25.0 ml of 15% NH4NO3 solution is diluted to 300.0 ml. What is the concentration of the diluted solution?
My Work:
Total molar mass of NH4CO3 = 80.052
.15 x 80.052= 12.0078
25.0 ml (12.00 M) = 300.0 ml (c2)
300.19/300.0ml =
The concentration is = 1.0M
I don't agree.
Why did you convert from % to M? I don't think you can convert to M without knowing the density.
I would keep it in % and use
15 x 25.0 = % * 300
% = ?
Or another way, you are diluting it by a factor of 300/25 or 12 times.
So 15%/12 = ?%
whoops I took 15% of the molar mass of NH4NO3
So .15 x 25=3.75
3.75 x 300= 1125 ?
To find the concentration of the diluted solution, we can use the equation:
C1V1 = C2V2
where C1 is the initial concentration, V1 is the initial volume, C2 is the final concentration, and V2 is the final volume.
In this case, the initial concentration (C1) is 15% NH4NO3 solution. To convert this to molarity, we first need to find the molar mass of NH4NO3. The molar mass of NH4NO3 is calculated by summing the atomic masses of the elements: nitrogen (N), hydrogen (H), and oxygen (O).
Molar mass of NH4NO3 = (14.01 g/mol) + (4 * 1.01 g/mol) + (3 * 16.00 g/mol) = 80.052 g/mol
Next, we can calculate the initial molarity (C1) by multiplying the percentage (0.15) by the molar mass of NH4NO3:
C1 = 0.15 * 80.052 g/mol = 12.0078 g/mol (rounded to four decimal places)
Given that the initial volume (V1) is 25.0 ml and the final volume (V2) is 300.0 ml, we can substitute these values into the equation to solve for the final concentration (C2):
12.0078 g/mol * 25.0 ml = C2 * 300.0 ml
Rearranging the equation to solve for C2:
C2 = (12.0078 g/mol * 25.0 ml) / 300.0 ml
C2 = 1.0006 g/mol (rounded to four decimal places)
Therefore, the concentration of the diluted solution is approximately 1.0006 M.