- #1

WeiLoong

- 35

- 0

Please post this type of questions in HW section using the template.

0.0125g of hydrogen gas produced was collected using water displacement method at 23.0°C and 99.99kNm^-2. If the pressure of water vapour at the same temperature is 2.81kNm^-2, calculate the volume of hydrogen gas collected

My calculation : PV = nRT

n = No. of moles of H2 = 0.0125/2 = 6.25 x 10^-3 mole

P = 99.99 -2.81 = 97.18 KN/m^2

T = 23degC = 296.15K

R = 8.314J/molK

V = nRT/P = (6.25 x 10^-3 x 8.314 x 296.15)/97180 = 1.5835 x 10^-4 m^3 = 158.35mL

but the answer given is 0.583L any mistakes did i make?

My calculation : PV = nRT

n = No. of moles of H2 = 0.0125/2 = 6.25 x 10^-3 mole

P = 99.99 -2.81 = 97.18 KN/m^2

T = 23degC = 296.15K

R = 8.314J/molK

V = nRT/P = (6.25 x 10^-3 x 8.314 x 296.15)/97180 = 1.5835 x 10^-4 m^3 = 158.35mL

but the answer given is 0.583L any mistakes did i make?