Question

Question #16856

Answer 1

3.61

Explanation:

its quite simple if you go by a simple method of finding density at STP
we got
`"Density"=("Mass")/("Volume")`
we know that mas of `HBr` is 81 units
and one mole of `HBr` weighs 81 `gm`
and we also know that the volume of one mole of any gas is 22.4 `L`hence , we have got our data .
upon substitution with the formal equation of density we get .
`d=81/22.4=3.61g m^(-3)`

Answer 2

The density of `"HBr"` at STP is 3.563 g/L.

Explanation:

Yes, there is a gas law equation for density.

It is a variation of the Ideal Gas Law.

`color(blue)(|bar(ul(color(white)(a/a)pV = nRTcolor(white)(a/a)|)))" "`

Since

`n = "mass"/"molar mass" = m/M`,

we can write

`pV = m/MRT`

or

`p = (m/V)(RT)/M`

And density `ρ = m/V`,

so

`p = ρ(RT)/M`

This gives

`ρ = (pM)/(RT)`

STP is defined as 1 bar and 0 °C.

In this problem,

`p color(white)(ll)= "1 bar"`
`M = "80.91 g·mol"^"-1"`
`Rcolor(white)(l) = "0.083 14 bar·L·K"^"-1""mol"^"-1"`
`T color(white)(l)= "273.15 K"`

∴ `ρ = (1 color(red)(cancel(color(black)("bar"))) × 80.91 "g"·color(red)(cancel(color(black)("mol"^"-1"))))/("0.083 14" color(red)(cancel(color(black)("bar")))·"L"·color(red)(cancel(color(black)("K"^"-1""mol"^"-1"))) × 273.15 color(red)(cancel(color(black)("K")))) = "3.563 g/L"`