Some Basic Concepts of Chemistry

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1 0 Some Basic Concepts of Chemistry Chapter 0: Some Basic Concept of Chemistry Mass of solute 000. Molarity (M) Molar mass volume(ml) mol L 3. (A) g atom of nitrogen 8 g (B) atoms of C has mass g atoms of C has mass g (C) mole of S has mass 3 g (D) 7.0 g of Ag So, lowest mass 6 g of C 4. 3A + B A 3 B 3 mol A + B mol mol A 3 B In option (A): 3 mol A + mol B 0.5 mol A 3 B (Excess reagent) (Limiting reagent) In option (B):.5 mol A + mol B 0.5 mol A 3 B (Limiting reagent) (Excess reagent) In option (C): 6 mol A + 3 mol B.5 mol A 3 B (Excess reagent) (Limiting reagent) In option (D): 3 mol A + mol B mol A 3 B In this case, none of the reactant is a limiting reagent. 7. Molecular mass of oxygen 3 g/mol Amount of oxygen 6 g Number of moles of oxygen mole 3 mole of oxygen molecules moles of oxygen molecules 8. mole of CH 4 contains 4 mole of hydrogen atom i.e., 4 g atom of hydrogen.

2 Chemistry Vol. (Med. and Engg.) molecules 8 g of CO molecules g of CO 0. 60g of urea contains g of carbon. 00g of urea contains 00 0% 60. % of X 75.8 % of Y g of CO Element % Composition Atomic ratio Simplest ratio X Y Thus, empirical formula of the compound is XY. 3. CaCO 3 CaO + CO (94) (56) (44) mol of CaCO 3 56 g of CaO 3 mole of CaCO g of CaO 5. 4Al + 3O Al O 3 (08 g) (96 g) (04 g) 08 g Al reacts with 96 g of O 7 g Al will react with g of O

3 0 Structure of Atom. The order in which the given orbitals are filled is as follows: 4s, 3d, 4p, 5s. Energy of one photon E h hc λ Js ms E m J As l (n ), hence for n, l cannot be. 4. The total number of nodes n For 3p orbital, n 3 Total no. of nodes 3 6. Mass of proton kg Mass of electron kg 8. Deuterium is an isotope of hydrogen. Deuterium (D) H 9. For M th shell, n 3; maximum number of electrons in M th shell n E E 4 E 3.6 ( 3.6) ev 34 h m 3 mv 0 00 Chapter 0: Structure of Atom 3. For s orbital l 0, m λ nm m

4 03 r Chapter 03: Classification of Elements and Periodicity in Properties Classification of Elements and Periodicity in Properties 8. s s p 6 3s is the electronic configuration of an element having atomic number. i.e. Magnesium. It can lose electrons easily and is electropositive in nature and hence, it is a metal (known as alkaline earth metal). 9. K Ca Sc Cl All have 8 electrons and therefore are isoelectronic. 0. Na : s s p 6 3s. The single electron can be easily removed. Hence, its I.P. will be the lowest. Mg : s s p 6 3s ; 3s orbital is completely filled. I.P. will be highest. 3Al : s s p 6 3s 3p, 3p orbital is partially filled. Electrons in sorbital are relatively closer to the nucleus than pelectrons. I st I.P. value of Mg is more than that of Al. Na < Mg > Al

5 04 Chemical Bonding and Molecular Structure Chapter 04: Chemical Bonding and Molecular Structure. F is formed from the atoms of same element. Therefore, electronegativity difference is zero. 4. Bond angles of BF 3, H O, NH 3 and CH 4 are 0, 0435, 078 and 098 respectively. 5. Cl possesses 0 electrons in ClF 3 i.e. expanded octet. In HF, F completes its octet and H is in duplet state. In BF 3, B possesses 6 electrons i.e. incomplete octet. In NH 3, N completes its octet and H is in duplet state. 8. Repulsion between different electron pairs is not the same in magnitude. The repulsion follows the order lone pair lone pair > lone pair bond pair > bond pair bond pair. 9. Bond order of He is () 0. As the bond order is zero, He molecule is unstable and hence, does not exist.. s-character decreases as the p-character increases : sp > sp > sp 3 s-character 50% 33.3 % 5%

6 05 States of Matter: Gases and Liquids Chapter 05: States of Matter: Gases and Liquids. At constant pressure, volume and temperature of a gas are related by V V T T VT V T V cm 3 Contraction in volume V V cm 3 53 cm 3 4. If V 0 is the volume of a given mass of gas at 0 C and V t is its volume at any temperature t C, then the volume, V t is given as t V t V V t Litre 5. The average kinetic energy of the gas molecules is directly proportional to the absolute temperature. Hence, at 98 K, the average kinetic energy of argon is same as that of helium. 6. Unit of viscosity kg m s Unit of surface tension kg s m s m s 9. Root mean square velocity. Since, CH 4 has the least molar mass (among the given Molar mass options), it has the highest root mean square velocity.. At STP, mol (44 g) of CO occupies.4 dm 3 volume.. litres CO mole of CO g

