94420 + Interview Questions in Chemistry in Class 12

1.	Calculate the pH of solution with H_3O^+ concentration in mol dm^(-3). (i) 10^(-4)
Answer» SOLUTION :`pH=-log[H_3O^+]` (i) `pH=-log[10^4]` `pH=log1-log10^4` `pH=4`

Discussion

2.	Calculate the pH of solution with H_3O^+ concentration in mol dm^(-3). (ii) 10^(-7)
Answer» SOLUTION :`pH=-log[H_3O^+]` (II) `pH=-log[10^(-7)]` `pH=log1-log10^(-7)` `pH=7`

Discussion

3.	Calculate the pH of solution obtained by mixing 10mL of 0.1M HCl and 40mL of 0.2M H_(2)SO_(4) take log3.4=0.53
Answer» Solution :Given is the CASE of a mixture of `2` strong acids. MILL moles of `H^(+)` form `HCl =10xx0.1=1` MILLI moles of `H^(+)` form `H_(2)SO_(4)=40xx0.2xx2=16` ,.TOTAL millimoles of `H^(+)` in solution =`1+16=17` `:.[H^(+)]=(17)/(50)=3.4xx10^(-1)( :.[H^(+)]_(f)=(("Millmoles"_("Total"))/(V))` `:.pH=-log[H^(+)]=-log0.34` `pH=0.47`

Discussion

4.	Calculate the ph of solution obtained by mixing 100 ml of 0.1 M HCl and 9.9 ml of 1.0m H_2SO_4.
Answer» 3.0409 3.4049 2.0409 None Answer :A

Discussion

5.	Calculate the pH of an aqueous solution of 1.0 M ammonium formate assumingcomplete dissociation. pK_aof formic acid = 3.8 and pK_bof ammonia = 4.8.
Answer» ANSWER :6.5

Discussion

6.	Calculate the pH of a solution obtained by mixing equal volume of 0.02MHOCl &0.2MCH_(3)COOH solutions Given that K_(a)(HOCl)=2xx10^(-4),K_(a)(CH_(3)COOH)=2xx10^(-5) Also calculate [OH^(-)],[OCl^(-)],[CH_(3)COOH] at equilibrium .Take log2=0.3
Answer» Solution :Volume of final solutions BECOMES double . So , concentrated becomes half so after mixing: `C_(1)=0.01M,C_(2)=0.1M` `[H^(+)]=sqrt(C_(1)K_(a_(1))+C_(2)K_(a_(2)))=sqrt(2xx10^(-4)xx0.01+2xx10^(-5)xx0.1)=sqrt(2xx10^(-6)+2xx10^(-6))=2xx10^(-3)M` `:.pH=3-log2=2.7` `[OCl^(-)]=(0.01xx2xx10^(-4))/(2xx10^(-3))=1xx10^(-3)M,[CH_(3)COO^(-)]=(0.1xx2xx10^(-4))/(2xx10^(-3))=1xx10^(-3)M`, `[OH^(-)]=(K_(w))/([H^(+)]=(10^(-14))/(2xx10^(-3))=5xx10^(-12)M`

Discussion

7.	Calculate the pH of a buffermixture which contains 7.5gms if acetic acid and 10.25 gms of sodium acetate in 1 litre of the solution. K_a for acetic acid is 1.8xx10^(-5).
Answer» Solution :`pH=pK_a+log.(["salt"])/(["acid"])` The CONCENTRATION of the salt and the acid should be in MOLES/lit. Number of moles of acetic acid `("weight of acetic acid")/("molecular weight of acetic acid")=(7.5)/(60)` Number of moles of sodium acetate `=("weight of sodium acetate")/("molecular weight of sodium acetate")=(10.25)/(82)` `pK_a=-log K_a` `=-log 1.8xx10^(-5)` `=log .(1)/(1.8xx10^(-5))` `=log 1-log 1.8-log 10^(-5)` `=5-0.2553=4.7447` `therefore pH=4.7447+log.(0.125)/(0.125)` `pH=4.7447.`

Discussion

8.	Calculate the pH of a buffer containing 0.08 mole of acetic acid and 0.12 mole of sodium acetate per dm_2 of the solution. The ionisation constant of acetic acid is 1.8xx10^-5.
Answer» SOLUTION :The pH of a buffer solution is GIVEN by the Henderson's equation . `pH=pK_a+log.(["SALT"])/(["ACID"])` `pK_a=-logK_a=-log[1.8xx10^(-5)]` `=log.(1)/(1.8xx10^(-5))` `=log 1-log1.8-log 10^(-5)` `=5-0.2553` `=4.7447` `pH=4.7447+log.(0.12)/(0.08)` `4.7447+log1.5` `=4.7447 +0.1761` `pH=.921.`

