CHEM J-2 June 2006 HCO 2. Calculate the osmotic pressure of a solution of 1.0 g of glucose (C 6 H 12 O 6 ) in 1500 ml of water at 37 C.

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Daniella McLaughlin
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1 CEM J-2 June 2006 Draw Lewis structures of ozone, 3, and the formate anion, C 2, including resonance hybrids where appropriate. 3 C 2 3 C C Calculate the osmotic pressure of a solution of 1.0 g of glucose (C ) in 1500 ml of water at 37 C. 4 The osmotic pressure π = crt where c is the concentration. The molar mass of glucose is: ( (C)) + ( ()) + ( ()) = g of glucose corresponds to mass 1.0 = = mol molar mass The concentration when this amount is dissolved in 1500 ml = 1.5 L is: c = number of moles = = M volume 1.5 ence, π = crt = (0.0037) ( ) ( ) = atm. Answer: atm Explain why a drip for intravenous administration of fluids is made of a solution of acl at a particular concentration rather than pure water. Blood plasma is isotonic with cells (same osmotic pressure). Using saline drip of same osmotic pressure as blood prevents haemolysis or crenation of red blood cells. Write down the ground state electron configuration of the iron atom. 1 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 6

2 CEM J-3 June 2006 The balanced equation for the complete oxidation of glucose to carbon dioxide and water is given below. C (s) (g) 6C 2 (g) (l) Calculate the mass of carbon dioxide produced by the complete oxidation of 1.00 g of glucose. 3 The molar mass of glucose is: ( (C)) + ( ()) + ( ()) = g of glucose corresponds to mass 1.00 = = mol molar mass From the chemical equation, oxidation of 1 mol of glucose leads 6 mol of C 2. ence the number of moles of C 2 produced is = mol. The molar mass of C 2 is (12.01 (C)) + ( ()) = Therefore, the number of mass of C 2 produced is: mass = number of moles molar mass = = 1.47 g Answer: 1.47 g Calculate the volume of this mass of carbon dioxide at 0.50 atm pressure and 37 C. The ideal gas law gives PV = nrt, hence: V = nrt (0.0333) ( ) (273+37) = = 1.69L P (0.50) Answer: 1.69 L Explain, in terms of chemical bonding and intermolecular forces, the following trend in melting points: C 4 < I 2 < acl < silica (Si 2 ) 3 There are only dispersion forces between the molecules in C 4 and I 2. The I atom is a large, many-electron atom so its electron cloud is more easily polarised than the C or in C 4 and therefore I 2 has stronger dispersion forces and the higher melting point. acl is an ionic compound with strong coulombic attraction between the a + ions and the Cl ions packed together in the solid. Silica is a covalent network solid. Melting it requires breaking of the very strong covalent Si bonds, so it has the highest melting point.

3 CEM J-4 June 2006 Draw a Lewis structure and thus determine the geometry of the ICl 4 ion. (The I is the central atom.) 2 Cl Cl I Cl Cl There are two lone pairs and 4 bonds around the iodine: the geometry is based on an octahedron with the lone pairs located opposite to one another to minimise repulsion between them. The geometry of the actual molecule is therefore square planar. Explain briefly, in terms of intermolecular forces, why an analogue of DA could not be made with phosphorus atoms replacing some nitrogen atoms, while still retaining a double-helical structure. 3 The double helical structure is held together by hydrogen bonding between the cytosine and guanine (C G) and the adenine and thymine (A=T) base pairs. o -bonding would occur if the electronegative atoms in these bases were replaced with P atoms. The solubility of nitrogen in water at 25 C and 1.0 atm is g L 1. What is its solubility at 0.50 atm and 25 C? 1 The equilibrium of interest is 2 (g) 2 (aq) with equilibrium constant: [ 2 (aq)] K = [ 2 (g)] From the perfect gas law, PV = nrt or concentration = n = P V RT. As a result, doubling the pressure doubles the concentration of 2 (g). As the temperature is unchanged, the equilibrium constant does not change and so [ 2 (aq)] must halve to ensure that K p is constant. The solubility is thus halved to ½ = g L -1 Answer: g L -1

