Question

A 0.428-gram mixture of gases contained in a vessel at 1.75 atm is found to be 15.6 % N 2 , 46.0% N 2 O
and 38.4 % CO 2 by mass. What is the partial pressure of each gas in the mixture

Answer 1

From the equation, we see that 1 mole of glucose gives 6 moles of CO₂

Now, 1 gram of glucose is equal to (1/180.16)moles of glucose

Since 1 mole glucose produces 6 moles of CO₂

Hence, (1/180.16)moles of glucose will produces 6 x (1/180.16) moles of glucose which is 0.0333 moles of CO₂

Assuming CO2 to be an ideal gas,

The volume of one mole of CO2 at Standard temperatue and pressure (273.15K and 1atm) is 22.4L

Hence for 0.0333 moles of CO2, volume = 0.0333 x 22.4 = 0.74592 L

Thus, we have to calculate the volume of CO2 at 37⁰C which is (37+273.15) = 310.15K

Using ideal gas equation,

P1 = 1 atm and P2 is also 1atm

T1 = 273.15K and T2 = 310.15K

V1 = 0.74592 L and V2 = ?

Thus,

Or V2 = 0.847 L

Thus, volume of CO2 is 0.847 L

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