Question

A gaseous mixture contains co2 and n2o in a2:5 ration by mass .the ratio of the no of molecules. of co2 and n2o

Answer 1

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Answer 2

Answer: The ratio of number of molecules of $CO_2$ and $N_2O$ is 2 : 5

Explanation:

It is given that the mass ratio of $CP_2$ and $N_2O$ is 2 : 5

Mathematically,

$\frac{W_{CO_2}}{W_{N_2O}}=\frac{2}{5}$

where,

$W_{CO_2}}$ = mass of carbon dioxide gas.

$W_{N_2O}}$ = mass of dinitrogen oxide gas.

To calculate the number of molecules, we use the equation:

$\text{Number of molecules}=\text{Number of moles}\times \text{Avogadro's Number}$

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

So,

$n_{CO_2}=\frac{W_{CO_2}}{M_{CO_2}}\times N_A$

where, $n_{CO_2}$ = number of molecules of carbon dioxide.

And, $n_{N_2O}=\frac{W_{N_2O}}{M_{N_2O}}\times N_A$

$M_{CO_2}=44g/mol\\M_{N_2O}=44g/mol$

Taking their ratio, we get:

$\frac{n_{CO_2}}{n_{N_2O}}=\left(\frac{\frac{W_{CO_2}}{M_{CO_2}}\times N_A}{\frac{W_{N_2O}}{M_{N_2O}}\times N_A}\right)\\\\\frac{n_{CO_2}}{n_{N_2O}}=\frac{W_{CO_2}}{W_{N_2O}}\times \frac{M_{CO_2}}{M_{N_2O}}\\\\\frac{n_{CO_2}}{n_{N_2O}}=\frac{2}{5}\times \frac{44}{44}\\\\\frac{n_{CO_2}}{n_{N_2O}}=\frac{2}{5}$

Hence, the ratio of number of molecules of $CO_2$ and $N_2O$ is 2 : 5