Question

find vapour density of mixture having 64% by mass of ch4 32% by mass of O2 and remaining H2

Answer 1

Answer : The vapour density of mixture is, 7.142

Solution : Given,

The mass of total mixture = 100 g

Mass of $CH_4$ = 64 g

Mass of $O_2$ = 32 g

Mass of $H_2$ = [100 - (64 + 32)] = 4 g

Molar mass of $CH_4$ = 16 g/mole

Molar mass of $O_2$ = 32 g/mole

Molar mass of $H_2$ = 2 g/mole

First we have to calculate the moles of $CH_4$,$O_2$ and $H_2$

$\text{Moles of }CH_4=\frac{\text{Mass of }CH_4}{\text{Molar mass of }CH_4}=\frac{64g}{16g/mole}=4moles$

$\text{Moles of }O_2=\frac{\text{Mass of }O_2}{\text{Molar mass of }O_2}=\frac{32g}{32g/mole}=1moles$

$\text{Moles of }H_2=\frac{\text{Mass of }H_2}{\text{Molar mass of }H_2}=\frac{4g}{2g/mole}=2moles$

The total number of moles of mixture = 4 + 1 + 2 = 7 moles

Now we have to calculate the molecular mass of mixture.

$\text{Molecular mass of mixture}=\frac{\text{Mass of mixture}}{\text{Moles of mixture}}=\frac{100g}{7mole}=147.28g/mole$

Now we have to calculate the vapour density of mixture.

$\text{Vapour density of mixture}=\frac{\text{Molar mass of mixture}}{2}=\frac{14.28g/mole}{2}=7.142$

Therefore, the vapour density of mixture is, 7.142

Answer 2

to find avg. molar mass we use ;- M1n1 + M2n2

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n1+n2