A solution of AgNO3 contains 29.66 g of solute in 100.0 mL of solution. What is the molarity of the solution?
Molecular weight os AgNO3 = 169.87
You have 29.66 g, therefore you have 29.66 / 169.87 moles = 0.174 moles. This is in 100 ml. Molarity is moles per liter, so, 0.174 x (1000/100) = 1.75. Concentration of the solution = 1.75 M. Worked out with chemCal for the iPhone/iPod Touch.
moles = grams/molar mass.
M = moles/L of solution.
29.66/100.0=.2966 molar mass
i don't know where to go from here.
The molar mass of AgNO3 is not 100. Add up atomic mass Ag, N, and three O atoms to get the molar mass (the periodic tab le will give you the atomic masss).
Then 29.66 g/molar mass = moles AgNO3.
Then moles AgNO3/liters of soln= M.
(They tell you the solution is 100 mL. That's 0.100 L.
To find the molarity of a solution, you need to know the moles of solute and the volume of solution in liters.
First, let's calculate the moles of AgNO3:
Moles = Mass / Molar mass
The molar mass of AgNO3 is the sum of the molar masses of its constituent elements: Ag (108 g/mol), N (14 g/mol), and O (16 g/mol):
Molar mass of AgNO3 = 108 + 14 + (16 × 3) = 169 g/mol
Now, we can calculate the moles of AgNO3:
Moles = 29.66 g / 169 g/mol = 0.175 moles
Next, we need to convert the volume of the solution from milliliters to liters:
Volume (in liters) = 100.0 mL / 1000 mL/L = 0.100 L
Finally, we can calculate the molarity (M) of the solution:
Molarity = Moles / Volume
Molarity = 0.175 moles / 0.100 L = 1.75 M
Therefore, the molarity of the AgNO3 solution is 1.75 M.