Assuming additive volumes, what will be the molarity of the solution that results when 35.5 mL of 2.00 M HNO3 is added to 214.5 mL of water?
0.284 M
2.00 M HNO3 x (35.5/(35.5 + 214.5) = ?
To determine the molarity of the resulting solution when two solutions are mixed, you can use the formula:
M1V1 + M2V2 = MFVF,
where M1 and V1 are the molarity and volume of the first solution, M2 and V2 are the molarity and volume of the second solution, and MF and VF are the molarity and volume of the resulting solution.
In this case, the first solution is 2.00 M HNO3 with a volume of 35.5 mL, and the second solution is water with a volume of 214.5 mL.
Using the formula, we have:
(2.00 M)(35.5 mL) + (0 M)(214.5 mL) = MF(35.5 mL + 214.5 mL)
Simplifying the equation, we get:
(2.00 M)(35.5 mL) = MF(250 mL)
Dividing both sides by 250 mL, we have:
(2.00 M)(35.5 mL) / 250 mL = MF
Solving the equation, we find:
MF = 0.284 M
Therefore, the molarity of the resulting solution is 0.284 M.
To calculate the molarity of the resulting solution, you can apply the formula:
M1V1 + M2V2 = M3V3
Where:
M1 and V1 are the molarity and volume of the first solution (2.00 M HNO3 in this case)
M2 and V2 are the molarity and volume of the second solution (water in this case)
M3 is the molarity of the resulting solution
V3 is the total volume of the resulting solution
Let's plug in the given values:
M1 = 2.00 M
V1 = 35.5 mL (convert to liters by dividing by 1000) = 0.0355 L
M2 = 0 M (since water has no molarity)
V2 = 214.5 mL (convert to liters by dividing by 1000) = 0.2145 L
M3 = ?
V3 = V1 + V2 = 0.0355 L + 0.2145 L = 0.25 L
Now, rearrange the formula to solve for M3:
M3 = (M1V1 + M2V2) / V3
M3 = (2.00 M * 0.0355 L + 0 M * 0.2145 L) / 0.25 L
M3 = (0.071 M * 0.00071 L) / 0.25 L
M3 = 0.000071 M / 0.25 L
M3 = 0.284 M
Therefore, the molarity of the resulting solution is 0.284 M.