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Exam is printed double-sided. Check every side of every page before turning in! Part I. Multiple Choice. Choose the best answer and write it on the line. 1. (3 pts) Assuming each solution is 0.100 M and behaves ideally, which solution has the largest van't Hoff factor? a) Na₂SO, b) Ba(CIO,)z c) C6H12O6 d) AICI, c) HCIO, Answer: 2. (3 pts) The vapor pressure of pure benzene (CH) is 0.1252 atm. Suppose 0.0587 mol of solid naphthalene (C₁0H₂) is added to 1.00 mol of benzene (CH). What is the vapor pressure of benzene above the solution? a) 0.1325 atm b) 0.1252 atm c) 0.1182 atm d) 0.00735 atm e) 0.00694 atm Answer: 3. (3 pts) The vapor pressure of pure acetone (propanone) is 266 torr. When a non-volatile solute is added, the vapor pressure of acetone above the solution falls to 232 torr. What is the mole fraction of the non-volatile solute in the solution? a) 0.87 b) 0.69 c) 0.32 d) 0.13 e) 0.045 Answer: 4. (3 pts) Select the reaction for which K, Kp. a) 2 COF₂ (g) = CO₂(g) + CF. (g) = 0 b) N₂(g) + 3H₂(g) = 2 NH₂(g) 2 c) NH HS(s) = NH₂(g) + H₂S(g) = d) 2 H₂(g) + CO(g) = CH₂OH(1) = -2 Answer: Kp= Ke(RT) Anges

5. (3 pts) Suppose that NOBr was allowed to decompose in a sealed container. When equilibrium was reached, Pros -7.68 atm, Po-0.526 atm, and Per, -1.59 atm. Calculate K, for this reaction. 2NOBr(g) + 2NO(g) + Bri(g) 2 3 a) 134 b) 91.8 Kp= [NOB-J² [7.68] [NO] [56] [154] c) 0.109 d) 0.0143 e) 7.46 × 10¹ Answer: Part II. Fill in the Blanks. 6. (10 pts) Consider this exothermic reaction at equilibrium: 4NH3 (g) +502 (g) = 4NO (g) + 6H₂0 (1) Predict how the following changes affect the equilibrium position (shift right, shift left, no change) and the equilibrium concentration of NO (g) (increase, decrease, no change) by completing the below table Change Effect on equilibrium position Effect on concentration of NO Increasing the volume We Removing H₂0 (1) from the mixture no change right Adding O₂ (g) to the mixture night Yes Adding 1 atm of argon gas no change NO Decreasing the temperature yes Part III. Problem Solving. Show all work including conversions and units. Answers should include correct sig. figs. Put your answers in the boxes. O 7. (6 pts) A 750.0 ml sample of pure water at 298 K is allowed to come to an equilibrium with pure nitrogen gas at a pressure of 1.70 atm. What is the mass, in mg, of nitrogen gas dissolved in the water? The Henry's law constant for nitrogen gas at 298 K is 6.1 x 10 M/atm. Seges Syos = H₁₂ Res M=1170 alm 2786.88 4 6₁1 X10-9 M/alm

-15.4. 8. (9 pts) An aqueous solution containing 22.4g sample of a purified protein (nonelectrolyte)has a volume of 100.0 mL at 15.0°C. The osmotic pressure of this solution is 78.1 mmHg at 15.0°C. What is the molar mass, in g/mol, of the protein? TT= MRT T= 15+273 = 288k R = 0.0821 L·alm/mol. K 18.1 IT = 78.1 mmig RT (0,0821) (288) 3.30 M = 22.44 M = mal L 0.330 ml 67.812/mal S.P. G81 mal = 0.100 x 3,30 = 0.330 mal 9. (10 pts) A particular brand of liquid hand soap solution is 21.0% glycerin (CH₂O) by mass. The density of the solution is 1.124 g/mL. and the molar mass of glycerin is 92.094 g/mol. Calculate the following: de 1.1243 2 a) molality of the solution My 0.222 mal of £348 Q1 y of Cz H 8 Oz x Didolo by Ⓒ کہ m = M = IT miel ky M = mal L 225 m b) molarity of the solution 2.2021244 9-1=68) 4100g-21= 592 Imel C₂ Hg 0₂ 92.0948 d = 0.228 mal 27/5m 182.45

10 417 10. (7 pts) Using the two given reactions, find the value of the equilibrium constant for the last equation Show all of your work. 2A(g) B B(g) +(8) K = 0.0334 Ke TAJ² Lagt Lys 63 D (g) 8(g) + ( K= 2,3 2A (g) + 3D (g) K=? K= [D]³ [A] [C] [A] [D]³ 11. (5) 11. (5 pts) Write the reaction quotient, Qe, for the following reaction: 2 H₂S(g) + S0₂(g) 3 S(s) + 2 H₂O(g) Q= [4₂0] [H₂5] [50₂] 5 12. (5 pts) The reaction below has a K, of 28.4 at 298 K. Determine the value for K, of this reaction at that temperature. 2-3-1 2 Br₂(g) +2 NO(g) 2 NOBr(g) - Ke = kp (RT) jes (RT) =(28.4)(0,0821) (298) 695 = 694.828 تر تھا you [6]3 = solve for [A]² (DJ³ 0.0334 = 2.3

13. (8 pts) At 800 °C, the equilibrium constant for the following reaction is K, 0,365. 2 50₂(g) + O₂(g) 250, (g) The reaction vessel initially contains 0.60 mol 50₂, 13 mol O₂, and 0.50 mol SO, in a 1.00 L flask 0.5 a Complete the ICE Table in terms of "x" 0.60 1.3. M L Concentration (M) [50₂] [0₂] [50₂] Initial 0.60M 1,3M 0.5M Change +2x 0-5 +xV -2x 0.5 +4 Equilibrium 0161X 13+X 0.5X b. Use the space below to show work supporting your ICE table "change" values. 14. (12 pts) A sealed 1.00 L flask at 58 °C initially contains 0.310 mol of 12(g) and 0.310 mol Cl₂(g). The reaction mixture is allowed to reach equilibrium according to the reaction: 1₂(g) + Cl₂ (g) 21C1(g) K₂= 78.8 Calculate the equilibrium concentration of ICL Show all work. [tci) [3] [4] 0.310 mat of I₂ =310M 0.310 mal of C1₂ = .310M LOBNOX 1₂ + Cl₂ = ZICI 0.310 310 20/+4 -X (0.310-x) (0.310-X) 2 = [s0₂² [s0₂] [₂] (4 6 C E 2(1137) = 2.75 IL 78.8 8.87= 2x Le 2.752 = 2X V = 1,33 ALK Shift right (1106)(1100) 6.73

4 15. (13 pts) Suppose that foul-smelling H₂S gas can react with chlorine gas according to the following equilibrium reaction at 125 °C. H₂S(g) + Cl₂(g) = S(s) + 2 HCl (g) Ke 3.10 x 10- At the start of the reaction, [H₂S] gas is 0.290 M and [Cl₂] is 0.340 M. Use the small x-approximation, if appropriate, to calculate the equilibrium concentration of gaseous HCL. You must show work to indicate whether the small-x approximation is justified or not. H₂S + Cl₂ = 2HC1 0.290 0.340 De 2X S 2,290-x 0.340-X ₁290-x Ć e X >400 ✓ v X <400 X K Сназа [H] [9₂] 2x (275² ✓ (0.240-x)(0.340- 0.290✓ > 400 3,10 110¹ x-approximation does not work. -End of Exam----