Answer:
The molar mass of the unknown is approximate 145g/mol
Explanation:
We can relate the speed of effusion of gases with the molar mass of the 2 compounds using Graham´s law:
[tex]\frac{v_a}{v_b} =\sqrt{ \frac{m_b}{m_a} }[/tex]
Where V is speed of the gas and m molar mass of a and b gases.
Molar mass of helium is 4g/mol. As the distance of the gases is the same, we can replace the speed of effusion with the time the gas takes to travel the distance. The equation will be:
[tex]\frac{t_b}{t_a} =\sqrt{ \frac{m_b}{m_a} }[/tex]
Replacing:
[tex]\frac{10.0min}{1.66} =\sqrt{ \frac{m_b}{4g/mol} }\\6.024 = \sqrt{ \frac{m_b}{4g/mol} }\\36.29 = \frac{m_b}{4g/mol}\\145 g/mol =m_b[/tex]
