Henry’s law constant for CO2 in water is 1.67 × 108 Pa at 298 K. What is the quantity of CO2 in 500 mL of soda water when packed under 2.5 atm CO2 pressure at 298 K?

Question

Henry’s law constant for CO2 in water is 1.67 × 108 Pa at 298 K. What is the quantity of CO2 in 500 mL of soda water when packed under 2.5 atm CO2 pressure at 298 K? (A) 1.8 g (B) 2.5 g (C) 4 g (D) 5 g

Tardigrade Admin · Accepted Answer

Given, KH= 1.67 × 108 Pa PCO2 = 2.5 atm = 2.5 × 101325 Pa Applying Henryâs law, PCO2 = KH × xCO2 ∴ xCO2 = (PCO2/KH) = (2.5 × 101325 Pa/1.67 × 108 Pa) = 1.517 × 10-3 i, e. xCO2 = (nCO2/nH2O + nCO2) = (nCO2/nH2O) = 1.517 × 10-3 [nCO2 is negligible] For 500 mL of soda water, volume of water = 500 mL; mass of water = 500 g, moles of water (500/18) = 27.78 i.e., nCO2 = 27.78 ∴ (nCO2/27.78) = 1.517 × 10-3 or nCO2 = 42 .14 × 10-3 mole Mass of CO2 = 42.14 × 10-3 × 44 g = 1.854 g

Q. Henry’s law constant for $C O_{2}$ in water is $1.67 \times 1 0^{8} P a$ at $298 K$ . What is the quantity of $C O_{2}$ in $500 m L$ of soda water when packed under $2.5$ atm $C O_{2}$ pressure at $298 K$ ?

$1.8 g$

$2.5 g$

$4 g$

$5 g$

Q. Henry’s law constant for CO2​ in water is 1.67×108Pa at 298K. What is the quantity of CO2​ in 500mL of soda water when packed under 2.5 atm CO2​ pressure at 298K?

1.8g

2.5g

4g

5g

Q. Henry’s law constant for $C O_{2}$ in water is $1.67 \times 1 0^{8} P a$ at $298 K$ . What is the quantity of $C O_{2}$ in $500 m L$ of soda water when packed under $2.5$ atm $C O_{2}$ pressure at $298 K$ ?

$1.8 g$

$2.5 g$

$4 g$

$5 g$