UNIT # 01 MOLE CONCEPT EXERCISE # H 2. : O 3 no. of atoms=2n A. : H e : O 2

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1 UNIT # 0 MLE CNCEPT EXERCISE #. No. of molecule () Mole N A N nn A 5. At STP or NTP volume of any gas (STP NTP ). L 6. gram ion mole ion N A ion mol Al + ion N A Charge (e) on mol Al + ion N A e columb. ( Al + ion N A e columb.) 7. No. of molecules () mole N A i.e., mole is equal then no. of molecules are also equal ( ) 8. Mole of Al t 5 mol At t 7 that is same for Mg atom (Mg ) So mol of Mg t t 8 g. 0. No. of oygen atom mole N A atomicity. ( N A ) (A) (B) (C) 6 N N A A 6 all are same. N A N A 6 8 N N A A 6. (NH ) P mol hydrogen atom contain atom of oygen ( ) mol hydrogen atom contain.8 mol hydrogen atom contain (.8 ).8.08 mole. Mass of e ( e ) 9. 0 kg kg g compound contain (00 g) 5.7 g Nitrogen () g Nitrogen H : H e : : no. of atomsn A : N A : N A : N A : : : 6. 6 Cu 65 Cu % abundance(% ) 00 Avg. mass () 6 65(00 ) M % 7. % by t. of H (H %) t. of H (H ) M Total t. of compound ( ) % Mol Simples ratio() C / 7. 7./7. H../../7. Emperical formal () CH PM DRT M DRT P n Molecular t.( ) 56 Et.( ) 56 Molecular formula () n E.F. CH C H 8 9. Element % Mole Simplest ratio () () C / 6 6/0 0 0 H 6. 6./ 6 6/0 0 0 N../ /6. E.F. C 0 H 0 N. M + 6F MF 6 Mole of M Mole of MF 6 t t Mole t Mol. t so element () is Mo

2 . NaH contain mole of atoms (NaH ) so mol of NaH ( NaH ) mol t. of NaH (NaH ) 0 0 g 0 % purity (% ) 00 % 000. Molarity of Cl (Cl ) MV MV Total vol.( ) X CH.5 X H.75 n C H.5 C H.5 6.5g 60 M 60 n H.75 % t of C H (C H % ) % 5. Mole of N P (N P ) m mol.008 mol Na P contain Na + ion (Na P Na + ) mol 7. Molality of H is 9 (H 9 ) i.e. 9 mole of H in kg solvent ( kg 9 H ) kg solvent contain 9 mole H ( kg 9 mole H ) kg solvent contain 9 98 t of H ( kg 9 98 t H ) 000 kg solvent contain (000 kg ) 9 98/ kg solvent contain (90 kg ) 80.6 g t. of solvent () 90 g t. of solution () % by t ( %) 7.6 g 8. R.D. Density ( Density ( ) of ) of at same temp. & pressure of density M M M ( M ( 9. A 0. H ) of ) of 8.5 t of H g Molality () moles of solute ( ) t. of solvent ( ) (H )in kg % by mass volume solution of KH (KH.8 %) i.e.,.8 g KH in 00 ml solution ( 00 ml.8 g KH) molarity ().8.5 M 56.. Molality of H (H ). mol/kg. mol H then t (. H ) g t. of solvent () kg 000 g t of solution () Molarity moles of solute ( ) vol of solution ( ) mol of solution (). Moles of solute () N A 0. mol 09.6 g ml concentration of solution() t of solute ( ) t of solution ( ) 00 moles vol

3 MLE CNCEPT EXERCISE #. 8.5% Ag solution 8.5% Ag Molarity () i.e. 8.5 g Ag contain in 00 g 00 g 8.5 g Ag. KCl KCl + mole.5 g mol moles of solute ( ) Vol. of solution mol L 08. ppm moles of solute ( ) mass of solution ( ) moles of solute ( ) mass of solution ( ) 00 Mass () % 0.0. Molarity () ( / ) d 0 Molar mass of solute ( ). 0 0 mol of solute ( ). 0 Vol 0 Vol.7 L Molarity () 8.9mol L 8 5. Molarity () M 6. Molality () 6 moles of solute ( ) mass of solvent in kg ( kg ) m XeF 6 + I 6 IF 7 + 7Xe 7 mol 6 mol m mol % Loss () C 6 NH +HN +HCl C 6 N + Cl + H C 6 N + Cl + KI C 6 I + N + KCl n P n r R R moles of C 6 I mole of C 6 NH R R t t. 0. g % yield of C 6 I (C 6 I % ) %. Let assume % of H is (H %, ) % of H (H %) % of C (C %) 6 % of N (N %) 7.5 Element % Ratio of mol Simplest () () () H / 6 C 6 6/ / N 7.5 F.F C H 6 N 7.5 atomic mass () 70 molar mass () 0 5. mole simple ratio () () % X /0 % Y /0 E.F X Y

