5,10g of pure silver nitrate is dissolved in 250cm³ of water. What is the concentration of the solution?
Calculations please
[Ag+] = 0.000742moles/130.0 *10^-3L
The answer is 0.00571m
molar mass of AgNO3 ... ≈ 170g/mol
(5.10g / 170g/mol) / .250 L = ? mol/L
AgNO3 is about 170 g/mol
5.10 g ( 1 mol/ 170 g) = 0.03 mols
in 1/4 liter of water
0.03 mol / 0.25 L = 0.12 mol/Liter
Ag = 108
N = 14
O = 16 so 16*3 = 48
108 + 14 + 48 = 170 grams / mol
Do you need a tie breaker?
Is that 5,10 g meant to be 5.10 grams AgNO3? Assume that is 5.10 grams AgNO3.
moles AgNO3 = grams/molar mass = 5.10/170 = 0.03
M = moles/L = 0.03 mols/0.250 L = 0.12 M.
HOWEVER,THIS IS NOT CORRECT. Let me point out a couple of problems. First, the definition of molarity is M = moles solute/L OF SOLUTION. The problem says 5.10 g is dissolve in 250 cc OF WATER. Therefore, the total volume of the SOLUTION is not 250 cc but a slightly larger volume. Technically, the molarity can't be calculated unless we know the density of the solution. I will grant that the volume of the solution will be, for all practical purposes, 250 cc, but that is only because the grams of AgNO3 is a relatively small number. If you're making a 5 M solution, for example, it will make a difference. Did you know that 50 mL water + 50 mL alcohol = approximately 95 mL (not 100 mL). So if you want to accurately calculate a "concentration" you can use 0.03 moles/0.250 kg water = 0.12 molal or 0.12 m. Yes, I'm being nit pitky; however, if one is training to be an analytical chemist it needs to be done right.