Question

True/False

If 10.3 g of Ne and 10.3 g of N2 are put into a 7.0 L container, the partial pressure of N2 will be less than the partial pressure of Ne in the container.

Answer 1

True.

Step-by-step explanation:

Moles =mass/molarmass

Moles Ne =mass/molarmass

`=\frac {10.3g}{(20.18g per mole)}=0.510mol`

Moles
`N_2` =mass/molarmass

`=\frac {10.3g}{(28g per mole)}=0.368mol`

Let Temperature be 273K (assumed)

`PV=nRT`

`P=\frac {nRT}{V}`

`P_(Ne)=\frac {0.510mol * 0.08206L atm K^(-1) mol^(-1) * 273K}{7.0L}`

`P_(Ne)=1.63atm`

`P_(N_2)=\frac {0.368mol * 0.08206L atm K^(-1) mol^(-1) * 273K}{7.0L}`

`P_(N_2)=1.17atm`

We see here partial pressure of
`N_2` (1.17atm) is less than that of partial pressure of Ne (1.63 atm).