Question

How many grams of AgNo3 are required to make 25 ml of .80M solution

Answer 1

Answer:

3.39 grams of AgNo3 are required to make 25 ml of .80M solution

Explanation:

Molarity , M = 0.80

Volume = 25 ml

$=\frac{25}{1000} L$

V = 0.025 L

$molarity = \frac{moles}{Volume\ of\ solution(L)}$

$M=\frac{n}{V}$

Here n = number of moles

$n=M\times V$

$n=0.8\times 0.025$

n = 0.02 mole

$moles = \frac{given\ mass}{Molar\ mass}$

Molar mass of AgNO3 = 169.87 g/mol

$mass =\frac{n}{Molar\ mass}$

$mass=\frac{0.02}{169.87}$

Mass = 3.397 g