How many liters of HCl gas, measured at 30.0 °C and 745 torr, are required to prepare 1.25 L of a 3.20-M solution of hydrochloric acid?

102 L

Explanation:

Data of the solution

Concentration: 3.20 M Volume: 1.25 L

The moles of HCl in the solution are:

1.25 L * 3.20 mol/L = 4.00 mol

The gas must contain 4.00 moles of HCl

Data of the gas

Temperature (T) : 30.0 °C + 273.15 = 303.2 K Pressure (P) : 0.980 atm

745 torr * (1 atm/760 torr) = 0.980 atm

Moles (n) : 4.00 mol

We can the volume (V) of HCl gas using the ideal gas equation.

P * V = n * R * T

V = n * R * T/P

V = 4.00 mol * (0.08206 atm. L/mol. K) * 303.2 K / 0.980 atm

V = 102 L