Discussion

9.	Calculate the pH of a 0.15 M aqueous solutions of AlCl_(3) Given :[Al(H_(2)O)_(6)]^(3+)(aq)+H_(2)O(l)hArr[(AlH_(2)O)_(5)OH]^(2+)(aq),K_(a)=1.5xx10^(-5)
Answer» `2.82` `5` `9` `11.18` Solution :`{:(,AlCl_(3),C=0.15M,(AlCl_(3))"is SAWB SALT",,),(,[AL(H_(2)O)_(6)]^(3+)+,H_(2)O(l)HARR,[Al(H_(2)O)_(5)OH]_(AQ)^(2+)+,H_(3)O_(aq)^(+),K_(a)=1.5xx10^(-5)),(At t=0,0.15,-,0,0,),(At t=t_(eq),0.15(1-alpha),-,0.15alpha,0.15alpha,):}`

Discussion

10.	Calculate the pH of a 0.10 N solution of H_2CO_3
Answer» 3.68 5.76 7.36 9.34 Answer :a

Discussion

11.	Calculate the pH of a 0.005 M Na_2 S solution. K_1 and K_2 for H_2S are 1xx10^(-7) and 1xx 10^(-14)respectively.
Answer» Solution : The first STEP of HYDROLYSIS, i.e., `S^(2-) + H_2O IFF HS + OH^-`is predominant and hence `K_2`value is used in the calculations. 11.70

Discussion

12.	Calculate the pH of 1.5xx10^(-3) M solution of Ba(OH)_2.
Answer» Solution :`{:(Ba(OH)_(2)" "RARR" "Ba^(2+)+2OH^(-)),(1.5xx10^(-3)M""2xx1.5xx10^(-3)M):}` `[OH^(-)]=3xx10^(-3)M` `[because pH+pOH=14]` `pH=14-pOH` `pH=14-(-LOG[OH^(-)])` `=14+log[OH^(-)]` `=14+log(3xx10^(-3))` `=14+log3+log10^(-3)` `=14+0.4771-3` `=11+0.4771` `pH=11.48`

Discussion

13.	Calculate the pH of 1.5 times 10^-3 M solution of Ba(OH)_2.
Answer» SOLUTION :`Ba(OH)_2toBa^(2+)+2OH^(-)` `1.5 times10^-3M 2 times1.5 times10^-5M` `[OH^-]=3 times10^-3M` `[becausepH+pOH=14]` `pH=14-pOH` `pH=14-(-LOG[OH^-])` `=14+log[OH^-]` `=14+log(3 times10^-3)` `=14+log3+log10^-3` `=14+0.4771-3` `=11+0.4771` pH=11.48

Discussion

14.	Calculate the pH of 0.5 M aqueous solution of NaCN, the pK_(b) " of " CN^(-) is 4.70
Answer» `4.70` `11.5` 7 6.5 Solution :NACN is a salt of STRONG base and WEAK acid , `pH =7+(1)/(2)pK_(a)+(1)/(2)log C` `pK_(a) "for " HCN=14-4.70=9.30` `therefore pH=7+(1)/(2)xx9.30+(1)/(2) log 0.5, pH=11.5`

Discussion

15.	Calculate the pH of 0.1M CH_3COOH solutoion. Dissociation constant of acetic acid is 1.8 times 10^-5.
Answer» Solution :`pH=-log[H^+]` For weak ACIDS, `[H^+]=sqrt(K_s times C)` `=sqrt(1.8 times 10^-5 times 0.1)` `=1.34 times 10^-3 M pH=-log(1.34 times 10^-3)` `=3- log1.34` `=3-0.1271` `=2.8729 APPROX 2.87`

Discussion

16.	Calculate the pH of 0.1 M solution of (i) NaHCO_(3) , (ii) Na_(2)HPO_(4) and (ii) NaH_(2)PO_(4). Given that : CO_(2)+H_(20O Leftrightarrow H^(+) + HCO_(3)^(-) K_(1)=4.2 xx 10^(-7) M HCO_(3)^(-) Leftrightarrow H^(+)+CO_(3)^(2-), K_(2)=4.8 xx 10^(-11)M H_(3)PO_(4) Leftrightarrow H^(+) +H_(2)PO_(4)^(-), K_(1)=7.5 xx 10^(-3)M H_(2)PO_(4)^(-) Leftrightarrow H^(+)+HPO_(4)^(2-), K_(2)=6.2 xx 10^(-8)M HPO_(4)^(2-) Leftrightarrow H^(+)+PO_(4)^(3-), K_(3)=1.0 xx 10^(-12)M
Answer» ANSWER :8.35,9.60,4.66

