Question 2 of 4 Sapling Learning 100 The IR (infrared) spectra of two pure compo

Question

Question 2 of 4 Sapling Learning 100 The IR (infrared) spectra of two pure compounds (0.020 M compound A in solvent, and 0.020 M compound B n solvent are shown to the right. The pathlength of the cell is 1.00 Note that the the spectra is 70 transmittance rather than absorption, so that the 8 60 wavenumbers at which there is a dip in the 50 curve correspond to absorption peaks. 2 40 A mixture of A and B in unknown concentrations 3030 cm gave a percent transmittance of 45.5% at 2976 cm and 44.6% at 3030 cm 10 3040 Wavenumber 0,020 M A 0.020 MB Unknown 446% 1 35.0% 93.0% 3030 cm 2976 cm 1 76.0% 42.0% 45.5% What are the concentrations of A and B in the unknown sample? Number Number M A 2976 cm 2990 Wavenumber (c M B 2940 Map Pure A. Pure B 2890

Explanation / Answer

Let Xa and Xb be the concentrations of A and B respectively

At 3030 cm-1, mixture of A and B gives 44.6 % transmittance

So, Xa*0.35 + Xb*0.93 = 0.446 (i)

At 2976 cm-1, mixture of A and B gives 45.5 % transmittance

So, Xa*0.76 + Xb*0.42 = 0.455 (ii)

From (i), Xa = (0.446-0.93*Xb)/0.35

Substituting Xa in (ii)

((0.446-0.93*Xb)/0.35) * 0.76 + 0.42*Xb = 0.455

0.968 - 2.019*Xb + 0.42*Xb = 0.455

Xb = 0.32 M , which is B concentration

Substituting Xb=0.32 in Xa

Xa=(0.446-0.93*0.32)/0.35

= 0.424 M   , which is A concentration

Related Questions

Navigate

• Browse (All)
• Browse Q
• Subjects

---

---

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