The quantity of CO ₂ in 500 ~mL of soda water when packed under 3.34 bar CO ₂ pressure at 298 ~K in g is

Step-by-step solution

Given, Quantity of CO ₂=500 ~mL Pressure of CO ₂=3.34 bar Temperature =298 ~K in Henry's law gathered p=K_ H X (x) X= p K_ H =3.34 bar =334000 ~Pa gathered = 3.34 10^1 1.67 10^8 =2.442 ~Pa But, we have 500 ~mL of soda water so that Volume of water =500 ~mL Density of water =1 ~g / mL Formula mass = volume density we get 500 ~mL of water =500 ~g of water Molar mass of water =18 ~g / mol ⁻¹ aligned 500 18 & =27.78 mole of water x & = n_ CO ₂ n_ CO ₂ +n_ H ₂ O aligned Value of moles fraction is very small so it is neg