4 6. 7 g Na contain salt (7 g Na )00g g 00 7 g 9 7. mole () atom ().8 % N.8.8/ N A 9.8 % S / N A 7 % Na 7 7/ A N 7 N A atom of Na contain (Na 7 N.8 N of N A ) A atom of Na contain (Na ) atom of N 7 N 9.8 N of S A atom of Na contain A (Na 7 N 9.8 N ) A S A atom of Na contain (Na ) 9.8 atom 7 8. C (NH ) NH C NH C NH + 6NH molecule molecule 0 molecule 0 mol mass X 6 X + X n X 6 n X 0 X amu t. of molecule ( ) volume occupied by its () 6 0 / mass ( ) density ( ) 9. 0 cc ml. Mn + HCl MnCl + Cl + H L.G molality () M d M M. + Al Al mol 7 5 g g % by t. of H (H %) t.of H (H ) 00 Total t.( ) / t. of carbon g g carbon contain 00 g cortisone (69.98 g00 g ) g carbon contain 00 g cortisone ( g 00 g ) g r. 7. no. of mol (7) 0 00

5 0. 0% (v/v) HCl 00 ml contain 0 ml HCl (00 ml 0 ml HCl) 0% (v/v) NaH i.e. 00 ml contain 0 ml NaH density () of NaH.5 density of HCl M V NaH.5 Resultant Basic (). C H y + a a + M V HCl y C + y H y a a ay y a 600. (a) In 00 ml (0 g) solution mass of solute (00 ml (0 g) ) g Mass of solute ( ) / 6 (b) 0 M Mass of solute () g (c) 00 g solution contain 5 g of solute mass of solute ] (00 g 5 g ) (d) 5 M Mass of solute ( ) / 6 6 / 000 Mass of solute () 0.58 g a + a y y y/ 7 5y / < 5 put the value () if 0 5y/ y 8 Ans is C H 8. B H PtCl 6 Pt PAc on Pt M B n 95 0, n diacidic org. base () Molarity () X(volume( ) ) M 000 m d 000 M M m H.8.5 moles in L solution ( L.5 ) d 65 g/l, mass () 65 solution () n H % t.of solute ( ) V volume of solution ( ) (ml ) % 8% V

6 MLE CNCEPT EXERCISE # CMPREHENSIN #. The cost of 000 gm KCl is 50 kg (000 gm KCl 50 kg ) The cost of 7.5 g KCl is (7.5 g KCl ) mol 000. the price of K (K ) Rs..8 kg. mole of K in KCl (KCl K ) mole of K form. mole of K (K K. ). mass of K (K ) gm 0.6 kg CMPREHENSIN #. amt mole fraction C H C a Simplest formula () C 6 H 0 Ca 6 Ca 6 C 6 H 0. Formula eight (). The molecular mass of lactate pentahydrate 08 ( 08) 8 gm anhydrous salt recovered 08 g lactate pentahydrate gm anhydrous salt recovered 08. gm 8 CMPREHENSIN #. 8 mole NaBr obtain from(8nabr ) mole Fe (Fe ) mole of Fe mole NaBr mass of Fe kg 0 8. mole of Fe Br 8 (Fe Br 8 ) mole of Fe mole FeBr mass of Fe mass of Fe kg. mole of C (C mole of NaBr CMPREHENSIN #. C g 0.5 mol H.5 g.5 mol. 8 C 0.5 mol 6 g H.5 mol.5 g 8 gm 0.5 mol E.F. CH let molar mass M M 08 M E.F. mass n 7 M.F. (CH ) C H 6 CMPREHENSIN # 5. Ba(H) + HN Ba(N ) + H 0. mole 0.mole In resultant sol. Ba(H ) is remaning, therefore nature of sol. basic. (Ba(H ) ). Vol. of Ba(H) (Ba(H) ) 600 ml 0.57 mole of H (H ) molarity of H (H )

