Question

ch4 diffuses two times faster than a gas X.The number of molecules present in 32 grams of gas X(in avagadro number)​

Answer 1

Answer:n÷2

Explanation:see the pic

Answer 2

Given that,

$CH_{4}$ diffuses two times faster than a gas X.

The number of molecules present in 32 grams of gas X

We need to calculate the molar mass of gas

Using graham's law

$\dfrac{r_{CH_{4}}}{r_{gas}}=\sqrt{\dfrac{m_{gas}}{m_{CH_{4}}}}$

Put the value into the formula

$2=\sqrt{\dfrac{m_{gas}}{16}}$

$4=\dfrac{m_{gas}}{16}$

$m_{gas}=64$

We need to calculate the number of molecules

Using formula of number of molecules

$\text{number of molecules}=\dfrac{w}{molar\ mass\ of\ gas}\times N$

Where, N = avogadro number

Put the value into the formula

$\text{number of molecules}=\dfrac{32}{64}\times N$

$\text{number of molecules}=\dfrac{N}{2}$

Hence, The number of molecules is $\dfrac{N}{2}$