3366 C 019 100 points How many moles of CONH 2 2 are present in100 g of water

3366 c 019 100 points how many moles of conh 2 2 are

• Notes
• 7

This preview shows page 5 - 7 out of 7 pages.

3366 C . 019 10.0 points How many moles of CO(NH 2 ) 2 are present in 100 g of water, if the freezing point of the solu- tion is - 4 . 02 C? k f (water) = 1 . 86 K · kg / mol. 1. None of these 2. 0 . 292 mol 3. 0 . 227 mol 4. 0 . 216 mol correct 5. 0 . 303 mol 6. 0 . 249 mol Explanation: k f = 1 . 86 K · kg / mol Δ T f = - 4 . 02 C = 4 . 02 K m solvent = 100 g = 0 . 1 kg Δ T f = k f m = k f n CO(NH 2 ) 2 m solvent n CO(NH 2 ) 2 = Δ T f m solvent k f = (4 . 02 K) (0 . 1 kg) 1 . 86 K · kg / mol = 0 . 216129 mol . 020 10.0 points Consider 0.01 m aqueous solutions of each of the following. a) NaI; b) CaCl 2 ; c) K 3 PO 4 ; and d) C 6 H 12 O 6 (glucose) Arrange the solutions in order of freezing point from lowest to highest. Assume that each compound behaves ideally. 1. c, b, d, a 2. b, a, d, c 3. a, b, c, d
estrada (gse77) – Homework 3 – Sutcliffe – (52410) 6 4. d, a, b, c 5. None of these 6. a, d, b, c 7. d, c, b, a 8. c, d, a, b 9. c, b, a, d correct Explanation: solvent is H 2 O m = 0.01 m solutes are NaI, CaCl 2 , K 3 PO 4 , C 6 H 12 O 6 FP depression of each solution = ? Δ t f = K f m a)For the solute NaI,NaI(s)-→Na++ I(aq)each formula unit of NaI yields two ions insolution:Δtf=-1.86C·kg H2Omol ions·0.01 mol NaIkg H2O·2 mol ionsmol NaI=-0.04Cb)For the solute CaCl2,CaCl2(s)-→Ca2++ 2 Cl(aq)each formula unit of CaCl2yields three ionsin solution:Δtf=-1.86C·kg H2Omol ions·0.01 mol CaCl2kg H2O·3 mol ionsmol CaCl2=-0.06Cc)For the solute K3PO4,-0.07C<-0.06C>-0.04C<-0.02C,so the order of increasing Δtfis c, b, a, d.02110.0 pointsEstimate the osmotic pressure associated with30.0 grams of an enzyme of molecular weight1.20×105dissolved in enough benzene tomake 1700 mL of solution at 313 K. Select theclosest answer.1.3.78×103atm.correct2.4.65×102atm.3.3.78×106atm.4.2.87 atm.5.6.43×103atm.6.6.43×106atm.Explanation:m = 30 gV= 1700 mLT= 313 Kπ=M R T=mol soluteL soln×R T=30 g enzyme1.700 L soln×1 mol enzyme1.2×105g enzyme×0.082 atm·LK·mol×313 K= 3.77×103atm K 3 PO 4 (s) -→ 3 K + + PO 3 4 (aq) each formula unit of K 3 PO 4 yields four ions in solution: Δ t f = - 1 . 86 C · kg H 2 O mol ions · 0 . 01 mol K 3 PO 4 kg H 2 O · 4 mol ions mol K 3 PO 4 = - 0 . 07 C d) For the solute C 6 H 12 O 6 (glucose), Δ t f = - 1 . 86 C · kg H 2 O mol ions · 0 . 01 mol C 6 H 12 O 6 kg H 2 O · 1 mol ions mol C 6 H 12 O 6 = - 0 . 02 C Arranging these from lowest to highest,
estrada (gse77) – Homework 3 – Sutcliffe – (52410) 7 022 10.0 points Consider the solutionsI) 1.0 M Na2SO4,II) 1.0 M NaCl, andIII) 1.0 M sugar.What answer gives the expected order of de-creasing (highest, next, lowest) osmotic pres-sure?