Discussion

17.	Calculate the pH of 0.1 M solution of H_2NCH_2 CH_2NH_2 , ethylenediamine (en). Determine the en H_2^(2+). concentration in the solution. K_(b_(1)) and K_(b_(2)) values of ethylenediamine are 8.5 xx 10^(-5) and 7.1 xx 10^(-8) respectively.
Answer» ANSWER :`PH =11.46 [en H_(2)^(2+)]=7.1 XX 10^(-8)`

Discussion

18.	Calculate the pH of 0.1 M CH_3 COOH solution.Dissociation constant of acetic acid is 1.8xx10^(-5) M.
Answer» Solution :For WEAK ACIDS, `[H^+]=sqrt(K_axxC)` `=sqrt(1.8xx10^(-5)xx0.1)=sqrt(1.8xx10^(-6))` `=1.34xx10^(-3)M` `therefore pH=-LOG[H^+]=log(1.34xx10^(-3))` `therefore pH=2.87`.

Discussion

19.	Calculate the pH of 0.1 M acetic acid if its ionisation constant K_a=1.8xx10^(-5).
Answer» Solution :Degree if dissociation `alpha=SQRT((K_a)(C )), C alpha = sqrt(K_a . C)` `THEREFORE pH=log.(1)/(sqrt(K_a. C))=log.(1)/(sqrt(1.8xx10^(-5)xx.1))` `1=log.(1)/(sqrt(1.8xx10^(-3)))=log.(10^3)/(sqrt(1.8))` `=3-(1)/(2)log1.8` `=3-(1)/(2)xx0.2553=3-0.1276=2.8724` Alternation, `[H^+]=sqrt(K_a xx C)` `=sqrt(1.8xx10^(-5)xx0.1)=sqrt(1.8xx10^(-6))` `=1.341xx10^(-3)` `therefore pH=-log[H^+]=-log(1.341xx10^(-3))` `=3-0.1277=2.8723`.

Discussion

20.	Calculate the pH of 0.04 M HNO_3 Solution.
Answer» SOLUTION :CONCENTRATION of `HNO_3=0.04M` `[H_3O^+]=0.04` MOL dm `pH=-log[H^3 O^+]` `=-log(4xx10^(-2))` =2-log4` `=2-0.6021` `=1.3979` `=1.40`

Discussion

21.	Calculate the pH of 0.02 MHCI.
Answer» Solution :( HCI strong ACID FULLY ionised) `pH=log.(1)/(H^+)=log.(1)/(01)` `=log. (100)/(2) = 1.6990`

Discussion

22.	Calculate the pH of 0.02 m Ba(OH)_2 aqueous solution assuming Ba(OH)_2 as a strong electrolyte.
Answer» SOLUTION :`BA(OH_2) to Ba^(2+)+2OH^-` `THEREFORE [OH^-]=2[Ba(OH)^2]` `=2xx0.02 =0.04 M` `therefore pOH=-log [OH^-]` `=1.398 - 1.40` `therefore pH=14-1.4=12.6`

Discussion

23.	Calculate the pH of 0.01 M NaOH.
Answer» Solution :(NaOH strong BASE fully ionised) `pOH=LOG.(1)/(OH^-)=log. (1)/(.01)` `=log.(100)/(1)=2.0` `pOH=14-pOH=14-2=12`.

Discussion

24.	Calculate the pH of 0.001M HCl solution HCl_(0.001M) leftrightarrow^(H_2O)H_3O_(0.001M)^(+)+Cl_(0.001M)^(-)
Answer» Solution :`H_3O^+` from the auto ionisation of `H_2O(10^-7M)` is negligible when compared from `10^-3 M HCL` HENCE `[H_3O^+]=0.001 MOL dm^-3` `pH=-log_10[H_3O^+]` `=-log_(10)(0.001)` `=-log_10(10^-3)=3`

