化學的摩爾數

Relative formula mass of N = 14.0, O = 16.0 and Al = 27.0

Molar mass of Al(NO3)3

= 27.0 + (14.0 + 16.0 X 3) X 3

= 213.0 g/mol

In 60 g of Al(NO3)3,

no. of mole of Al(NO3)3

= mass / molar mass

= 60 / 213

= 20 / 71 mol



a. The no. of mole of the cations, Al3+

= 20 / 71 mol

= 0.282 mol


b. The no. of mole of the anions, NO3-

= 20 / 71 X 3

= 60 / 71 mol

= 0.845 mol


For the following calculations, they are based on the Avodagro's number. i.e., there are 6.02 X 10²³ particles in one mole of substances.


c. no. of cations

= no. of mole X Avodagro's no.

= 20 / 71 X 6.02 X 10²³

= 1.70 X 10²³


d. no. of anions

= no. of mole X Avodagro's no.

= 60 / 71 X 6.02 X 10²³

= 5.09 X 10²³