Question

at 300k ,40ml of O3(g) dissolves in 100g of water at 1atm. what mass of ozone dissolved in 400g of water at a pressure of 4atm at 300K? (ans 4.8g)

Answer 1

From the given statement we get,m α P ⇒40×4 ml of ozone dissolves in water of 400 g at 300 K, 4 atm⇒dissolved number of mole of ozone=PV/RT=(4×160×10−3)/(0.0821×300)=0.025hence mass of ozone = 48 gmol/×0.025 mol =1.2 g

Answer 2

Hello dear,

● Answer -
W = 4.989 g

● Explanation -
Moles of O3 required to dissolve in 400 g water can be given by -
n = PV/RT × W1/W2
n = (4 × 4×40×10^-3)/(0.0821 × 300) × (400/100)
n = 0.104 mol

Mass of ozone dissolved is calculated as -
W = n × M
W = 0.104 × 48
W = 4.989 g

Therefore, mass of ozone dissolved is 4.989 g

Thanks for asking..