Answer:
0.00018M
Explanation:
Mass of the aspirin tablet =0.4g.
% by mass of ASA=68.2
mass of in the tablet=68.2/100×0.4g
=0.2748g
mass conc of ASA=mass of ASA/Volume of diluent(250ml)
=0.2748/0.25
=1.0992g/dm3
molar conc=mass conc/molar mass
=1.0992/180.16
=0.006M
Conc of the stock solution C1=0.006M
Initial volume for making the diluted solution(V1)=3ml
final volume of the diluted solution=100ml
from dilution principle
no of mole=conc × volume
initial conc(C1)×initial volume(V1)=final conc(C2)×final volume(V2)
C2=0.006M×3ml/100ml
=0.00018M
When an aspirin tablet that contains 68.2% acetylsalicylic acid (ASA) is dissolved in hot NaOH to a volume of 250 mL and then 3.00 mL of this solution is dissolved to 10 mL in a FeCl₃ solution, the ASA concentration in the diluted solution is 1.82x10⁻⁴ mol/L.
To find the ASA concentration in the diluted solution, we need to follow the next steps.
1. Calculation of the initial mass of ASA
The initial mass of ASA can be calculated with the following equation:
[tex] \%_{m/m} = \frac{m_{ASA}}{m_{a}} \times 100 [/tex]
Where:
[tex] m_{ASA}[/tex]: is the mass of the acetylsalicylic acid (ASA) =?
[tex] m_{a}[/tex]: is the mass of aspirin = 0.400 g
[tex]\%_{m/m}[/tex]: is the mass percent = 68.2%
Solving equation (1) for [tex] m_{ASA}[/tex] we have:
[tex] m_{ASA} = \frac{\%_{m/m}}{100}*m_{a} = \frac{68.2 \%}{100}*0.400 g = 0.273 g [/tex]
2. Calculation of the initial number of moles of ASA
With the above mass we can find the initial number of moles:
[tex] n = \frac{m_{ASA}}{M} = \frac{0.273 g}{180.16 g/mol} = 1.52 \cdot 10^{-3} \:moles [/tex]
3. Calculation of the initial ASA concentration
When the student dissolves the tablet in hot NaOH and then dilutes the solution to 250 mL, we have the following concentration:
[tex] C_{i} = \frac{n}{V} = \frac{1.52 \cdot 10^{-3} \:moles}{0.250 L} = 6.08 \cdot 10^{-3} mol/L [/tex]
4. Calculation of the diluted ASA concentration
After the student takes 3.00 mL of the above solution and dilutes it to 100 mL with FeCl₃, we can find the concentration of ASA in the diluted solution with the equation:
[tex] C_{i}V_{i} = C_{f}V_{f} [/tex]
Where:
[tex] C_{i}[/tex]: is the initial ASA concentration = 6.08x10⁻³ mol/L
[tex] V_{i}[/tex]: is the initial ASA volume = 3.00 mL
[tex] C_{f}[/tex]: is the final ASA concentration =?
[tex]V_{f} [/tex]: is the final volume = 100 mL
The concentration of ASA in the diluted solution is:
[tex] C_{f} = \frac{C_{i}V_{i}}{V_{f}} = \frac{6.08 \cdot 10{-3} mol/L*3.00 mL}{100 mL} = 1.82 \cdot 10^{-4} mol/L [/tex]
Therefore, the concentration of ASA in the diluted solution is 1.82x10⁻⁴ mol/L.
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