48 L of dry N, is passed through 36 g H20 at
27°C and this results in a loss of 1.20 g in
water. The vapour pressure of water is
(a) 1.03 atm
(b) 0.021 atm
(c) 0.034 atm
(d) 0.66 atm
Answers
answer : option (c) 0.034 atm
it is given that 48L of dry N, is passed through 36g H₂O at 27°C and this results in a loss 1.20 g in water.
we have to find the vapor pressure of water
The water vapor occupied by the volume of N₂ gas i.e., V = 48 L
loss of weight of water , w = 1.2 g
molecular weight , M = 18 g/mol
no of moles , n = w/M = 1.2/18 = 0.067
R = 0.082 L.atm/mol.K
T = 27°C = 27 + 273 = 300K
using formula, PV = nRT
⇒P × 48 = 0.067 × 0.082 × 300
⇒P = (0.067 × 0.082 × 300)/48
= 0.0343375 ≈ 0.034 atm
therefore, The vapor pressure of water is 0.034 atm.
The vapour pressure of water is 0.0341 atm.
Explanation:
The ideal gas equation is:
PV = nRT
Where,
P = Pressure = ?
V = Volume = 48 L
n = Number of molecules = (Weight lost by water)/(molecular mass of water) = w/m = 1.2/18
R = Gas constant = 0.082 L.atm/mol.K
T = Temperature = 27°C = 273 + 27 = 300 K
On substituting the values, we get,
P × 48 = 1.2/18 × 0.082 × 300
P = 1.2/18 × 0.082 × 300 × 1/48
∴ P = 0.0341 atm