If 11.50 mL of an unknown gas effuses through a small hole in a heated vessel in the same time it takes 9.63 mL of argon to effuse through the hole under the same conditions, what is the molecular weight of the unknown gas. (Use 3 sig figs and amu for the units of the answer.)

Answer : The molecular weight of the unknown gas is, 57.0 amu

Explanation :

According to the Graham's law, the rate of effusion of gas is inversely proportional to the square root of the molar mass of gas.

And the relation between the rate of effusion and volume is :

From this we conclude that,

where,

= rate of effusion of unknown gas  = 11.50 mL

= rate of effusion of argon gas = 9.63 mL

= molar mass of unknown gas

= molar mass of argon gas = 40 amu

Now put all the given values in the above formula, we get:

Therefore, the molecular weight of the unknown gas is, 57.0 amu

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"