# How many moles of gas are contained in 890.0 ml at 21.0 C and 125.3 mm Hg pressure?

Using the ideal gas law:

PV = nRT

or, n = PV/RT

where, P is the pressure, V is the volume, n is the number of moles, R is the universal gas constant and T is the temperature.

Here, P = 125.3 mm Hg = 125.3 mm Hg/ 760 mm...

Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Using the ideal gas law:

PV = nRT

or, n = PV/RT

where, P is the pressure, V is the volume, n is the number of moles, R is the universal gas constant and T is the temperature.

Here, P = 125.3 mm Hg = 125.3 mm Hg/ 760 mm Hg/atm = 0.165 atm

V = 890 ml = 0.89 L

R = 0.0821 L atm/mol/K

T = 21 degree Celsius = (21 + 273) K = 294 K

Using the ideal gas law:

n = PV/RT = 0.165 atm x 0.89 L / (0.0821 L atm/mol/K x 294 K)

= 0.0061 moles.

Thus, the gas at given conditions contain 0.0061 moles.

While doing such calculations, always remember to use the right set of units. For example, in this case, R determines the units of other parameters. Since it has units of L atm/mol/K, volume has to be converted to liters, temperature to Kelvin scale and pressure to atm (atmosphere) from mm Hg.

Hope this helps.

Approved by eNotes Editorial Team