7 MLE CNCEPT. Ist ep. Cu.75 gm Cu.098 gm 0.77 gm IInd ep. Cu.79gm Cu.76 gm gm Cu.968 Cu In both the cases ratio of Cu/ is same (Cu/ ) Y r X 0. Y r X 0.7 r.996 r so satisfy la of multiple proposition. () gm molecular () N A nnacl 6. np 06.5 mole 06.5 N. of atom of (N ) Na N A Cl N A N A P mole N 0.75 N.5 0 molecules P A N molecules nna 0.5 mole 8. (a) gm (b) 5.5 gm (c) 6.5 gm 9. m Hg gm n Hg m Hg /00 68 mole 0. CaC + H P Ca (P ) + H + C 50/00mole 70/98 mole Limiting reactant () 0.5 m CaC M Ca (P ) 5.66 gm m M.gm HP HP EXERCISE # [A]. ClNH + NH N H + NH Cl 000 mole mole ecess () % yield () % C + S ecess() 50 8% CS + C 7.0 Kmole Kmole 9.09 kg. Ba + Ca [5] + y [56] 8...(I) Ba + HCl BaCl + H Ca + HCl CaCl + H y y + y (II). 5. % of Ba gm 0.95 (7 / 98) M (00 /.) (n ) 6. C n H n % nc + (n + ) H (n ) / 7 6n 7n n C n H 6 7. C H C + H 5 C H 0 + C + 5H y y 5y ( + 5y) y () ( + 5y) y 0....(II) + y 0....(I) y 0.0 % by mass of C H 0 (C H 0 %) %.86

8 8. y C H y + C + y v ecess () y v y v v y y +v v v v.5 v H y.5 y 6 v v 9. Molar mass () C H gm 0. M ( cm) ( cm) cm / gm gm M gas.7 M M gm gas air. Y A (a) y.95%, Al.7%,.% (b) 7.98 gm n mole [ ]. 6LiH + 8BF 6LiBF + B H Al + HCl AlCl + H.5 + y (I) mole y Mg + HCl MgCl + H 9 + y (II) y mole y 6. % Al % Mg 5.% CaCl + NaCl Na C CaC Ca.6 ncacl n Ca m g CaCl m NaCl gm % NaCl 67.9% 7. n n C mole n n C C C + C n n n...(i) C C nc nc nc...(ii) nc 8 n C elem en t ma s s per 0 0 g m m o le s im plest r atio C / 5 H.8.8 / N / E.F. C 5 H N 0 M.F. (77.5.5) nn mole mn gm % N % 0.. (a) M / (b) M 5. / (c) M 0.65 / X ethanol 6 / 6 6 / 6 5 / C + C / Cl + + C + H () y y y y (I) + y 7...(II) + y + z 0...(III) 5 ml y ml z ml Increase in volume()0 000ml

9 MLE CNCEPT. Empirical formula: KAlS 8 Al K S Elements Mass percentage Mole ratio 8 Simple ratio Empirical formula eight ) 58 From eight loss information : 5. g anhydrous salt 5.6 g H (: 5. g 5.6g H ) 58 g anhydrous salt 6.6 g mol H (58 g 6.6 g H ) Empirical formula of hydrated salt KAlS 8. H KAlS 8. H )..0 mole of KCl.0 mole of Zn mole KCl mole of Zn.5 8. g Zn. Apply conservation of moles of silver before and after precipitate echange reaction as : ( :) here, is mass of AgBr in mied precipitate. (, AgBr ).06 Also, moles of CuBr moles of AgBr ) (,CuBr AgBr Mass of CuBr (on substituting ) (CuBr ( )) Mass % of CuBr (CuBr %).8. Moles of NaCl in sample 0.0 moles of AgCl from NaCl in precipitate ( NaCl 0.0 NaCl AgCl ) EXERCISE # [B] Total moles of AgCl precipitate ( AgCl ) Moles of AgCl from KCl moles of KCl (KCl AgCl KCl ) Mass of KCl in sample g (KCl g) Mass % of KCl in the sample 9.8 (KCl %) 5. Let the miture contain g Cu. 5H. ( g Cu. 5H. ) Mass percentage of Cu.5H (Cu.5H ) Mass % of Ca (Ca %) Mass % of S (S %) Mass % of N (N %) Mass % of C (C %) 5.8 No : Elements () C a S N C Mass % ( %) Mol ratio () Simple ratio () Empirical formula () CaC N S, Empirical formula eight () 56 Hence, molecular formula(,)cac N S 7. Working in backard direction( ) In the last step moles of(agbr+agi) moles of AgI ( (AgBr + AgI) AgI ) Mass % of NaI (NaI ) g No subtracting mass of AgI from st and nd precipitate gives ( st nd AgI ) :

