Question

A sample of neon effuses from a container in 72 seconds. the same amount of an unknown noble gas requires 147 seconds. you may want to reference (pages 442 - 444) section 10.9 while completing this problem. part a identify the second gas.

Answer 1

The   second gas  is  identified  as  follows

by  Graham law  formula
let  the  unknown gas  be  represented  by  letter y

=time of  effusion of Neon/ time  of effusion of  y =   sqrt (molar mass of neon/molar  mass of y)

= 72  sec/ 147  sec =  sqrt( 20.18  g/mol/ y   g/mol)

square the  both  side  to  remove  the  square  root  sign

72^2/147^2  = 20.18 g/mol/y g/mol

=0.24 = 20.18g/mol/y g/mol

multiply  both side  by  y  g/mol
= 0.24  y g/mol = 20.18g/mol
divide both side  by  0.24 
y =  84  g/mol

y  is  therefore Krypton  since  it  is  the one  with a molar mass  of  84 g/mol

Answer 2

Answer: The unknown noble gas is Krypton.

Explanation:

Rate of effusion is defined as the amount of volume displaced per unit time.

$\text{Rate of effusion}=\frac{V}{t}$

To calculate the rate of diffusion of gas, we use Graham's Law.

This law states that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molar mass of the gas. The equation given by this law follows the equation:

$\text{Rate of diffusion}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}$

We are given:

Time taken by neon gas = 72 s

Time taken by unknown gas = 147 s

Molar mass of neon gas = 20.18 g/mol

By taking their ratio, we get:

$\frac{\frac{V}{t_{Ne}}}{\frac{V}{t_{\text{unknown gas}}}}=\sqrt{\frac{M_{Ne}}{M_{\text{unknown gas}}}}\\\\\\\frac{147}{72}=\sqrt{\frac{M_{\text{unknown gas}}}{20.18}}\\\\M_{\text{unknown gas}}=84.11g/mol$

The noble gas having molar mass of 84.11 g/mol is Krypton

Hence, the unknown noble gas is Krypton.