Question

At 1.00 atm pressure and 273 K one mole of gas should occupy 22.4 L. What volume would one mole of gas occupy at 21 ^oC and 1.00 atm?

In the Molar Mass of Magnesium experiment:

a) At the end of the reaction, the temperature was 21.5 ^oC (294.65 K), 99.2 mL of gas were formed, and the pressure was 1.00 atm in the gas syringe. The Gas Constant has a value of 0.08206 L atm/ (mol K) (Pay attention to these units!!!).

How many moles of Mg reacted?

b) What Molar Mass did you calculate for Mg?

Answer 1

1) Given ,

Pressure, P1 = 1.0 atm

Temperature, T1 = 273 K

Volume , V1 = 22.4 L

no. of moles, n1 = 1

question asked, volume = ??

Pressure, P2 = 1.0 atm

Temperature, T2 = 21 ⁰C = ( 21 + 273 ) Kelvin = 294 K

Volume , V2 = ?

no. of moles, n2 = 1

applying ideal gas law in both the cases , (assuming the gas is behaving ideally )

PV = nRT

where, R - > universal gas constant

so here, both P1 = P2

and n1 = n2 = 1

also R is constant in both the cases

so we can write ,

as P1 = P2 = 1 atm

2 ) 2nd question

a) Given ,

Temerature , T = 21.5 ⁰C = (21.5 + 273.15) Kelvin = 294.65 K

Volume , V = 99.2 mL = 99.2 * 10^-3 L

Pressure , P = 1 atm

Gas constant , R = 0.08206 L* atm / mol * K

Applying the ideal gas law, ( assuming the gas is behaving ideally )

n = no. of moles reacted

the no. of moles of Mg reacted = 4.10 * 10 ^-3 = 0.00410 moles

b ) we know that,

no. of moles, n = mass reacted (m) / molecular mass (M) -------(i)

molecular mass or molar mass is always constant for a perticular element

molecular mass or molar mass of Mg = 24 g , { molar mass means mass of 1 mole of element or compound , so its constant }

=> M = 24g

and from part a ,

no. of moles , n = 0.00410

from equation (i)

so the mass of the magnesium reacted = 0.0984 g

and the molecular mass or molar of Magnesium = 24 g