Answer:
1214 mg
Explanation:
Data given
Volume of NaOH solution (V) = 14.85 mL
Molarity of NaOH solution (M) = 0.100 M NaOH.
Amount of acetylsalicylic acid = ?
Molecular weight = 180.2 g/mol
Solution:
It is a acid base titration
As equal amount of NaOH will neutralize equal amount of acetylsalicylic acid.
First find the moles of NaOH
As we know
M = moles of solute / 1 L x volume of solution
Rearrange and modify the above equation
moles of NaOH = M of NaOH x 1 L / volume of NaOH solution
Put values in above equation
moles of NaOH = 0.100 M x 1 L / 14.85 mL
moles of NaOH = 0.0067 moles
Now find to find the weight of acetylsalicylic acid
As
equal amount of NaOH will neutralize equal amount of acetylsalicylic acid.
moles of acetylsalicylic acid = moles of NaOH.......... (1)
As we know
no. of moles = mass in g / molar mass
So,
The modified form of equation 1 will be
mass in g / molar mass (acetylsalicylic acid) = moles of NaOH
Put values in above equation
mass in g / 180.2 g/mol = 0.0067 moles
Rearrange the above equation
mass of acetylsalicylic acid = 0.0067 moles x 180.2 g/mol
mass of acetylsalicylic acid = 1.214 g
Convert the grams to milligrams
1 g = 1000 milligram
So,
1.214 g = 1214 mg
