Question

# A) A mixture of He, Ar, and Xe has a total pressure of 2.80 atm ....

A) A mixture of He, Ar, and Xe has a total pressure of 2.80 atm . The partial pressure of He is 0.400 atm , and the partial pressure of Ar is 0.300 atm . What is the partial pressure of Xe?

B) A volume of 18.0 L contains a mixture of 0.250 mole N2 , 0.250 mole O2 , and an unknown quantity of He. The temperature of the mixture is 0 ∘C , and the total pressure is 1.00 atm . How many grams of helium are present in the gas mixture?

C)Imagine that you have an ideal gas in a 9.00 L container, and that 2150 molecules of this gas collide with a square-inch area of the container at any given instant.

If the volume is increased to 36.0 L at constant temperature, how many collisions will occur per square inch of this larger container?

D)What volume of O2 at 798 mmHg and 37 ∘C is required to synthesize 14.5 mol of NO?

A) Clearly total pressure of the mixture=partial pressure of the each gas.

now partial pressure of He=0.400 atm

partial pressure of Ar=0.300 atm

so the partial pressure of Xe= total pressure - (partial pressure of He+ partial pressure of Ar)

=2.80 atm - (0.400+0.300)atm

=2.10 atm

B)We know, in N.T.P molar volume of a gas=22.4 Lmol-1

For 0.250 mole of N2 the volume the gas= 5.6 L

similarly for 0.250 mole of O2 the volume the gas= 5.6 L

So, the volume of He= total volume -(volume of N2+ volume of O2)

= 18.0 L - (5.6+5.6) L = 6.8 L

In S.T.P 22.4 L of a gas= 1 mole

6.8 L of a gas=0.3035 mole

now 1 mole of He= 4 gm

so, 0.3035 mole of He=(4*0.3035) gm

=1.214 gm

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