Solveeit Logo

Question

Chemistry Question on Solutions

Henry's law constants for aqueous solution of CO,O2,CO2CO , O _{2}, CO _{2} and C2H2C _{2} H _{2} gases are respectively at 25C25^{\circ} C as 58×103,43×103,1.61×10358 \times 10^{3}, 43 \times 10^{3}, 1.61 \times 10^{3} and 1.34×1031.34 \times 10^{3}. The solubility of these gases decreases in the order

A

CO>O2>CO2>C2H2 CO>O_{2}>CO_{2}>C_{2}H_{2}

B

O2>CO2>CO>C2H2O_{2}>CO_{2}>CO>C_{2}H_{2}

C

C2H2>CO2>O2>COC_{2}H_{2}>CO_{2}>O_{2}>CO

D

O2>CO2>C2H2>COO_{2}>CO_{2}>C_{2}H_{2}>CO

Answer

C2H2>CO2>O2>COC_{2}H_{2}>CO_{2}>O_{2}>CO

Explanation

Solution

p=KHXp=K_{H} X; Higher the value of KHK_{H} at a given pressure, the lower is the solubility of the gas in the liquid.
Therefore, the solubility of the given gases decreases in the order
C2H2>CO2>O2>COC _{2} H _{2}> CO _{2}> O _{2}> CO