7 Chemistry Vol. (Med. and Engg.). u av of O 8RT 6 u rms of N u ofo 3RT 4 av urms of N 8RT RT 3. The temperature below which the gas can be liquefied by the application of pressure alone is called critical temperature. 4. Average speed 0.93 Root mean square speed , cm s 5. Volume of Helium: V nrt WRT P MP 4.0R P R 300 P Volume of hydrogen required V WRT MP W R 473 P Now, V V R300 W R 473 P P W g 473

8 06 Thermodynamics. The required equation is: P (s) + 5 Cl (g) PCl 5(s) P (s) + 3Cl (g) PCl 3(l) ; H x kj () PCl 3(l) + 3Cl (g) PCl 5(s) ; H y kj () Dividing equation () by and adding it to equation (), P (s) + 3 Cl (g) PCl 3(l) ; H x kj PCl 3(l) + Cl (g) PCl 5(s) ; H y kj P (s) + 5 Cl (g) PCl 5(s) ; H x + ( y) 4. P.06 atm V 500 cm 3.5 L w PV L atm L. atm 4. cal w cal (x + y) 5. For mol AgCl, H 65.5 kj i.e., for 43.5 g of AgCl, H 65.5 kj For 9.00 g of AgCl, H kj 6. G H TS 3.0 H 500(0.030) H.0 kcal H U + nrt n (3 0) 3.0 U U kcal Chapter 06: Thermodynamics 8. Work is not a state function as during a process its value depends on the path followed, whereas the values of enthalpy, internal energy and entropy depend on the initial and final states of the system and not on the path followed, hence these are state functions. 9. U U products U reactants When U is negative then, U reactants > U products i.e. heat is evolved in the reaction.

9 Chemistry Vol. (Med. and Engg.) 0. V 500 cm dm 3 V m 3 V 5 dm 3 V m 3 w PV (0.5 5) Joules. H H H N N H 4N H bonds are present Their energy So the bond energy of N N in N H kj mol. Sublimation is the process of conversion of a substance from solid state to the gaseous state directly without passing through the liquid state. 3. The equation for photosynthesis is the reverse of combustion of glucose. 6CO (g) + 6H O (g) C 6 H O 6(s) + 6O (g) ; H 7 kcal/mol Molecular mass of glucose 80 g 80g 7 kcal g kcal. 4. Given that, C (s) + 3H (g) C H 6(g) ; H 30. kcal C (s) + H (g) C H 4(g) ; H +.5 kcal Subtracting eq. (ii) from (i), we get C H 4(g) + H (g) C H 6(g) ; H 4.7 kcal 5. H U + nrt For the given reaction, n H U 3RT H < U (i) (ii)

10 07 Equilibrium Chapter 07: Equilibrium. Blood consists of H CO 3 + HCO 3 buffer solution.. K p K c (RT) n n (for the given equilibrium) Now, K p K c (RT) n (RT) (0.08 T) T K 3. K c K c K c K c for the backward reaction ( K ) (0.408) ph log K b + log [salt] [acid] 5 log log [salt] [acid] log [salt] [acid] log [salt] [acid] [salt] antilog 0 :. [acid] c 5. A (g) 3C (g) + D (s) For this reaction, n g number of moles of gaseous products number of moles of gaseous reactants n g 3 K K p K c [RT] p or K RT or K K p c RT c 6. It is not a mixture of weak acid or base and their strong salt. 7. K [HI] (80 ) [H ](0.5 0 ) 3 [H ][I ], 3 3 [H ] M 8. Reaction: A + B C + D Initial moles: 4 4 Moles at equilibrium: K

11 Chemistry Vol. (Med. and Engg.) 0. CH 3 COONa + H O CH 3 COOH + NaOH Weak acid Strong Base Aqueous solution of CH 3 COONa contains strong base, hence the resulting solution is alkaline (basic).. Acceptor of electron pair is known as Lewis acid. All can accept an electron pair so answer is (D). 3. [OH ] 0 5 poh log[oh ] 5 ph + poh 4 ph AgBr Ag (aq) + Br (aq) K sp [Ag + ] [Br ] [Ag + ] S, [Br ] S K sp [Ag + ][Br ] SS or S S mol L 5. A x B y xa y+ + yb x S xs ys K sp (xs) x (ys) y S x+y. x x. y y

I. Multiple Choice Questions (Type-I) is K p

Unit 7 EQUILIBRIUM I. Multiple Choice Questions (Type-I) 1. We know that the relationship between K c and K p is K p K c (RT) n What would be the value of n for the reaction NH 4 Cl (s) NH 3 (g) + HCl