Answer:
3.39 grams of AgNo3 are required to make 25 ml of .80M solution
Explanation:
Molarity , M = 0.80
Volume = 25 ml
[tex]=\frac{25}{1000} L[/tex]
V = 0.025 L
[tex]molarity = \frac{moles}{Volume\ of\ solution(L)} [/tex]
[tex]M=\frac{n}{V}[/tex]
Here n = number of moles
[tex]n=M\times V[/tex]
[tex]n=0.8\times 0.025[/tex]
n = 0.02 mole
[tex]moles = \frac{given\ mass}{Molar\ mass} [/tex]
Molar mass of AgNO3 = 169.87 g/mol
[tex]mass =\frac{n}{Molar\ mass}[/tex]
[tex]mass=\frac{0.02}{169.87}[/tex]
Mass = 3.397 g
