Chemistry, asked by gvedant2104, 8 months ago

The vapour density of a mixture containing NO2 and N2O4 is 38.3 at 27C. Which of the following is/are correct for 100 moles of mixture ? (A) Moles of NO2 in mixture is 33.48 (B) Moles of N2O4 in mixture is 66.52 (C) Weight of NO2 in mixture is 1540 g (D) Weight of N2O4 in mixture is 1540 g multiple ans type question...

Answers

Answered by Anonymous

Answer:

The vapour density of a mixture containing NO2 and N2O4 is 38.3 at 27°C. Calculate the mole of NO2 in 100 gram of mixture?

OnePlus Nord available on Amazon India.

Let the total number of moles = 1

let the number of moles of NO2 = X

so number of moles of N2O4 = 1 -x

molar mass of mixture = 2 x vapour density

= 2 × 38.3 = 76.6

Now x × 46 + ( 1-x) 92 = 76.6

46 x + 92 -92 X = 76.6

46 x = 15.4

x = 15.4 / 46 = 0.335

% age of NO2 = 33.5 %

% age of N2O4 = 66.5 %

Number of moles NO2 in 100 g of mixture =0.335 × 100 / 76.6 = 0.437

Answered by mehek2440

Answer:

꧁༒ÂⱠÊӾ ᴳᵒᵈ༒꧂

⌛⏳FOLLOW Me ✔️⚡

$<body bgcolor=yellow><maruee><fontcolor=blue>$

See the attachment for detailed solution to the question ...answer taken from nageshwar70❤

#TRUE BRAINLIAN#BE BRAINLY♻️

❌✖️FREE FORM SPAM✖️❌

✔️MARK MY ANSWERS BRAINLIEST

Attachments:

Previous Question

Next Question