Discussion

25.	Calculate the pH at which the following conversion (reaction) will be at equilibrium in basic medium ? I_2(s)hArr I^(-)(aq)+IO_(3)^(-)(aq) When the equilibrium concentrations at 300 K are [I^(-)] =0.10 M and [IO_3^(-)]=0.10 M {Given that DeltaG_f^@(I^(-)aq)=-50 kJ/mole , DeltaG_f^(@)(IO_3^(-),aq)=-123.5 kJ/mole , DeltaG_f^(@)(H_2O,l)=-233 kJ/mole , DeltaG_f^(@)(OH^(-),aq)=-150 kJ/mole , Ideal gas constant =R=25/3 J "mole"^(-1)K^(-1),log e =2.3 , deltaG_f^@(element, standard state )=0, DeltaG_r^@(reaction)=sumv_pDeltaG_f^@(products)-sumv_rDeltaG_f^@(reactants), where v_p and v_r are the stochiometric coefficients in the balanced chemiclal equation. }
Answer» SOLUTION :Balanced equation will be `3I_2(s)+6OH^(-)hArr5I^(-)(AQ)+IO_(3)^(-)(aq)+3H_2O(l)` `DeltaG^(@)=-172.5 "kJmole"^(-1)=-25/3xx300xx2.3xx10^(-3)`log k log k=30 `10^(30)=(10xx^(-5)xx10^(-1))/([OH^(-)]^(6))` so `[OH^-]=10^(-6)` and therefore `[H^+]=10^(-8)` so, pH=8

Discussion

26.	Calculatethe pH at whichan acidindicator with K_(a)= 1xx 10^(5) changescolourwhenindicator concentrationis 1 xx 10^(-3)M . Alsorport the pH at whichcoloured ionsare 80 %present.
Answer» SOLUTION :Forindicatordissociation equilibrium `HlnhArr H^(+)+In^(-)` ColourAColour B ` K_(In) = ([H^(+)][In^(-)])/([HLN]) ` `rArr1 xx 10^(-5) = ([H^(+)]xx 80//100)/(20//100)` ` :.[ H^(+)] = 0.25 xx 10^(-5)` `:. PH = 5.6020`

Discussion

27.	Calculate the pH at the equivalence point when a solution of 0.01 M CH_(3)COOHis titrated with a solution of 0.01 M NaOH.pKa of CH_(3)COOH is 4.74
Answer» Solution :`CH_(3)COOH + NAOH Let acid be = V ML VmL of 0.01 M `CH_(3)COOH` will require V ml of 0.01 mL NaOH. But `CH_(3)COONa` formed will make solution alkaline due to hydrolysis `CH_(3)COONa = 0.01/2 = 0.005` M USING EQUATION for PH of salt of weak acid and strong base. `pH = 7 + (pK_(a))/2 + (log C)/2 = 7 + (4.74)/2 + (log 0.005)/2` = 8.22

Discussion

28.	Calculate the percentage of water of crystallisation in the sample of washing soda, Na_(2)CO_(3).10H_(2)O.
Answer» SOLUTION :`62.94%`

Discussion

29.	Calculate the percentage of water of crystallisation in the sample of blue vitriol (CuSO_(4).5H_(2)O).
Answer» Solution :`"MOL. MASS of "CuSO_(4).5H_(2)O=63.5+32+4xx16+5xx18=249.5` `"No. of PARTS by mass of "H_(2)O=5xx18=90""therefore""%" of "H_(2)O=(90)/(249.5)xx100=36.07%.`

Discussion

30.	Calculate the percentage of water of crystalline in Blue vitrial CuSO_4.5H_2O (Atomic masses aregiven as Cu = 63.5, S = 32, O = 16, H = 1
Answer» 0.072 0.36 0.5642 0.64 Answer :B

Discussion

31.	Calculate the percentage of vacant space in a Si unit cubic cell. The unit-cell content for Si is 8 and r=(sqrt(3)a)/8.
Answer» ANSWER :`66%`

Discussion

32.	Calculate the percentage of the naturally occurring isotopes .^(35)Cl and .^(37)Cl that accounts for the atomic mass of chlorine taken as 35.45.
Answer» Solution :Suppose `.^(35)Cl` present `=x%`. Then `.^(37)Cl` present `=(100-x)%` `therefore"AVERAGE atomic MASS"=(x xx35+(100-x)xx(37)/(100))=35.45"(Given)"` `"or"35x+3700-37x=3545 or 2x=155 or x=77.5%` Thus, `.^(35)Cl=77.5% and .^(37)Cl=100-77.5=22.5%.`

