How much water would be needed to completely dissolve 1.52 L of the gas at a pressure of 730 torr and a temperature of 21 ∘C?

The correct answer is 0.4 L.

Explanation:

The mentioned question can be solved by using the equation,  

C = K × Pgas(i)

Here K is the Henry law constant whose value is 0.158 mol/L/atm, C is the concentration of the gas in liquid state, and Pgas is the partial pressure of the gas.  

Now to find the volume of water, the formula to be used is,  

PV = nRT(ii)

Here P is the pressure of the gas, V is the volume, R is the universal gas constant whose value is 0.082 Latm/mol/K and T is the temperature.  

PgasV = nRT

Pgas = nRT/Vgas

The value of Pgas is inserted in equation (i) we get,  

C = K × nRT/Vgas

It is to be noted that C = n/V, here n is the no. of the moles and V is the volume of liquid.  

n/Vliquid = K × nRT/Vgas

1/Vliquid = KRT/Vgas

Vliquid = Vgas/KRT(iii)

Based on the given information, the volume of the gas is 1.52 L, the value of K is 0.158 mol/L/atm, the value of R is 0.082 Latm/mol/K and value of T is 21 degree C or 273 + 21 = 294 K.  

Now putting the values in equation (iii) we get,  

Vliquid = 1.52 L / 0.158 × 0.082 × 294

Vliquid = 1.52 / 3.809

Vliquid = 0.399 or 0.4 L

Hence, the volume of water required to dissolve 1.52 L of gas is 0.4 L.  

0.88m hope i am correct : )