Please answer all three parts in the picture. A student prepared a solution cont
ID: 512983 • Letter: P
Question
Please answer all three parts in the picture. A student prepared a solution containing 0.07973% (w/v) of nickel(II) sulfate and measured its absorbance, The experimental absorbance was 0.928. Calculate the experimental molar absorptivity constant for the nickel(II) sulfate solution if the path of length of light was determined to be 1.20cm. Show ALL your work. Your final value should have the correct unit and number of significant figures. Molecular weight for N_iSO_4 = 154.75%g/mol. Refer to w/v examples shown on bottom of this page. Suppose you want to prepare a second solution containing nickel(II) sulfate that has an absorbance of 0.278 starting with the solution in question notequalto 1. Given a choice of 10-mL. and 5-mL pipets and 50-mL and 100-mL volumetric flasks, explain how you would procced in preparing the new diluted solution in ONE SINGLIE DILUTION step. The concept for this question is related to SRM02 reading material (pages 140 -144) posted on CCLE Calculate the concentration in ppm for the new diluted solution based on the choice of glassware in question notequalto 2. SHOW ALL YOUR WORK Your final value should have the correct unit and number of significant figures. Refer to ppm definition listed on bottom of this page.Explanation / Answer
Ans. 1. 0.07973% w/v means that there is 0.07973 g NiSO4 in 100.0 mL of solution.
Mass of NiSO4 in 1000 mL (= 1.0 L) solution = 0.7973 g
Moles of NiSO4 in 1.0 L solution = Mass/ Molar mass
= 0.7973 g/ (154.75 g/ mol) = 0.0051522 mol
Molarity of NiSO4 = Moles per liter of solution = 0.0051522 mol/ L = 0.0051522 M
Therefore, the molarity of 0.07973% w/v solution = 0.0051522 M
Now, using Beer-Lambert’s Law, A = e C L - equation 1,
where,
A = Absorbance
e = molar absorptivity at specified wavelength (M-1cm-1)
L = path length (in cm)
C = Molar concentration of the solute
0.928 = e x 12.0 cm x 0.0051522 M = e x 0.00618264 M cm
Or, e = 0.928 / (0.00618264 M cm) = 150.0977 M-1 cm-1
Therefore, extinction coefficient, e = 150.0977 M-1 cm-1
#2. Suppose, the required volume of solution 2 = 100.0 mL
Note that absorbance of a solution is directly proportion to its concentration.
So, Abs of soln. 1 / Abs of soln. 2 = Conc. of soln. 1 / Conc. of soln. 2
Or,
0.928 / 0.278 = 0.0051522 M / C2
Or, C2 = 0.0051522 M / 3.3381 = 0.00154345 M
Therefore, the concertation of solution 2 = 0.00154345 M
Now, using C1V1 = C2V2
C1= Concentration, and V1= volume of initial solution 1 ; Original solution
C2= Concentration, and V2 = Volume of final solution 2 ; Solution 2
Or, 0.0051522 M x V1 = 0.00154345 M x 100.0 mL
Or, V1 = (0.00154345 M x 100.0 mL) / 0.0051522 M = 29.96 mL
Thus, V1 = 29.96 mL
Preparation of solution 2: Transfer 29.96 mL of solution 1 (original solution) to 100.0 mL standard volumetric flask using 10.0 mL graduated pipette. Make the final volume upto mark with distilled water. It is the desired solution 2.
Note: The use of 10.0 mL pipette for 3 times is inevitable because we need to pipette 29.9 mL solution 1 for 100.0 ml solution 2. Even if we choose 50.0 mL flask, we would need to transfer 14.98 mL (29.96 mL/ 2 = 14.98 mL) of solution 2. Therefore, we need to reuse the pipette.
#3. Concentration of solution 2 = 0.00154345 M
[NiSO4] in terms of mass = Molarity x Vol. of solution x Molar mass of NiSO4
= 0.00154345 M x (154.75 g/ mol)
= 0.2388488875 g / L
= 238.85 mg/ L
= 238.85 ppm ; [1 ppm = 1 mg/ L]