Discussion

33.	Calculate the percentage of (i) SO_(4)^(2-) (ii) H_(2)O in pure crystals of Molar salt, viz., FeSO_(4).(NH_(4))_(2)SO_(4).6H_(2)O.
Answer» Solution :`"Mol. MASS of "FeSO_(4).(NH_(4))_(2)SO_(4).6H_(2)O=56+32+64+36+32+64+108=392` `%" of "H_(2)O=(108)/(392)xx100=27.55%,%" of "SO_(4)^(2-)=(2(32+64))/(392)xx100=48.98%`

Discussion

34.	Calculate the percentage of free volume available in 1 mole gaseous water at 1 atm pressure and 373 K.
Answer» ANSWER :`99.94%`

Discussion

35.	Calculate the percentage of hydrolysis in 0.003 M aqueous solution of NaOCN. K_a for HOCN = 3.33 xx 10^(-4) M .
Answer» SOLUTION :`OCN^(-) + H_2O IFF HOCN + HO^(-)` 0.01%

Discussion

36.	Calculate the percentage of free volume available in 1 mol of gaseous water at 1.0 atm and 100^(@)C. Density of liquid H_(2)O at 100^(@) is 0.958 g//mL
Answer» ANSWER :`99.9386`

Discussion

37.	Calculate the percentage of free SO_(3)in an oleum ( considered as a solution of SO_(3)in H_(2)SO_(4)) thatis labelled109 %H_(2)SO_(4) .
Answer» SOLUTION :109 % `H_(2)SO_(4)` refersto the total massof pure `H_(2)SO_(4)`i.e ., 109 G that will be formed when 100 g ofoleum is dilutedby 9g of `H_(2)O ` which`(H_(2)O)` COMBINES with all the free `SO_(3)` present in OLEUM to form `H_(2)SO_(4)` `H_(2)O+SO_(3) to H_(2)SO_(4)` 1 mole of `H_(2)O` combines with 1 moleof `SO_(3)` or 18 g of `H_(2)O ` combines with 80 g of `SO_(3)` or 9 gof `H_(2)O `combines with 40 g of `SO_(3)` Thus , 100 g of oleum contains 40 gof `SO_(3)`or oleum 40 % of free `SO_(3)`

Discussion

38.	Calculate the percentage of carbon, hydrogen and oxygen in ethanol (C_(2)H_(5)OH).
Answer» SOLUTION :`C=52.17%,H=13.04%,O=34,78%`

Discussion

39.	Calculatethe percentageefficiencyof packing in case of bodycenteredcubiccrystal.
Answer» SOLUTION :Packingefficiency: inbodycenteredcubicarrangementthe spheresare touchingalongtheleadingdiagonalof the cubeas shownin the FIGURE. `AC^(2) - AB^(2) + BC^(2)` `AC = sqrt(a^(2) +a^(2)) = sqrt(2a^(2))= sqrt(2a)` In `Delta ACG` `AG^(2) = AC^(2) +CG^(2)` `AG = sqrt((2a)^(2) +a^(2))` i.e., `sqrt(3) a= 4r` Volumeof THESPHERE withradiusr Numberof spherebelongto aunitcell inbccarrangement is equalto twoandbencethe totalvolumeof allspheres Packin fraction= `("Totalvolumeoccupiedbyspheresin a unitcell ") /("Volumeof theunitcell") xx 100` Packingfraction`=((sqrt(3) pi a^(3))/( 8))/((a^(3))) xx 100` `=1.732 xx 3.14xx 12.5=68%`

Discussion

40.	Calculate the percentage efficiency of packing in case of body centered cubic crystal.
Answer» Solution :Packing efficiency `:` Here, the spheres are touching along the leading DIAGONAL of the cube as shown in the figure. In `Delta ABC` `AC^(2) = AB^(2) + BC^(2)` `AC = sqrt( AB^(2) + BC^(2))` `AC = sqrt(a^(2) +a^(2)) = sqrt(2a^(2)) = sqrt(2) a` In `Delta ACG` `AG^(2) = AC^(2) + CG^(2)` `AG= =sqrt( AC^(2) +CG^(2))` `= AG = sqrt((sqrt(2a))^(2) + a^(2))` `AG = sqrt( 2a^(2) + a^(2)) = sqrt( 3a^(2))` `AG =sqrt(3) a` i.e., `sqrt(3) a = 4r` ` r(sqrt(3) )/( 4) a` `:.` Volume of the sphere with radius 'r' `= ( 4)/( 3) pir^(3)` `= ( 4)/( 3) pi ((sqrt(3))/( 4) a)^(3)` `= ( sqrt(3))/(16) pi a^(3)`....(1) Number of spheres belong to a unit cell in bcc arrangement is EQUAL to TWO and hence the total volume of all spheres. `=2 XX ((sqrt(3) pia^(3))/(16))= (sqrt(3) pia^(3))/( 8)` Dividing (2) by (3) Packing fraction `= ((sqrt(3) pi a^(3)))/((a^(3))) xx 100` `= ( sqrt(3)pi)/( 8) xx100` `= sqrt(3) pi xx 12.5 ` `= 1.732 xx 3.14 xx 12.5` `= 68%` i.e., 68% of the available volume is occupied. The available space is used more efficiently than in simple cubic packing.

