Answer:
0.2M
Step-by-step 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