Question

A gaseous mixture contains 4.0g of H2

and 56.0g of N2. The mole fraction of H2 in the mixture is​

Answer 1

$\bigstar$ Answer $\bigstar$

$\checkmark$ Given:

⇒ A gaseous mixture contains 4.0 g of H₂ and 56.0 g of N₂

$\checkmark$ To Find:

⇒ The mole fraction of H₂ in the mixture

$\checkmark$ Solution:

We know that,

Mole Fraction (χ)

Mole fraction is defined as the number of moles of a component divided by the total number of moles of a mixture

$\purple{\boxed{\boxed{\chi_{A}=\dfrac{n_{A}}{n_{A}+n_{B}} }}}$

Mole (n)

$\green{\boxed{\boxed{\sf Mole(n)=\dfrac{Mass \ (g)}{Molar \ Mass \ (g \ mol^{-1})} }}}$

Given,

A gaseous mixture contains 4.0 g of H₂ and 56.0 g of N₂

• Given Mass of H₂ = 4 g
• Given Mass of N₂ = 56 g
• Molar Mass H₂ = 2 g
• Molar Mass N₂ = 28 g

According to the Question,

We are asked to find the

The mole fraction of H₂ in the mixture

Therefore,

$\orange{\longrightarrow \rm \chi_{H_{2}} = \dfrac{n_{H_{2}}}{n_{H_{2}}+n_{N_{2}}}}$

$\pink{\longrightarrow \rm \chi_{H_{2}}=\dfrac{\dfrac{4}{2} }{\dfrac{4}{2}+\dfrac{56}{28} }}$

$\green{\longrightarrow \rm \chi_{H_{2}}=\dfrac{2}{2+2}}$

$\blue{\rm \longrightarrow \chi_{H_{2}}=\dfrac{1}{2} }$

$\red{\rm \longrightarrow \chi_{H_{2}}=0.5}$

Hence,

The mole fraction of H₂ in the mixture is 0.5

$\pink{\large{\bold{\boxed{\boxed{\chi_{H_{2}}=0.5}}}}}$