4 CEM J-5 June 2006 Glucose is a common food source. The net reaction for its metabolism in humans is: C (s) (g) 6C 2 (g) (l) Calculate for this reaction given the following heats of formation. f (C (s)) = 1274 kj mol 1, f (C 2 (g)) = 393 kj mol 1 and f ( 2 (l)) = 285 kj mol 1 5 Using o o o rxn f f = m (products) n (reac tan ts) : rxn = [6 f (C 2(g)) + 6 f (2(l))] [ f (C612 6(g))] 1 = [(6 393) + (6 285)] [( 1274)] = 2794kJ mol (ote that f ( 2(g)) = 0 as it is an element in its standard state). Answer: kj mol -1 If the combustion of glucose is carried out in air, water is produced as a vapour. Calculate the for the combustion of glucose in air given that 2 (l) 2 (g) = +44 kj mol 1 As vaporising liquid water requires energy (+44 kj mol -1 ), the combustion enthalpy is reduced. Six moles of 2 are produced in the combustion so the enthalpy of combustion is reduced to: (6 +44) = kj mol -1 Answer: kj mol -1 Will S be different for the two oxidation reactions? If so, how will it differ and why? As gaseous molecules have higher entropy than liquid phase molecules, oxidation to produce 2 (g) will lead to a higher value for S than oxidation to produce 2 (l).

5 CEM J-6 June 2006 Butyric acid, C 3 C 2 C 2 C, is found in rancid butter and parmesan cheese. The pk a of butyric acid is (a) What is the p of a 0.10 M water solution of butyric acid? 6 As pk a = -logk a = 4.83, K a = Denoting butyric acid as A, the initial concentration of [A(aq)] = 0.10 M. The reaction table is then: [A(aq)] [ + (aq)] [A - (aq)] t = change -x +x +x equilibrium 0.10 x x x ence, K a = + - [ (aq)][a (aq)] [A(aq)] = (x)(x) 0.10-x 2 x = 0.10-x As K z is small, the amount of dissociation, x, is also small so 0.10 x ~ x Using this approximation, K z = =10 hence x = M As x = [ + (aq)], p = -log[ + (aq)] = -log( ) = 2.92 Answer: 2.92 (b) Calculate the p of the solution formed when mol of a(s) is added to 1.0 L of 0.10 M butyric acid. As a is a strong base, it will dissociate completely and each mole of - will react with butyric acid to form one mole of A - (aq). 1.0 L of 0.10 M A contains 0.10 mol. After addition of mol of -, the number of moles of A = ( ) = 0.05 mol and the number of moles of A - = mol. As 1.0 L of solution is present, [A(aq)] = 0.05 M and [A - (aq)] = 0.05 M. Substituting into the expression for K a gives: K a = + - [ (aq)][a (aq)] [A(aq)] + [ (aq)] (0.05) = = (0.05) so [ + (aq)] = 1.5 M ence, p = -log[ + (aq)] = 4.83 Answer: 4.83

6 (c) Using equations, comment on how the final solution in (b) will respond to additions of small amounts of acid or base in comparison to 1 L of water. Solution (b) consists of a mixture of a weak acid and its conjugate base: it is a buffer system and will resist changes in p. If acid is added, the system can respond by removing it using A - : + (aq) + A - (aq) A(aq) If base is added, the system can respond by removing it using A: - (aq) + A(aq) 2 (l) + A - (aq)

