Question

Question 49 5 pts A 1.00 L solution of 0.10 M NH3 and 0.4 M NH4Cl has 0.17 moles of KOH added. What is the pH of the resulting solution? K) (NH3) = 1.8 x 10-5. Report your answer to the nearest whole number Question 50 5 pts Ethanol has a heat of vaporization of 38560 J/mol and a normal boiling point of 352 K. What is the vapor pressure (in torr) of ethanol at 314 K? Report your answer to the nearest whole number and do NOT include units in your answer, only type in the number.

Answer 1

49 no. of moles of NH₃ = 0.1X 1000 = 0.1 no. of moles of NHAU 2 0.4 X 1000 = 0.4 1000 KOH Now 0.17 moles of KoH are added KOH NH CO NH₃ + thot ke. Imole of the KOH rearts with Imote of NHAU 0,17 moles of which reacts with 0.17 motes of NH CC. Also, I mole of KoH forems I mole of NH₃ * 0.17 motes of KoH forums 0.17 moles of NH3 ne of new no. of moles of NHA CU = 0.4-0.17 0.23 new no. of moles of NH₃ 20.17 0.17 0. 27 new (NHA Ce] = 0.23 x 1000 = 0.23M new (Nth] = 0, 27 X 1000 - 0.27 M 1000 NH₂ (aq) + tho (1) - NH OH ko (NHAT LOH = 0.23% [o] [Ntb] 0.27 1000 1.8 x 10-580.27 - 0.23 lort

[H] = 2.11 X10 M pOH = log 2.1110 - 4.675 pH o nama 14-4 1675 TpH = 9.324 5 Stwap - 38560 J/mol | T, = 352k P, z Po e apeti P, = 1 atm = 760 torer P22 Po e atuak eta T2 = 314K Poe het (at) torer PA ? : -38560 ( 352 314) so, 760 - € 8.314 P2 38560 (352 314) x 3 1314 >> P2 - 760 e 7 P 760 e 13-257 -1.6 P2 = 760 e & 1P₂ 2 156.44 torer The vapour pressure of ethanol at 314K is 156.44 torer