Original molarity was 1.7 moles of NaCl
Final molarity was 0.36 moles of NaCl
Given Information:
Original (concentrated) solution: 25 g NaCl in a 250 mL solution, solve for molarity
Final (diluted) solution: More water is added to make the new total volume 1.2 liters, solve for the new molarity
1. Solve for the molarity of the original (concentrated) solution.
Molarity (M) = moles of solute (mol) / liters of solution (L)
Convert the given information to the appropriate units before plugging in and solving for molarity.
Molarity (M) = 0.43 mol NaCl solute / 0.250 L solution = 1.7 M NaCl (original solution)
2. Solve for the molarity of the final (diluted) solution.
Remember that the amount of solute remains constant in a dilution problem; it is just the total volume of the solution that changes due to the addition of solvent.
Molarity (M) = 0.43 mol NaCl solute / 1.2 L solution
Molarity (M) of the final solution = 0.36 M NaCl
I hope this helped:))
Answer:
= 0.030 M
Explanation:
We can take x to be the concentration in mol/L of Ag2SO4 that dissolves
Therefore; concentration of Ag+ is 2x mol/L and that of SO4^2- x mol/L.
Ksp = 1.4 x 10^-5
Ksp = [Ag+]^2 [SO42-]
= (2x)^2(x)
= 4x^3
Thus;
4x^3 = 1.4 x 10^-5
= 0.015 M
molar solubility = 0.015 M
But;
[Ag+]= 2x
Hence; silver ion concentration is
= 2 x 0.015 M
= 0.030 M
