L/mol Kelvin

PV=nRT(ideal gas equation)

On substituting the values, we get volume ,V :

$V=\frac{nRT}{P}=\frac{0.25\times 0.0821 atm L/mol Kelvin\times 273 Kelvin}{1atm}=5.60 L$

2) For $SO_x$ gas

Given that at STP, same volume of 5.60 L will be occupied by the 20 grams of $SO_x$ gas.(Value of P and T will remain same at STP)

$PV=\frac{\text{Gives mass of gas} SO_x}{\text{molecular mass of gas}SO_x}RT$

$\text{molecular mass of gas} SO_x}=\frac{16g\times 0.0821atm L/mol Kelvin\times 273 Kelvins}{1 atm\times 5.60 L}=64.03 gram/mol$

Molecular mass $SO_x=64.03g/mol$

(molar mass of sulfur)+ x(molar mass of oxygen) = 32 +x(16) = 64.03g/mol

x = 2.00

Two atoms of oxygen are contained in given amount of $SO_2$.