Answer:
see explaination
Explanation:
Case 1) When we consider vapor pressure of H2O;
Let the pressure is 1 atm OR 760 torr.
As H2 is collected over water, we have to consider the vapor pressure of H2O as well.
Using data i.e. vapor pressure of H2O at 40° C = 55.365 torr
So, pressure of H2 = P = 760 - 55.365 = 704.635 torr = 704.635/760 = 0.9272 atm
Volume of H2 = 80 ml = 0.08 liter
Temperature (T) = 40 + 273 = 313 K
Gas constant (R) = 0.0821 L atm/mol K
Let n is moles of H2. Applying ideal gas equation;
PV = nRT
n = PV/RT = 0.9272 * 0.08 / 0.0821 * 313
n = 0.00289 moles
Mass of H2 = moles * molar mass = 0.00289 * 2.016 = 0.00582 grams
OR 5.8*10^-3 grams ...Answer
----> Case 2) When we don't consider vapor pressure;
Pressure of H2 = 1 atm, all other parameters will remain same as in case 1.
So, mass of H2 = 6.3*10^-3 grams
But Case 1) is correct approach as in question it is mentioned that H2 is collected over water.