Discussion

41.	Calculate the percentage composition of calcium nitrate.
Answer» SOLUTION :The FORMULA of calcium nitrate is `Ca(NO_(3))_(2)`. Thus, the formula mass or molecular mass =At. Mass of Ca+2 X at. Mass of N+6X at. Mass of oxygen. `=40+2xx14+6xx16=164` % of `Ca=(40)/(164)xx100=24` % of `N=(28)/(164)xx100=17` % of `O=100-(24+17)=59`

Discussion

42.	Calculate the percentage amount of oxalate in a given sample of oxalatesaltwhen 0.3 g of salt was dissolved in 100 mL and 10 mLof which required 8 mLof N/20 KMnO_(4)solution.
Answer» SOLUTION :Let the AMOUNT of oxalate in the sample be x g . Now , m.e of `KMnO_(4) = 1/20 xx 8 = 0.4 "" ` …(Eqn.1) ` :. ` m.e of 10 mL of oxalate salt solution = 0.4..(Eqn.2) ` :. ` m.eof 100 mL of oxalatesolution = 4.0 but m.e of oxalate = m.e of oxalate salt = 4 ...(Eqn . 7) ` :. ` equivalent of oxalate = `4/1000= (0.04) ` ....(Eqn.3) Wt . of oxalate ` = (0.004 xx 44 ) g ` ` = 0.176 g` ` {" eq. wt of " C_(2)O_(4)^(2-)= 88/2 = 44}` Percentage amount of oxalate = `(0.176 )/(0.3) xx 100 = 58.67 ` %

Discussion

43.	Calculate the percent loss in weight after complete decomposition of a pure sample of potassium chlorateKClO_(3)(s) to KCl + O_(2)(g)
Answer» ANSWER :`39.18%`

Discussion

44.	Calculate the per cent dissociation of H_2S(g) if 0.1 mole of H_2Sis kept in a 0.4-litrevessel at 1000 K.For the reaction2H_2S (g) iff 2H_2(g) + S_2(g)the value of K_c is 1.0 xx 10^(-6)
Answer» ANSWER :0.02

Discussion

45.	Calculate the per cent error in hydronium ion concentration made by neglecting the ionisation of water in 1.0 xx 10^(-6) M NaOH
Answer» Solution :Because `[H^(+)] = 1.0 XX 10^(-8) =10 xx 10^(-9)` Which is very SMALL and so NEGLIGIBLE if ionisation of `H_(2)O` is not neglected, then `{:(H_(2)O,hArr,H^(+)+,H^(-)),(,,a,(10^(-6)+a)):}` `:. a xx (10^(-6) + a) = 10^(-14) "" a = 9.9 xx 10^(-9)` `:.` % error `= (10 xx 10^(-9) - 9.9 xx 10^(-9))/(9.9 xx 10^(-9)) xx 100 = 1%`

Discussion

46.	Calculate the per cent concentration of a 9.28N NaOH solution of density 1.31g//mL.
Answer» ANSWER :`28.3%`

Discussion

47.	Calculate the partial vapour pressure of C_(2)H_(4)Br_(2) at 85^(@)C for an ideal solution with mole fraction of 0.25. Vapour presure of pure C_(2)H_(4)Br_(2) at 85^(@)C is 170 mm Hg.
Answer» <P> SOLUTION :`P_(A)=P_(A)^(@)X_(A)=170mmxx0.25=42.5 MM HG`.

