16.99 g of silver nitrate are dissolved in water, then diluted to 500 ml what is the molarity of this solution

0.2M

Explanation:

First, we need to calculate the numbers of mol AgNO3 present in 16.99g. This is achieved as follows:

Molar Mass of AgNO3 = 108 + 14 + (16x3) = 108 + 14 + 48 = 170g/mol

Mass of AgNO3 = 16.99g

Number of mole = Mass / Molar Mass

Number of mole of AgNO3 = 16.99/170 = 0.1mol

Volume = 500mL = 500/1000 = 0.5L

Molarity = ?

Molarity = mole / Volume

Molarity = 0.1/0.5

Molarity = 0.2M