Question

600 ml of a mixture of o3 and o2 weighs 1g at ntp. find volume of ozone

Answer 1

here is your ans........

Answer 2

Hey dear,

◆ Answer-
Volume of ozone = 200 ml

◆ Explanation-
# Given-
V = 600 ml = 0.6 L
W = 1 g

# Solution-
Let's consider x = volume of ozone.

1 mole of gas at NTP is 22.4 L.

Therefore, mass of x volume of ozone(M=48) is
W1 = 48 × x / 22.4
W1 = 2.143x

Also, mass of oxygen(M=32) is -
W2 = 32 × (0.6-x) / 22.4
W2 = 1.43(0.6-x)

Given,
W1 + W2 = W = 1
2.143x + 1.43(0.6-x) = 1

Solving this, you'll get,
x = 0.2 L = 200 ml

Volume of ozone in the mixture is 200 ml.

Hope this helps...