Discussion

48.	Calculate the partial pressure of carbon monoxide from the following data’s CaCO_(3)(s) overset(Delta)to CaO(s) + CO_(2)(g), K_(p) = 8 xx 10^(-2)CO_(2)(g) + C(s) rarr 2CO(g), K_(p) = 2
Answer» <P>0.2 0.4 1.6 4 Solution :`CaCO_(3)(s) overset(Delta)to CAO(s) + CO_(2)(g)` `CO_(2)(g) + C(s) rarr 2CO(g)` `K_(p) = p^(2)_(CO)/p_(CO_(2)) = (2X)^(2)/0.08`

Discussion

49.	Calculate the partial pressur of carbon monoxide from the following CaCO_(3(s))overset(Delta)toCaO_((s))+CO_(2)uarr,K_(p)=8xx10^(-2) CO_(2(g))+C_((s))to 2CO_((g)),K_(p)=2
Answer» `0.2` `0.4` `1.6` 4 Solution :Given, `CaCO_(3(s))OVERSET(Delta)toCaO_((s))+CO_(2(G))uarr` `C_((s))+CO_(2(g))hArr2CO_((g)), K_(p1)=pCO_(2)` `K_(p2)=([pCO]^(2))/([pCO_(2)]),pCO=sqrt([Kp_(1)xxKp_(2)])` `pCO=sqrt([8XX10^(-2)xx2])=sqrt(16xx10^(-2))=4xx10^(-1)=0.4`

Discussion

50.	Calculate the packing factor for spheres occupying (a) a body-centred cubic structure, and (b) a simple cubic structure, where closest neighbours in both cases are in contact.
Answer» ANSWER :(a) 0.68, (B) 0.524

Discussion

Explore topic-wise InterviewSolutions in Current Affairs.

Calculate the pH of solution with H_3O^+ concentration in mol dm^(-3). (i) 10^(-4)

Calculate the pH of solution with H_3O^+ concentration in mol dm^(-3). (ii) 10^(-7)

Calculate the pH of solution obtained by mixing 10mL of 0.1M HCl and 40mL of 0.2M H_(2)SO_(4) take log3.4=0.53

Calculate the ph of solution obtained by mixing 100 ml of 0.1 M HCl and 9.9 ml of 1.0m H_2SO_4.

Calculate the pH of an aqueous solution of 1.0 M ammonium formate assumingcomplete dissociation. pK_aof formic acid = 3.8 and pK_bof ammonia = 4.8.

Calculate the pH of a solution obtained by mixing equal volume of 0.02MHOCl &0.2MCH_(3)COOH solutions Given that K_(a)(HOCl)=2xx10^(-4),K_(a)(CH_(3)COOH)=2xx10^(-5) Also calculate [OH^(-)],[OCl^(-)],[CH_(3)COOH] at equilibrium .Take log2=0.3

Calculate the pH of a buffermixture which contains 7.5gms if acetic acid and 10.25 gms of sodium acetate in 1 litre of the solution. K_a for acetic acid is 1.8xx10^(-5).

Calculate the pH of a buffer containing 0.08 mole of acetic acid and 0.12 mole of sodium acetate per dm_2 of the solution. The ionisation constant of acetic acid is 1.8xx10^-5.

Calculate the pH of a 0.15 M aqueous solutions of AlCl_(3) Given :[Al(H_(2)O)_(6)]^(3+)(aq)+H_(2)O(l)hArr[(AlH_(2)O)_(5)OH]^(2+)(aq),K_(a)=1.5xx10^(-5)

Calculate the pH of a 0.10 N solution of H_2CO_3

Calculate the pH of a 0.005 M Na_2 S solution. K_1 and K_2 for H_2S are 1xx10^(-7) and 1xx 10^(-14)respectively.

Calculate the pH of 1.5xx10^(-3) M solution of Ba(OH)_2.

Calculate the pH of 1.5 times 10^-3 M solution of Ba(OH)_2.

Calculate the pH of 0.5 M aqueous solution of NaCN, the pK_(b) " of " CN^(-) is 4.70

Calculate the pH of 0.1M CH_3COOH solutoion. Dissociation constant of acetic acid is 1.8 times 10^-5.

Calculate the pH of 0.1 M solution of H_2NCH_2 CH_2NH_2 , ethylenediamine (en). Determine the en H_2^(2+). concentration in the solution. K_(b_(1)) and K_(b_(2)) values of ethylenediamine are 8.5 xx 10^(-5) and 7.1 xx 10^(-8) respectively.

Calculate the pH of 0.1 M CH_3 COOH solution.Dissociation constant of acetic acid is 1.8xx10^(-5) M.

Calculate the pH of 0.1 M acetic acid if its ionisation constant K_a=1.8xx10^(-5).

Calculate the pH of 0.04 M HNO_3 Solution.

