Question

2. 20.83 g. of a gas occupies 4.167 L at 79.97 kPa at 30.0 °C. What is its molecular weight?

Answer 1

Answer:

f. wt. = 158 g/mol

Explanation:

Given:

mass = 20.83 g

Vol = 4.167 L

Pressure = 79.97 KPa x 0300987 Atm/KPa = 0.78924 Atm

T = 30°C + 273 = 303K

R = 0.08206 L·Atm/mol·K

PV = nRT = (mass/f.wt.)RT => f. wt. = mass x R x T / P x V

f. wt. = 20.83g x 0.08206 L·Atm/mol·K x 303K / 0.78924 Atm x 4.167L

= 158 g/ mol

Answer 2

20.83 g of a gas occupies 4.167 L at 79.97 kPa at 30.0°C. Its molecular weight is 158 g/mole.

Ideal Gas Law

The ideal gas law states that the pressure of gas is directly proportional to the volume and temperature of the gas.

PV = nRT

where,

P = Pressure

V = Volume

n = number of moles  

R = Ideal gas constant

T = Temperature

How to find the molecular weight ?

Molecular Weight = $\frac{\text{Given Mass}}{\text{Number of moles}}$

Convert kPa into atm

atm = $\frac{kPa}{101.3}$

So, 79.97 kPa = $\frac{79.97\ kPa}{101.3}$

                       = 0.7894 atm

Convert Celsius into Kelvin

Kelvin = 273 + Celsius

Kelvin = 273 + 30.0

          = 303 Kelvin  

Now, first we have to find the number of moles.

Use the Ideal Gas formula to find the number of moles.

PV = nRT

n = $\frac{PV}{RT}$

n = $\frac{0.7894\ \text{atm} \times 4.167 L}{0.0821\ \text{L atm / mole K} \times 303\ K}$

n = $\frac{3.289}{24.87}$

n = 0.1322 mole

Now put the value of number of moles in above formula to find the molecular weight

Molecular Weight = $\frac{\text{Given Mass}}{\text{Number of moles}}$

                              = $\frac{20.83}{0.1322}$

                              = 158 g/mole

Thus, we can say that 20.83 g of a gas occupies 4.167 L at 79.97 kPa at 30.0°C. Its molecular weight is 158 g/mole.

Learn more about the Ideal Gas Law here: https://brainly.com/question/25290815
#SPJ2