Question

The solubility of a gas is 0.14 g/L at STP. Its solubility at 50.°C and the

same pressure can be:

Answer 1

0.12 g/L

Step-by-step explanation:

We first calculate its molar mass M = solubility at STP × molar volume = 0.14 g/L × 22.4 L/mol = 3.14 g/mol

Using the ideal gas law, PV = mRT/M where P = pressure at STP = pressure at 50°C = 1 atm, V = volume of gas, m = mass of gas, R = molar gas constant = 0.082 atm L/mol-K, T = temperature of gas = 50°C = 273 + 50 = 323 K and M =molar mass of gas = 3.14 g/mol

So, PM/RT = m/V = solubility of gas at 50°C

So, PM/RT = 1 atm × 3.14 g/mol/(0.082 atm L/mol-K × 323 K)

= 3.14 atm g/mol ÷ 26.486 atm L/mol

= 0.119 g/L

≅ 0.12 g/L