Answer:

(a) Mass of H2S in 258 mL at STP (0 oC, 1 atm) can be calculated using the ideal gas law: PV = nRT. At STP, 1 mole of any gas occupies 22.4 L, so 258 mL is equal to 0.258/1000 L or 0.000258 L. The temperature at STP is 273.15 K and the molar mass of H2S is 34.08 g/mol. Therefore, using the ideal gas law, the number of moles of H2S gas in 258 mL at STP is:

n = PV/RT = (1 atm)(0.000258 L)/(0.08206 L atm/mol K)(273.15 K)

n = 1.07 × 10^-5 mol

To calculate the mass of dissolved H2S in 100 g of water, we use the solubility data:

0.258 mL H2S /100 g H2O = 1.07 × 10^-5 mol H2S / x g H2O

Solving for x, we get:

x = 41.4 g

Therefore, the mass percent of dissolved H2S is:

(mass of H2S / mass of solution) × 100% = (1.07 × 10^-5 mol × 34.08 g/mol) / (100 g + 1.07 × 10^-5 mol × 34.08 g/mol) × 100%

= 0.034 %

(b) Molality is defined as the number of moles of solute per kilogram of solvent. In this case, the number of moles of H2S gas in 258 mL at STP is 1.07 × 10^-5 mol, and the mass of water corresponding to 258 mL is:

100 g H2O / 0.258 mL H2S = 38,759.69 g H2O/L

Therefore, the molality of the H2S solution is:

molality = 1.07 × 10^-5 mol / (38,759.69 g H2O/kg H2O) = 2.76 × 10^-7 mol/kg

(c) To calculate the solubility of H2S gas in water at 20 oC and 255 torr, we can use the modified Henry's law equation:

[H2S] = kH × PH2S

where [H2S] is the molar concentration of H2S in solution, kH is the Henry's law constant, and PH2S is the partial pressure of H2S gas above the solution.

We can assume that the solubility of H2S gas at 20 oC and 255 torr is proportional to the solubility at 20 oC and 1 atm. Therefore, we can use the following proportion:

0.258 mL H2S/100 g H2O at 0 oC, 1 atm = x mL H2S/100 g H2O at 20 oC, 255 torr

Solving for x, we get:

x = 0.258 mL H2S/100 g H2O × (255 torr / 760 torr) × (293.15 K / 273.15 K)

x = 0.099 mL H2S/100 g H2O

Therefore, the solubility of H2S gas at 20 oC and 255 torr is 0.099 mL H2S/100 g H2O.