Question

23. Calculate the ThOD for wastewater containing 40 mg/l. sucrose C.H0.1. (atomic weights: C: 12 H:1 0:16) a 192 b 180 40.0 42.7 d e 85.4 other, write you answer- 24. When drinking water is disinfected with chlorine, an undesired byproduct, chloroform (CHCb). may be formed. Suppose a 70-kg person drinks 2 L of water every day for 70 years with a chloroform concentration of 0.04 mg/L (the drinking water standard). Potency factor for chloroform- 6.1 x 10 (mg/kg-day) The cancer risk for this individual is most nearly a. 1.14 x10 b. 6.97x10 c. 8.63x10 d. 1.14x101 e. 6.97x101 f. other, write your answer . The concentration of 1,2-dichlorobenzene in a water supply is 1.2 ug/L. Assume that an adult drinks 2L /day water. The chronic daily intake (mg/Kg-d) for a 70 kg adult is most nearly: a. 3.43 x10 b. 3.43 x103 c. 2.40x10 d. 1.4x103 1.03x10 f other, write your answer A case-control study of 1700 participants looked at the association between Tamoxifen and uterine cancer. The study reported the following data. The odds ratio of developing uterine cancer for the Tamoxifen users would be most nearly: Uterine Cancer Yes Tamoxifen use 0 58 139 es 953 No 550 s a 4.15 b 2.40 c 1.45 d 0.41 e 0.25

Answer 1

Answer 23) sucrose is C6H12O6 having concentration 40 mg/l thus milli moles of sucrose=40/(molecular weight of sucrose) = 40/(72+12+96) = 40/180

Thus milli moles of O2 = 6*40/180 = 240/180

Thus ThOD = 32*240/180 = 42.67 say 42.7 . Thus option D is correct answer.

Answer 24) cancer risk = exposure dose x potency factor

Exposure dose = 0.04*2 /70 = 1.143*10^-3 mg/kg-d

Thus risk = 1.143* 10^-3 x 6.1*10^-3 = 6.972 * 10^-6 . Thus option B is correct answer.

Answer 25) chronic intake = 1.2*10^-3 *2/70 = 3.428 *10^-5 mg/kg-d . Thus option A is correct answer.