85.0 mL of 1.50 mol/L ammonium chloride solution is added to 250 mL of water. Calculate the molar concentration of the final solution
The answer on the sheet says 0.381 mol/L but I keep getting 0.51
M1V1=M2V2
(1.50M)(85mL) = M2 (335)
M2 = 0.381
you have to take into account the volume of the total solution, 85+250 mL added to get a new molarity.
To calculate the molar concentration of the final solution, you need to consider the fact that the volume of the solution after mixing is the sum of the initial volumes of the ammonium chloride solution and water.
Given:
Initial volume of ammonium chloride solution (V1) = 85.0 mL
Molar concentration of ammonium chloride solution (C1) = 1.50 mol/L
Volume of water added (V2) = 250 mL
The total volume of the final solution (Vf) is calculated by adding V1 and V2:
Vf = V1 + V2
Vf = 85.0 mL + 250 mL
Vf = 335 mL
Next, we can use the equation for calculating molar concentration (C = n/V) to find the molar concentration of the final solution.
First, let's find the moles of ammonium chloride in the initial solution:
n1 = C1 * V1
n1 = 1.50 mol/L * 85.0 mL
n1 = 127.5 mmol
Since the moles of solute (ammonium chloride) do not change after dilution, we can use this value to find the molar concentration of the final solution:
Cf = n1 / Vf
Cf = 127.5 mmol / 335 mL
Cf = 0.381 mol/L
Therefore, the molar concentration of the final solution is 0.381 mol/L, which matches the answer on the sheet.