25 ml of 3M hno3 are mixed with 75ml of 4.0M HNO3 if volumes are additive then find molarity
3.25
molarity=moles/volume= (molenitric+molesnitric)/volume
= (3*.025 + 4*.075)/.100
To find the final molarity after mixing the two solutions, we need to consider the volume and concentration of each solution. We can use the following formula:
M1V1 + M2V2 = MFVF
where M1 and M2 are the initial molarities of the solutions, V1 and V2 are the volumes of the solutions, and MF and VF are the final molarity and volume, respectively.
Given the information:
M1 = 3.0 M (molarity of the first solution)
V1 = 25 ml (volume of the first solution)
M2 = 4.0 M (molarity of the second solution)
V2 = 75 ml (volume of the second solution)
We can substitute these values into the formula:
(3.0 M)(25 ml) + (4.0 M)(75 ml) = MF(VF)
75 + 300 = MF(VF)
375 = MF(VF)
To find the final molarity (MF), we need to know the final volume (VF). However, the problem does not specify the final volume. If the volumes are additive, we can assume the final volume is the sum of the volumes of the two solutions:
VF = V1 + V2
= 25 ml + 75 ml
= 100 ml
Substituting the final volume into the equation:
375 = MF(100 ml)
To find MF, we can rearrange the equation:
MF = 375 / 100
= 3.75 M
Therefore, the final molarity after mixing the solutions would be 3.75 M.
To find the molarity of the final solution, we need to consider the volumes and concentrations of the two solutions being mixed.
Let's first calculate the moles of HNO3 in each solution:
moles HNO3 in 25 ml of 3M solution = (0.025 L) x (3 mol/L) = 0.075 mol
moles HNO3 in 75 ml of 4.0M solution = (0.075 L) x (4 mol/L) = 0.3 mol
Next, we add the moles of HNO3 from the two solutions:
Total moles of HNO3 = 0.075 mol + 0.3 mol = 0.375 mol
Now, we calculate the total volume of the solution:
Total volume of the solution = 25 ml + 75 ml = 100 ml = 0.1 L
Finally, we can calculate the molarity of the final solution using the formula:
Molarity (M) = Moles of solute / Volume of solution
Molarity (final solution) = 0.375 mol / 0.1 L = 3.75 M
Therefore, the molarity of the final solution is 3.75 M.