Question

The temperature at which the rate of effusion of o, is 1.6 times of so, at 127°C is (mx128)K. The value of mis​

Answer 1

Answer:

Explanation:

The rate of effusion is directly proportional to the square root of temperature and inversely proportional to the square root of molecular weight.

r ∝ U ;U= $\sqrt{} \frac{3TM}{M}$

​ r ∝ =  $\sqrt{} \frac{T}{M}$

∴ ​$\frac{r N 2}{rSO2} = \sqrt{} \frac{xTN2MSO2}{TSO2MN2}$

 T2 =373K

At 373K, the rate of effusion of N  2

​

 would be 1.625 times than that of SO  2  at 50  o C.

​