First, we have to get the molar mass of aniline. Just get a periodic table and add the individual masses of the elements in the chemical formula:
C6H5NH2: 6*12 + 5*1 + 1*14 + 2*1 = 93 g/mol
Then we get the mass of aniline using the given density and volume. Recall that density is just
d = m/V, or
m = V*d
We also convert the liters to milliliters. Thus,
m = (1.00 L)(1000 mL / 1 L) * (1.02 g/mL)
m = 1020 g
Then we divide this mass to the molar mass of aniline, to get the number of moles:
1020 / 93 = 10.968
Finally, recall that 1 mole of any substance is equivalent to 6.022*10^23 representative particles (can be atoms, molecules, ions, etc.) Therefore,
10.968 * 6.022*10^23 = 6.60 x 10^24 molecules
hope this helps~ :3
the density aniline c6h5nh2 is 1.02g/mL how many molecules are present in 1.00Lsample
1 answer