Calculate the pH of 0.02 MHCI.

Calculate the pH of 0.02 m Ba(OH)_2 aqueous solution assuming Ba(OH)_2 as a strong electrolyte.

Calculate the pH of 0.01 M NaOH.

Calculate the pH of 0.001M HCl solution HCl_(0.001M) leftrightarrow^(H_2O)H_3O_(0.001M)^(+)+Cl_(0.001M)^(-)

Calculatethe pH at whichan acidindicator with K_(a)= 1xx 10^(5) changescolourwhenindicator concentrationis 1 xx 10^(-3)M . Alsorport the pH at whichcoloured ionsare 80 %present.

Calculate the pH at the equivalence point when a solution of 0.01 M CH_(3)COOHis titrated with a solution of 0.01 M NaOH.pKa of CH_(3)COOH is 4.74

Calculate the percentage of water of crystallisation in the sample of washing soda, Na_(2)CO_(3).10H_(2)O.

Calculate the percentage of water of crystallisation in the sample of blue vitriol (CuSO_(4).5H_(2)O).

Calculate the percentage of water of crystalline in Blue vitrial CuSO_4.5H_2O (Atomic masses aregiven as Cu = 63.5, S = 32, O = 16, H = 1

Calculate the percentage of vacant space in a Si unit cubic cell. The unit-cell content for Si is 8 and r=(sqrt(3)a)/8.

Calculate the percentage of the naturally occurring isotopes .^(35)Cl and .^(37)Cl that accounts for the atomic mass of chlorine taken as 35.45.

Calculate the percentage of (i) SO_(4)^(2-) (ii) H_(2)O in pure crystals of Molar salt, viz., FeSO_(4).(NH_(4))_(2)SO_(4).6H_(2)O.

Calculate the percentage of free volume available in 1 mole gaseous water at 1 atm pressure and 373 K.

Calculate the percentage of hydrolysis in 0.003 M aqueous solution of NaOCN. K_a for HOCN = 3.33 xx 10^(-4) M .

Calculate the percentage of free volume available in 1 mol of gaseous water at 1.0 atm and 100^(@)C. Density of liquid H_(2)O at 100^(@) is 0.958 g//mL

Calculate the percentage of free SO_(3)in an oleum ( considered as a solution of SO_(3)in H_(2)SO_(4)) thatis labelled109 %H_(2)SO_(4) .

Calculate the percentage of carbon, hydrogen and oxygen in ethanol (C_(2)H_(5)OH).

Calculatethe percentageefficiencyof packing in case of bodycenteredcubiccrystal.

Calculate the percentage efficiency of packing in case of body centered cubic crystal.

Calculate the percentage composition of calcium nitrate.

Calculate the percentage amount of oxalate in a given sample of oxalatesaltwhen 0.3 g of salt was dissolved in 100 mL and 10 mLof which required 8 mLof N/20 KMnO_(4)solution.

Calculate the percent loss in weight after complete decomposition of a pure sample of potassium chlorateKClO_(3)(s) to KCl + O_(2)(g)

Calculate the per cent dissociation of H_2S(g) if 0.1 mole of H_2Sis kept in a 0.4-litrevessel at 1000 K.For the reaction2H_2S (g) iff 2H_2(g) + S_2(g)the value of K_c is 1.0 xx 10^(-6)

Calculate the per cent error in hydronium ion concentration made by neglecting the ionisation of water in 1.0 xx 10^(-6) M NaOH

Calculate the per cent concentration of a 9.28N NaOH solution of density 1.31g//mL.

Calculate the partial vapour pressure of C_(2)H_(4)Br_(2) at 85^(@)C for an ideal solution with mole fraction of 0.25. Vapour presure of pure C_(2)H_(4)Br_(2) at 85^(@)C is 170 mm Hg.

Calculate the partial pressure of carbon monoxide from the following data’s CaCO_(3)(s) overset(Delta)to CaO(s) + CO_(2)(g), K_(p) = 8 xx 10^(-2)CO_(2)(g) + C(s) rarr 2CO(g), K_(p) = 2

Calculate the partial pressur of carbon monoxide from the following CaCO_(3(s))overset(Delta)toCaO_((s))+CO_(2)uarr,K_(p)=8xx10^(-2) CO_(2(g))+C_((s))to 2CO_((g)),K_(p)=2

Calculate the packing factor for spheres occupying (a) a body-centred cubic structure, and (b) a simple cubic structure, where closest neighbours in both cases are in contact.