10 Mass of (AgCl + AgBr) 0.87 g ((AgCl + AgBr) 0.87 g) and mass of AgBr 0.98 g ( AgBr 0.98 g) y Again y y 0.5g Mass % of NaCl (NaCl ) Mass % of NaBr (NaBr ) Weight loss is due to conversion of NaHC into Na C : g eight is lost per mole of NaHC. (NaHC Na C : NaHC g ) 0. g t. loss from 0. mol of NaHC producing 0. 6 moles of Na C. ( 0. NaHC Na C ) 0. g 0. 6 Total moles of carbonate()5 0 Moles of carbonate in original sample ( ) Mass of Na C in original sample ( Na C ).06. % Na C 9. If M is molar mass of (CH ) AlCl y ( M, (CH ) AlCl y ) m(ch ) 0.6 M and m(agcl) 0.6 y M 6 0. dividing () :, y Also M y y Mass of AgCl g hich is % of total ppt. (AgCl g % ) Total mass of precipitate().85g and mass of impurity ( ) 0.57 g Mass of NaCl + KCl 5.9 g (NaCl + KCl 5.9 g) g NaCl,.96 g KCl m (Na ).558 g m% (Na ).6 m (K ).867 g m% (K ) 7.. In order to obtain maimum yield from a reaction, the reactants must be supplied in stoichiometric amount so that no reactant should be left unreacted. ( ) The balanced chemical reaction is,(,) Pb(N ) + KI PbI + KN Let g of KI is taken ( g KI ) moles of KI 66 moles of Pb(N ) present g mass of PbI 60.6 g. Mass of uranium in the sample ( ) g Mass % of uranium in the sample 89. (% 89.) U (N ) +Na C +H U (C ) H m mol NaN Here Na C is the limiting reagent, therefore, m mol of U (C ).H formed is.985. ( Na C, U (C ).H m mol.985 ) M(U (C )).H Volume of smallest cell r l ( cm) ( cm) cm ( r l ( cm) ( cm) cm ) mass of one smallest cell () g Molar mass of mother cell ( ) amu

11 . Let the sample contain () g Mohr's salt () [Fe (NH ). 6H ] Solving 0. g Mohr's salt %, (NH ) 5% ( 0. g %, (NH ) 5%) Also moles of Fe in 0.g sample.7 0 ( 0. g Fe ) mass of Fe obtained on ignition of 0. sample mg (0. Fe ) 5. Smallest volume of AgN ould be required hen the entire mass is due to highest molecular eight constituent. (AgN ) Hence, for smallest volume, the hole mass should be of BaCl. H (BaCl. H ) m mol of BaCl.H m mol m mol of AgN required.9.58 ( AgN m mol) Volume of AgN required ml 0.5 (smallest) ( AgN ml ()) Largest volume of AgN ould be required hen entire mass is due to loest molecular eight constituent, i.e., NaCl. (AgN NaCl ) m mol of NaCl m mol 58.5 of AgN required AgN m mol) (NaCl m mol 0. Volume of AgN required ml 0.5 ( l a r g e st ) (AgN ml 0.5 ()) 6. Miture(N,N,N )has mean molar mass55.. ((N, N, N ) 55..) y z Given : N N z z 8 6(y z) 55. y z mean molar mass t. mole Total mole Given : + y + z (mole) so (y + z)...() 8 6(y z) 9. y z 9.6 ( + y + z) (y + z) From eq () & + y + z or 9.6 ( + z) 59. or + z or z 0. from eq. () (y + z) z 0. put () y y () + y + z + y + 0. ( z 0.) + y () eq. ()... eq. () y y y.8 y 0. + y + z 0.5

