Calculate the volume produce of hydrogen in water displaceme

[WeiLoong]

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?

 

[Borek]

Your answer looks OK to me*. My bet is there is a typo in the answer given - note how 583 digits appear both in both the correct and the given answer.

*Actually it doesn't - if all the data is given with no less than 3 significant digits, molar mass of hydrogen should be accurate as well, not just 2 g/mol, but 2.02 g/mol (or even 2.016 g/mol to avoid rounding errors). That means the volume of 157 mL.