Questions (1-3 go together) 1. As aspirin tablet is dissolved in 250.0 ml. of wa

Question

Questions (1-3 go together) 1. As aspirin tablet is dissolved in 250.0 ml. of water. 10.00 mL. and this diluted solution has an absorbance reading of O.1 calibrated and the equation for the standard curve is found to where Y is the absorbance at What is the concentration A 0.10 ml sample o the solution is diluted to Y 1538 X-0.01 be: 530 nm and X is the molarity of an acetyl salicylic acid solution. 2. Remembering th ml in order to read the absorbance, what is the molarity of the original solution? at the solution in the previous question was formed by dilution of 0.10 mL to 10.00 3. What is the total weight (mg) of acetylsalicylic acid in the aspirin tablet? The original solution had a volume of 250.0 mL

Explanation / Answer

given Absorbance (A) is related to concentration as A= 1538* concentration-0.01

given A= 0.11, hence concentration from the calibration chart is    0.11= 1538* concentration-0.01

concentration =(0.11+0.01)/1538, concentration =7.8*10-5 moles/L

let x= mass of acetylsalicyclic acid , molar mass = 180 g/mole, moles of acetylsalicyclic acid =x/180

tihs is dissolved in 250ml, 1000ml= 1L, concentration of original solution = moles/volume in L= x/(180*0.250) =0.022x

moles present in 0.1ml solution 0.1ml= 0.1/1000L

moles =0.022x*0.1/1000 , this is further diluted to 10ml,

concentration =0.02x*0.1/1000/10/1000 =0.0022x= 7.8*10-5, x=0.035 gm

molarity of original solution = 0.022*0.035 =0.00077 moles/L

moles in original solution = 0.00077*0.25 = 0.000193

mass of acid in original solution = 0.000193*180 gm =0.035 gm

Related Questions

Navigate

• Browse (All)
• Browse Q
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.