Question

1. 88.45 kL of nitrogen dioxide gas is collected in a vessel under a pressure is 0.746 atm. On the next day, the pressure in the vessel was changed to 1.24 atm. What will the volume of the NO2 gas be on the second day if the gas was kept at 56.4 °F?

2. What is the volume of 62.3 g of iodine gas (I2) at STP?

3. At STP, 2.561 μL of a gas weights 0.689 cg. Calculate the molar mass of the gas?

4. What is the mass of 7.654 x 10-11 GL of acetylene gas (C2H2) if it is held at 2356.9 mmHg and 89.51 °C?

Answer 1

1] given terms

pressure p1=0.746atm=75588.45pascal

pressure p2=1.24atm =125643pascal

volume v1=88.45Kl

volume v2=?

from ideal gas law

p1/p2=v1/v2

i.e v2=(p1/p2)*v1

=(75588.45/125643)*88.45

=53.212Kl

2] given terms

weight of iodine=62.3g

molucular weight of iodine=126.90447 u

volume=?

PV=nRT

V=nRT/P

wher n=m/M

=62.3/126.90447

=0.4909

therfore

V=(0.4909*0.0821*273.15)/1atm

V=11L

3] given terms

volume=2.561 micro liters=2.561e^-6 L

weight=0.689cg=0.00689grams

pv=nrt

molar mass/molecular weight=0.00689*0.0821*273.15/2.561e^-6*1

=24.34g/mol

4] given terms

volume=7.654*10-11GL=28973.54L

molecular weight of C2H2=26.04g/mol

pressure=2350.9mmHg=3.093atm

temperature=89.51oc+273K=362.51K

therfore pv=nRT

mass=(3.093*289.735*26.04)/(0.0821*362.51)

=784.07699grams