If 10-4 dm3 of water is introduced into a 1.0 dm3 flask to 300 K, how many moles of water are in the vapour phase when equilibrium is established ? (Given: Vapour pressure of H2O at 300 K is 3170 Pa ; R = 8.314 J K-1 mol-1)

Question

If 10-4 dm3 of water is introduced into a 1.0 dm3 flask to 300 K, how many moles of water are in the vapour phase when equilibrium is established ? (Given: Vapour pressure of H2O at 300 K is 3170 Pa ; R = 8.314 J K-1 mol-1) (A) 5.56 × 10-3 mol (B) 1.53 × 10-2 mol (C) 4.46 × 10-2 mol (D) 1.27 × 10-3 mol

Tardigrade Admin · Accepted Answer

n = (PV/RT) = 128 × 10-5 moles = (3170 × 10-5 atm × 1L/0.0821L atm k-1 mol-1× 300K) ≈ 1.27 × 10-3 mol

Q. If $1 0^{- 4} d m^{3}$ of water is introduced into a $1.0 d m^{3}$ flask to $300 K$ , how many moles of water are in the vapour phase when equilibrium is established ?
(Given : Vapour pressure of $H_{2} O$ at $300 K$ is $3170 P a; R = 8.314 J K^{- 1} m o l^{- 1}$ )

$5.56 \times 1 0^{- 3} m o l$

$1.53 \times 1 0^{- 2} m o l$

$4.46 \times 1 0^{- 2} m o l$

$1.27 \times 1 0^{- 3} m o l$

$n = \frac{P V}{RT}$
$= 128 \times 1 0^{- 5}$ moles
$= \frac{3170 \times 1 0 ^{- 5} a t m \times 1 L}{0.0821 L a t m k ^{- 1} m o l ^{- 1} \times 300 K} \approx 1.27 \times 1 0^{- 3} m o l$

Q. If 10−4dm3 of water is introduced into a 1.0dm3 flask to 300K, how many moles of water are in the vapour phase when equilibrium is established ? (Given : Vapour pressure of H2​O at 300K is 3170Pa;R=8.314JK−1mol−1)

5.56×10−3mol

1.53×10−2mol

4.46×10−2mol

1.27×10−3mol