Question

Calculate the density of a gaseous mixture which consists of 3.01 * 10²4

Answer 1

Answer:

Density of gaseous mixture is 1.2g/lit

Answer 2

The question is incomplete, here is the complete question:

Calculate density of a gaseous mixture which consist of 3.01 x 10²⁴ molecules of N₂ and 32 g of O₂ gas at 3 atm pressure and 860 K temperature (Gives R = 1/2 atm L/mole K) : (a) 0.6 g / L (b) 1.2 g / L (c) 0.2 g / L (d) 12 g / L​

The density of gaseous mixture is 0.2 g/L

Explanation:

• For nitrogen gas:

We are given:

Number of molecules = $3.01\times 10^{24}$

According to mole concept:

$6.022\times 10^{23}$ number of molecules are present in 1 mole of a compound

So, $3.01\times 10^{24}$ number of molecules will be present in = $\frac{1}{6.022\times 10^{23}}\times 3.01\times 10^{24}=5mol$ of nitrogen molecule

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$    .....(1)

Moles of nitrogen gas = 5 moles

Molar mass of nitrogen gas = 28 g/mol

Putting values in equation 1, we get:

$5mol=\frac{\text{Mass of nitrogen gas}}{28g/mol}\\\\\text{Mass of nitrogen gas}=(5mol\times 28g/mol)=140g$

• For Oxygen gas:

Given mass of oxygen gas = 32 grams

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

$\text{Moles of oxygen gas}=\frac{32g}{32g/mol}=1mol$

• For mixture:

To calculate the volume of mixture, we use the equation given by ideal gas equation:

PV = nRT

P = Pressure of the gas = 3 atm

V = Volume of the gas = ?

n = number of moles of mixture = [5 + 1] = 6 moles

R = Gas constant = $\frac{1}{2}\text{ L atm }mol^{-1}K^{-1}$

T = Temperature of the gas = 860 K

Putting values in above equation:

$3atm\times V=6mol\times \frac{1}{2}\text{ L atm }mol^{-1}K^{-1}\times 860K\\\\V=\frac{6\times 1\times 860}{3\times 2}=860L$

To calculate the density of gas, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Mass of mixture = [140 + 32] = 172 g

Volume of mixture = 860 L

Putting values in above equation, we get:

$\text{Density of mixture}=\frac{172g}{860L}\\\\\text{Density of mixture}=0.2g/L$

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