untuk membuat 500ml larutan timbal (ii) nitrat, pb(no3)2 0,1 m berapa gram timbal (ii) nitrat yang diperlukan? (ar pb = 27, n=14, o =16)​

Penjelasan:

dik : V pb(no3)2 = 500 mL = 0.5 L

M pb(no3)2 = 0.1 M

dit : m = ?

jb :

– cari Mr pb(no3)2 :

Mr = ( 1 * Pb) + ( 2 * N) + ( 6 * O)

= ( 1* 207.2 ) + ( 2 * 14 ) + ( 6 * 16)

= 207.2 + 28 + 96

= 331.2

– cari m pb(no3)2 :

M = n / V

M = m /mr/V

M = m / mr * V

0.1 M = m / 331.2 * 0.5 L

0.1 M = m / 165.6

m = 0.1 M * 165.6

m = 16.56

semoga membantu