Question

here is your question​

Answer 1

SOLUTION

For this question we need to use:

. N204------2NO2

Initial 1 mole. 0mole

Final. (1-0.2) mol. 0.4 mol

Molecular mass of the mixture

$= > \frac{n(n2o4) \times mm(n2o4) + (n)(no2) \times mm(no2}{n(n2o4)n(no2)} \\ \\ = > \frac{(0.8 \times 92) + (0.4 \times 46)}{(0.8 + 0.4)} \\ = > 76.66g \: per \: mol$

Now using the ideas gas Equation

PV = nRT

so,

PV = m/M RT

thus density,

d= m/V= 1atm× 76.66g/mol/300k× 0.0821l atm/K mol

=) 3.1162 g/L

hope it helps ✔️

Answer 2

⭐⭐Here is your answer⭐⭐

⚡⚡Mahir⚡⚡