12 y z 7. C H y z + C + y H Given vol. 0mL + 00mL () y z After reaction () y z y z y z...() Property of KH has to absorbed all C. (KH C ) 0 0 V.D. of compound (C H y z ) ( (C H y z ) ) V.D. M 6 M 6 + y + 6z 6 + y + 6z 6 y + 6z...() from eq. () & () y z y + 6z 8z 8 z, y 6 Molecular formula () C H 6. M 8. (CH ) n + ncof (CF ) n + nhf + ncof...() CoF + F CoF t. F 9, C, Co 59, M t. (CF ) n 50n from eq. () (CF ) n ncof E n 000 n 97 50n g (CoF ) CoF + F CoF 97 8 E n g.5kg. 97 HF H + F (CF ) n moles moles of (CF 50n n ) n 0 0 n HF 0 n 0 mol n kg 9.(a) A + B A B A + B A B Initial - After 0 A B + A A B A, A B (b) A + B A B A + B A B Initial After A B 0.5, B (c) A + B A B A + B A B Initial.5 After A B + A A B A B A B (a) L KMn 79% (/v) i.e. 00 ml solution contain 79 g KMn (L KMn 79% (/v) 00 ml 79 g KMn ) t. moles of KMn M (KMn 79 M ) Molarity ) (M) M 00 HCl0%(/)i.e.00 g solution contain 0g HCl (HCl 0% (/) 00 g 0 g HCl ) D.85 g/ml V M 00 D

13 Molarity ) M KMn +6HClKCl + MnCl + 8H + 5Cl M V M V Cl mol. (b) KMn +6HCl KCl+MnCl + 8H + 5Cl (c) L vol.of ater treated vol.of total feed. 00 ml milk ml fat (00mL ml ) L milk 0 ml fat ( L 0 ml ) density of fat 875 kg/m g/ml ( 875 kg/m g/ml) mass of fat g ( g) fat free milk mass g ( g) Vol.() ml g/ml 5 vol. of KMn. D.0 g/cm HCl % NaCl 00 g solution contain.8 g NaCl. (.8% NaCl 00 g.8 g NaCl ) 00 V L L g 00 moles M V M V V V 9 0 so ater evaporated () L. Let free S g ( S g) S in form of H (H S ) so total () ater required g % oleum % ( g % %). H S S S S.5% a b a + b + c g c so, C 0.05 g S a + b g H + NaH Na a 98 S + NaH Na + H b 80 S + NaH Na S + H a b a b a +.5 b.75 a + b b 0.5 b 0.6 g S a 0.5 M g H Combined () S g

14 5. Volume() m 0.5 cm g/cm mass () 0.5 g n(ch ) SiCl +nh ncl +nh + [(CH ) Si] n 7. CaCl 5M 555 g in L solution or in 050g solution (CaCl 5M L 050g 555g) t. of (solvent + MgCl ) g 9 {7 n} 9n ((+ MgCl ) g) g MgCl 5 m 6. CH + C + H a a a C n H n + n nc + (n ) H 000 g solvent 5 mol of MgCl (000 g MgCl 5 ) g MgCl (0 a) n (0 a) n(0 a) i.e., 75 (solvent + MgCl ) 75g MgCl For methane( a + n (0 a ) 0...() For oygen() n 00 a (0 a) 0 a + n (0 a) 60 a + 0 n.5na a 60.5 a.5na + 0 n 70.5 a.5n (a 0) 70.5 a +.5n (0 a) 70...() from () & () a 0 n (, 75 ( + MgCl ) 75g MgCl ) 95 (solvent + MgCl ) 59. g MgCl (95 ( + MgCl ) moles of MgCl (MgCl ) Total moles of Cl (Cl ) ( ).56 volume of solution () L C H % composition % 50% Molarity of Cl (Cl ).56 M

REDOX REACTION EXERCISE # 1

REDOX REACTION EXERCISE # 1 REDX REACTIN EXERCISE #. No. of equivalent mole n-factor ( n-) S 3 + H S + H + + e...() n-factor for R N () is () 0. n. n. n Final oidation state will be (3 ) ((3 ) ). Meq.() of K Cr 7 Meq.() of ABD n-factor