This set up of a conversion table should show you that if you multiply the grams of BeI2 times .02 moles, it equals <span>5.256 g (your answer) </span>

8 0

3 years ago

The rate of effusion of an unknown gas was measured and found to be 11.9 mL/min. Under identical conditions, the rate of effusio

iren2701 [21]

Answer : The correct option is, (B) $CO_2$

Solution :

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.

$R\propto \sqrt{\frac{1}{M}}$

or,

$(\frac{R_1}{R_2})=\sqrt{\frac{M_2}{M_1}}$ ..........(1)

where,

$R_1$ = rate of effusion of unknown gas = $11.9\text{ mL }min^{-1}$

$R_2$ = rate of effusion of oxygen gas = $14.0\text{ mL }min^{-1}$

$M_1$ = molar mass of unknown gas = ?

$M_2$ = molar mass of oxygen gas = 32 g/mole

Now put all the given values in the above formula 1, we get:

$(\frac{11.9\text{ mL }min^{-1}}{14.0\text{ mL }min^{-1}})=\sqrt{\frac{32g/mole}{M_1}}$

$M_1=44.2g/mole$

The unknown gas could be carbon dioxide $(CO_2)$ that has approximately 44 g/mole of molar mass.

Thus, the unknown gas could be carbon dioxide $(CO_2)$

5 0

4 years ago