The liquid CHBr3 has a density of 2.89 g dm-³. What volume of this liquid should be measured to contain a total of 4.8×10²⁴ molecules of CHBr3 ?

Hello!

The molar mass of `C H Br_3` (bromoform) is about

`12 + 1 + 3*80 = 253 (g/(mol)).`

Therefore one mole of this substance has a mass of about `253 g.`

One mole of any substance contains `N_A approx 6*10^(23)` molecules (this constant is called Avogadro's constant). Hence the given number of molecules represents  `(4.8*10^(24))/(6*10^(23)) = 8` (moles), and from the above paragraph their mass is about `8*253 = 2024 (g).`

Finally, volume may be computed as mass divided...

Hello!

The molar mass of `C H Br_3` (bromoform) is about

`12 + 1 + 3*80 = 253 (g/(mol)).`

Therefore one mole of this substance has a mass of about `253 g.`

One mole of any substance contains `N_A approx 6*10^(23)` molecules (this constant is called Avogadro's constant). Hence the given number of molecules represents  `(4.8*10^(24))/(6*10^(23)) = 8` (moles), and from the above paragraph their mass is about `8*253 = 2024 (g).`

Finally, volume may be computed as mass divided by the density, because `rho = m/V.` In this case it is

`V = (2024 g)/(2.89 g/((cm)^3)) approx 700 (cm)^3.`

This is the same as 0.7 `dm^3.`

Note that the density of `C H Br_3` is incorrectly stated in the question as `2.89 g/((dm)^3).` Actually it is `2.89 g/((cm)^3).` This liquid is much more dense than water (about `1 g/(cm^3)` ).