Question

1.) Mass (in grams) of 1.50×10−2 mol of CdS.

2.) Number of moles (n) of NH4Cl in 86.6 g of this substance.

3.) Number of molecules in 8.487×10⁻² mol  C6H6.

4.) Number of O atoms in 7.25×10⁻³ mol Al(NO3)3.

Answer 1

1)

Molar mass of CdS,

MM = 1*MM(Cd) + 1*MM(S)

= 1*112.4 + 1*32.07

= 144.47 g/mol

use:

mass of CdS,

m = number of mol * molar mass

= 1.5*10^-2 mol * 1.445*10^2 g/mol

= 2.167 g

Answer: 2.17 g

2)

Molar mass of NH4Cl,

MM = 1*MM(N) + 4*MM(H) + 1*MM(Cl)

= 1*14.01 + 4*1.008 + 1*35.45

= 53.492 g/mol

mass(NH4Cl)= 86.6 g

use:

number of mol of NH4Cl,

n = mass of NH4Cl/molar mass of NH4Cl

=(86.6 g)/(53.49 g/mol)

= 1.619 mol

Answer: 1.62 mol

3)

use:

number of molecules = number of mol * Avogadro’s number

number of molecules = 8.487*10^-2 * 6.022*10^23 molecules

number of molecules = 5.111*10^22 molecules

Answer: 5.111*10^22 molecules

4)

1 mol of Al(NO3)3 has 9 moles of O

So,

moles of O = 9*number of moles of Al(NO3)3

= 9*7.25*10^-3 mol

= 0.06525 mol

use:

number of atoms = number of moles * Avogadro’s number

= 6.525*10^-2 * 6.022*10^23 atoms

= 3.929*10^22 atoms

Answer: 3.93*10^22 atoms