the density aniline c6h5nh2 is 1.02g/mL how many molecules are present in 1.00Lsample
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
To determine the number of molecules present in a 1.00 L sample of aniline (C6H5NH2), we need to follow these steps:
1. Calculate the mass of aniline in the 1.00 L sample:
Density = Mass / Volume
Rearranging the equation:
Mass = Density * Volume
Mass = 1.02 g/mL * 1000 mL (since 1 L = 1000 mL)
Mass = 1020 g
2. Determine the molar mass of aniline (C6H5NH2):
C: atomic mass of carbon = 12.01 g/mol (6 carbon atoms)
H: atomic mass of hydrogen = 1.01 g/mol (5 hydrogen atoms)
N: atomic mass of nitrogen = 14.01 g/mol (1 nitrogen atom)
Total molar mass = (12.01 g/mol * 6) + (1.01 g/mol * 8) + 14.01 g/mol
Total molar mass = 93.13 g/mol
3. Calculate the number of moles of aniline:
Moles = Mass / Molar mass
Moles = 1020 g / 93.13 g/mol
Moles ≈ 10.96 mol
4. Convert the number of moles to molecules using Avogadro's number:
1 mole of a chemical compound = 6.022 x 10^23 molecules
Molecules = Moles * Avogadro's number
Molecules = 10.96 mol * (6.022 x 10^23 molecules/mol)
Molecules ≈ 6.60 x 10^24 molecules
Therefore, there are approximately 6.60 x 10^24 molecules of aniline (C6H5NH2) present in a 1.00 L sample.