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.
Using the formula C(Concentration)=n(moles)/v(Volume measured in litres.)
How am I supposed to solve this question given two volumes?
use M1V1=M2V2
the C=M
That formula is just telling you what concentration is.
In the formula i mention M1 is the initial concentration and V1 is the initial volume. M2 is the final concentration. V2 is the final volume. You are solving for M2.
Oh, that makes sense. Thanks!
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
How would I do this, given 2 volumes?
To solve this question, you need to consider the initial volume of the ammonium chloride solution and the final volume of the solution after the addition of water.
First, convert the given volumes to liters:
Initial volume = 85.0 mL = 85.0 mL ÷ 1000 mL/L = 0.085 L
Final volume = 250 mL = 250 mL ÷ 1000 mL/L = 0.250 L
Next, determine the moles of ammonium chloride in the initial solution. To do this, multiply the initial volume by the initial concentration:
Initial moles = Initial volume × Initial concentration
= 0.085 L × 1.50 mol/L
= 0.1275 mol
Now, calculate the moles of ammonium chloride in the final solution. Since there is no ammonium chloride added or removed during the dilution process (water is added only), the moles of ammonium chloride remain the same:
Final moles = Initial moles = 0.1275 mol
Finally, calculate the molar concentration of the final solution using the final moles and the final volume:
Final concentration = Final moles / Final volume
= 0.1275 mol / 0.250 L
= 0.51 mol/L
Thus, the molar concentration of the final solution is 0.51 mol/L.