7 CEM J-7 June 2006 Consider the reaction of 2 (g) with I 2 (g) at 298 K to give I(g). 2 (g) + I 2 (g) 2I(g) K p = 2.24 If partial pressures of 0.20 atm of all three gases are mixed, in which direction will the reaction proceed? 3 The reaction quotient Q = 2 2 (p I ) (0.20) = =1.0 (p )(p ) (0.20) (0.20) 2 I2 As Q < K p, the reaction will proceed towards I, to increase the partial pressure of I and decrease the partial pressure of 2 and I 2, until Q = K. Calculate G for this reaction at 298 K. Answer: towards products Using G = -RTlnK p : G = -(8.314) (298) ln(2.24) = J mol -1 = kj mol -1 Answer: J mol -1 = kj mol -1

8 CEM J-8 June 2006 Draw the constitutional formula(s) of the major organic product(s) of the following reactions Cr 2 7 / S S + AD / S + AD + excess C 3 / 3 C C 3 Br + Br 2 Br Cl + excess 2 C 3 C 3 What are the requirements for a molecule to be aromatic? Give one example of an aromatic heterocycle. 2 The molecule must be cyclic. The ring system must have a conjugated π bond system. The number of electrons in the π bond system must be 4n+2 where n=integer. All atoms in the ring system must be sp 2 hybridised. Examples of heterocyclic heterocycles include: S

10 CEM J-9 June 2006 Adenine and thymine have the structures shown below C adenine Draw a tautomer of the shown structure of adenine. thymine In DA, adenine forms a base pair with thymine. Explain what is meant by base pair and indicate the point(s) of interaction between adenine and thymine. Four different bases are found in DA chains. The two strands in the double helix are held together by 3 hydrogen bonds between guanine (G) and cytosine (C), and by 2 hydrogen bonds between adenine (A) and thymine (T). The complementary bases C and G are called a base pair. A and T are another base pair. adenine = A thymine = T

11 CEM J-10 June 2006 The open chain form of D-glucose has the structure shown. C C 2 5 Draw the aworth projection of β-d-glucopyranose. Draw the major organic product of the reaction of D-glucose with the following reagents. 1. ab 4 2. / 2 C 2 C 2 [Ag( 3 ) 2 ] / C 2 C 2 Would you expect D-glucose to be water soluble? Why? D-glucose will be water soluble as it has numerous alcohol functional groups which can hydrogen bond with the water molecules.

12 CEM J-11 June 2006 L-Tyrosine is a naturally occurring amino acid with the following side-chain. 6 C 2 The pk a values of tyrosine are 2.20 (α-c), 9.19 (α- 3 ) and (sidechain). Draw the Fischer projection of L-tyrosine indicating the correct charge state at physiological p. C 2 3 C C 2 What is the absolute stereochemistry of L-tyrosine? Write (R) or (S). S What is the value of the pi of L-tyrosine? pi = ½ ( ) = 5.70 What does pi represent? pi represents the isoelectric point - the p at which there is no net charge on the molecule.

13 Account for the difference in acidity of the carboxylic acid group and the phenol. Acid strength is dependent on the stability of the conjugate base. The carboxylate anion is resonance stabilised, with the charge being spread over the electronegative atoms: R C R C The phenoxide anion is resonance stabilised also, but the charge in the resonance contributers is spead over the C atoms in the ring. C is not as electronegative as, so these contributers are not as significant as that with the charge on the. Resonance stabilisation is not as great as for carboxylate and therefore phenol is weaker acid than carboxylic acid.

14 CEM J-12 June 2006 Draw the products of acid hydrolysis of the following peptide, indicating the correct charge state under these conditions. 3 3 C C 3 C C C 2 C 2 C 2 C 2 3 C C 3 C C C 3 2 C C 2 C

CHEM J-3 June Calculate the osmotic pressure of a 0.25 M aqueous solution of sucrose, C 12 H 22 O 11, at 37 C

CHEM J-3 June Calculate the osmotic pressure of a 0.25 M aqueous solution of sucrose, C 12 H 22 O 11, at 37 C CHEM1405 003-J-3 June 003 Calculate the osmotic pressure of a 0.5 M aqueous solution of sucrose, C 1 H O 11, at 37 C The osmotic pressure for strong electrolyte solutions is given by: Π = i (crt) where