Calculate the number of molecules of HF which are contained in gaseous form due to hydrogen bonding if the density of HF in gaseous phase at 1 atm , 300 K is 3.248 g / L.

Question

Calculate the number of molecules of HF which are contained in gaseous form due to hydrogen bonding if the density of HF in gaseous phase at 1 atm , 300 K is 3.248 g / L. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

It vapour state of HF be (HF)n P × mol. wt. =d R T 1 × Mol. wt =3.248 × 0.0821 × 300 n(20)= mol. w t=79.9 ≈ 80 n=4 hence vapour state is represented as ( HF )4.

Q. Calculate the number of molecules of $H F$ which are contained in gaseous form due to hydrogen bonding if the density of $H F$ in gaseous phase at $1 a t m, 300 K$ is $3.248 g / L$ .

It vapour state of $H F$ be $(H F)_{n}$
$P \times$ mol. wt. $= d RT$
$1 \times$ Mol. wt $= 3.248 \times 0.0821 \times 300$
$n (20) =$ mol. $wt = 79.9 \approx 80$
$n = 4$
hence vapour state is represented as $(H F)_{4}$ .

Q. Calculate the number of molecules of HF which are contained in gaseous form due to hydrogen bonding if the density of HF in gaseous phase at 1atm,300K is 3.248g/L.

It vapour state of HF be (HF)n​ P× mol. wt. =dRT 1× Mol. wt =3.248×0.0821×300 n(20)= mol. wt=79.9≈80 n=4 hence vapour state is represented as (HF)4​.

Q. Calculate the number of molecules of $H F$ which are contained in gaseous form due to hydrogen bonding if the density of $H F$ in gaseous phase at $1 a t m, 300 K$ is $3.248 g / L$ .

It vapour state of $H F$ be $(H F)_{n}$
$P \times$ mol. wt. $= d RT$
$1 \times$ Mol. wt $= 3.248 \times 0.0821 \times 300$
$n (20) =$ mol. $wt = 79.9 \approx 80$
$n = 4$
hence vapour state is represented as $(H F)_{4}$ .