The solubility of a gas in water at 300 K under a pressure of 100 atmospheres is 4 × 10-3 kg / L. Therefore, the mass of the gas in kilogram dissolved in 250 mL of water under a pressure of 250 atm and 300 K is:

Question

The solubility of a gas in water at 300 K under a pressure of 100 atmospheres is 4 × 10-3 kg / L. Therefore, the mass of the gas in kilogram dissolved in 250 mL of water under a pressure of 250 atm and 300 K is: (A) 2.0 × 10-3 (B) 2.5 × 10-3 (C) 1.25 × 10-3 (D) 5.0 × 10-3

Tardigrade Admin · Accepted Answer

(S1/S2)=(P1/P2) (4 × 10-3/S2)=(100/250) ∴ S2=(4 × 10-3 × 250/100)=10-2 kg L -1 Amount of gas dissolved in 250 water =(10-2/1000) × 250=2.5 × 10-3 kg

Q. The solubility of a gas in water at $300 K$ under a pressure of $100$ atmospheres is $4 \times 1 0^{- 3} k g / L$ . Therefore, the mass of the gas in kilogram dissolved in $250 m L$ of water under a pressure of $250 a t m$ and $300 K$ is:

$2.0 \times 1 0^{- 3}$

$2.5 \times 1 0^{- 3}$

$1.25 \times 1 0^{- 3}$

$5.0 \times 1 0^{- 3}$

$\frac{S _{1}}{S _{2}} = \frac{P _{1}}{P _{2}}$
$\frac{4 \times 1 0 ^{- 3}}{S _{2}} = \frac{100}{250}$
$∴ S_{2} = \frac{4 \times 1 0 ^{- 3} \times 250}{100} = 1 0^{- 2} k g L^{- 1}$
Amount of gas dissolved in $250$ water
$= \frac{1 0 ^{- 2}}{1000} \times 250 = 2.5 \times 1 0^{- 3} k g$

Q. The solubility of a gas in water at 300K under a pressure of 100 atmospheres is 4×10−3kg/L. Therefore, the mass of the gas in kilogram dissolved in 250mL of water under a pressure of 250atm and 300K is:

2.0×10−3

2.5×10−3

1.25×10−3

5.0×10−3