Question

Question #50761

Answer 1

`30.072 L`

Explanation:

We know that,

The atomic mass of `He` is `4 g=1 mol`

At `STP,`

the molar volume of every gasses of `1mol` is `22.4L`

Then,at `STP`

The volume of `4 g He` is`22.4L`

`:.`The volume of `5.37 gHe` is
`=(5.37 gxx22,4L)/(4 g)=30.072 L`

Answer 2

See the simple steps to take in solving the question above;

Explanation:

We are dealing on two given information..

Volume and Mass..

Using this we can get the required Volume..

Recall;

`n = v/(mV) = m/(mM)`

Where;

`n = "no of moles"`

`v = "volume"`

`m = "mass"`

`mV = "molar volume"`

`mM = "molar mass"`

So we have;

`v/(mV) = m/(mM)`

`v = ?`

`mV = 22.4dm^3mol^-1or Lmol^-1`

N/B `rArr dm^3 = L`

`m = 5.37g`

`mM = 4gmol^-1`

Substitute the values into the fomula

`v/(22.4dm^3mol^-1) = (5.37g)/(4gmol^-1)`

`v/(22.4dm^3cancel(mol^-1)) = (5.37cancelg)/(4cancelgcancel(mol^-1))`

`v/(22.4dm^3) = (5.37)/(4)`

Cross multiplying..

`v xx 4 = 22.4dm^3 xx 5.37`

`4v = 22.4dm^3 xx 5.37`

`4v = 120.288dm^3`

Divide both sides by `4`

`(cancel4v)/cancel4 = (120.288dm^3)/4`

`v = (120.288dm^3)/4`

`v = 30.072dm^3 or 30.072L`