Question

a 374 ml container holds 0.146g of Ne and an unknown amount of argon at 35°C total pressure =638mmHg calculate moles of Ar present

Answer 1

Answer – We are given, mass of Ne = 0.146 g , volume = 374 mL = 0.374 L, total pressure = 638 mm Hg , T = 35 +273.15 = 308.15 K

We need to calculate the total number of moles using the formula Ideal gas law

PV = nRT

We know, 760 mm Hg = 1 atm

So 638 mm Hg = ?

= 0.839 atm

So, n = RT/PV

          = 0.839 atm * 0.374 L / 0.0821 L.atm.mol^-1.K^-1 * 308.15 K

          = 0.0124 moles

Now moles of Ne = mass of Ne / moral mass of Ne

                              = 0.146 g / 20.18 g.mol^-1

                             = 0.00723 moles

Total moles = moles of Ne + moles of Ar

So, moles of Ar = total moles – moles of Ne

                          = 0.0124 moles – 0.00723 moles

                          = 0.00518 moles

So, 0.00518 moles of Ar are present.