Question: A solution of Silver nitrate, AgNO3, contains 1.08 grams in 250 cm3. What is its molarity?
Solution:
Molar mass of Silver nitrate = 108 g/mole.
n(AgNO3) = 1.08 grams of AgNO3 x 1 mole of AgNO3108 grams AgNO3 = 0.01 mole.
Number of moles in 1 dm3 = 0.01 mole x 1000/250 = 0.04 moles/dm3
So, Molarity of solution = 0.04M.
This is an example, if 0.04 mol is right then replace 250 with 40 and that's your answer.
To 40cm^3 of 1mol AgNO3 is added 20cm^3 of 0.500 AlCl3. What is the molar concentration of the resulting AgNO3 solution.
The reaction, if needed, is
3AgNO3 + AlCl3 ==> 3AgCl + Al(NO3)3
I worked out the number of moles present in AgNO3 which is 0.04 mol, and the number of moles of AlCl3 which is 0.01 mol.
What do I do next though?
Thanks in advance! Any help is greatly appreciated
1 answer
1 answer