The solubility of a gas in water at 300 K under a pressure of 100 atmospheres is 4 × 10 -3 kgL Therefore, the mass of the gas in kg dissolved in 250 mL of water under a pressure of 250 atmospheres at 300 K is

Question

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

Tardigrade Admin · Accepted Answer

As 4 × 10-3 kg of gas is dissolved in 1 L of water, According to Henry' s law, m = K H ρ Where, m =4 × 10-3 kg and ρ=100 atm ∴ K H =( m /ρ)=(4 × 10-3/100) =4 × 10-5 kg L -1 atm -1 Now, if the pressure is increased to 250 atm the mass of gas (m)= K H P =4 × 10-5 × 250 =10-2 kg L -1 and the mass of gas present in 250 ml =(250/100) × 10-2 =2.5 × 10-3 kg

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

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

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

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

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

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

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

2.5×10−3

2.0×10−3

1.25×10−3

5.0×10−3