94420 + Interview Questions in Chemistry in Class 12

1.	Calculate the number of alpha and beta particles emitted in the conversion of 90^(Th^(232) to 82^(pb^(208)).
Answer» Solution :Leta and b be the NUMBER of `alpha beta` particlesemitted duringthe CHANGE comparingthe mass numbers 232 +208 +4a+b x0 4a=232-208 a=6 number of `alpha`PARITCLE emitted =6 numberof `beta` particles emitted =4

Discussion

2.	The end product of decay of ""_(90)Th^(232) is ""_(82)Pb^(208) . The number of alpha and beta particles emitted are respectively
Answer» SOLUTION :LETA and b be the NUMBER of `alpha beta` particlesemitted duringthe change comparingthe mass numbers 232 +208 +4a+b x0 4a=232-208 a=6 number of `alpha`paritcle emitted =6 NUMBEROF `beta` particles emitted =4

Discussion

3.	Calculate the normality of the solution obtained by mixing (ii) 100 cc of 0.1 N H_(2)SO_(4) with 100 cc of 0.2 N NaOH.
Answer» Solution :100 cc of of 0.1 N `H_(2)SO_(4)=100xx0.1` meq = 10 meq , 100 cc of 0.2 N NaOH `=100xx0.2" meq = 20 meq"` 10 meq of `H_(2)SO_(4)` will NEUTRALIZE 10 meq of NaOH `THEREFORE` meq of NaOH left after neutralisation = 10 meq TOTAL volume of the solution `=100+100="200cc " therefore" NORMALITY of NaOH in the solution "=("10 meq.")/("200 cc")=0.05N.`

Discussion

4.	Calculate the normality of the solution obtained by mixing 100 cc of 0.2 N H_(2)SO_(4) with 50 cc of 0.1 N HCl.
Answer» Solution :100 cc of 0.2 N `H_(2)SO_(4)=100x0.2" meq. = 20 meq."` `"50 cc of 0.1 N HCL cotains HCl "=50xx0.1" meq.= 5 meq."` Total volume of the solution of the solution after MIXING `=100+50="150 cc"` `therefore"Normality of the final solution"=("25 meq")/("150 cc")="0.167 N."` Alternatively `""N_(1)V_(1)+N_(2)V_(2)=N_(3)V_(3)` `0.2xx100+0.1xx50=N_(3)(100+50) or 150 N_(3)=20+5=25"or"N_(3)=(25)/(150)="0.167 N."`

Discussion

5.	Calculate the normality of the solution obtained by mixing (iii) 100 cc of 0.1 M H_(2)SO_(4) with 100 cc of 0.1 M NaOH.
Answer» SOLUTION :100 cc of 0.1 M `H_(2)SO_(4)=100x0.1xx"2 meq , 100 cc of 0.1 M NAOH "=100xx0.1" meq = 10 meq"` 10 meq of NaOH will NEUTRALIZE 10 meq of `H_(2)SO_(4) therefore H_(2)SO_(4)` left after NEUTRALISATION = 10 meq Volume of the solution `=100+100="200 cc " therefore " Normality of "H_(2)SO_(4)" in the solution "=("10 meq")/("200 cc")="0.0 N"`

Discussion

6.	Calculatethe normality of a solution of FeSO_(4).7H_(2)O containing 2.4g/100 mL (Fe=56,S=32,O=16,H=1) which converts to ferric form in a reaction .
Answer» Solution :Weight per ML =2.4 g Equivalents/100 mL =`(2.4)/(278) =0.0086 (" eq.wt .of" FeSO_(4).7H_(2)O=278)` `{:{("As "Fe^(2)toFe^(3+)),("eq.wt of "FeSO_(4).7H_(2)O=("MOLECULAR wt")/("change in ON")=278/1):}}`. Thus m.e per 100 mL =`0.0086xx1000=8.6 " "...(Eqn.3)` Normality of solution `=(v.e)/("VOLUME in mL") ""...(Eqn.1)` `=(8.6)/100` `= 0.086` N

Discussion

7.	Calculate the number of particles present per unit cell in a B.C.C unit cell
Answer» SOLUTION :No. of ATOMS present in a body centered cube = (No. of corner atoms) `xx` (Share of each corner atoms per unit cell) + (No. of atoms at the center of the body) `xx` (Share of body centered atom per unit cell) `= 8 xx (1)/(8) + 1` `= 1+1 = 2` atoms per unit cell. DETAILED Answers. No. of particles present per unit cell in B.C.C. unit cell `= 8 xx (1)/(8) + 1 = 2`.

Discussion

8.	Calculate the no. of particles (atoms) per unit cell in a FCC crystal lattice:
Answer» Solution :(1) A face centre cube contains 8 lattice points at the eight corners and 6 lattice points at the centres of six faces. + (2) /A PARTICLE PRESENT at the corner SHARES 1/8 of that particle to each unit cell (3) A particle present at the centre of axx fxxa=ce provides a share of 1/2 of that particle to each unit cell. (4) The number of particles present in unit cell of fcc `=1/8xx8+1/2xx6=1+3=4`

Discussion

9.	Calculate the no. of electrons lost or gained during elelctrolysis of 3.55 gm of Cl^- ions
Answer» ANSWER :A::B::C

Discussion

10.	Calculate the no. of electrons lost or gained during elelctrolysis of 2.7 gm of Al^(3+) ions
Answer» ANSWER :A::B::C::D

Discussion

11.	Calculate the no. of electrons lost or gained during elelctrolysis of 1 gm Cu^(2+) ions
Answer» ANSWER :A::B::C::D

Discussion

12.	Calculate the no. of disintegrations which 1 g of radioactive element ""^(200)X undergoes per sec. t_(1//2) of X =1000 years.
Answer» `3.66 xx 10^(10)` DPS `6.6 xx 10^(10) ` dps `10^(11)` dps `1.0 xx 10^(10)` dps Solution :`-(dN)/(dt) = KN` `= (0.693)/(1000 xx 365 xx 24 xx 60 xx 60) xx (6.02 xx 10^(23))/(200)` `-6.6 xx 10^(10)` dps.

Discussion

13.	Calculate the momentum of a particle which has a de-Broglie wavelength of 1Å. [ h = 6.626 xx 10^(-34) kg m^2 s^(-1) ]
Answer» Solution :`6.63 xx 10^(-24) KG MS^(-1)`

Discussion

14.	Calculate the momentum of a particle which has a de Broglie wave length of 0.1 mm.
Answer» SOLUTION :`6.6256xx10^(-24)KGMS^(-1)`

Discussion

15.	Calculate the molecular weight of cellulose acetate if its 0.2% (wt./vol.) solution in acetone (sp.gr.0.8) shows an osmotic rise of 23.1 mm against pure acetone at 27^(@)C.
Answer» Solution :`0.2`per cent solution means `0.2g` of CELLULOSE acetate DISSOLVED in 100mL of solution. Osmotic PRESSURE `=2.31cm` of acetone `p=2.31xx(0.80)/(13.6)cmHg=0.136cmHg` `p=(0.136)/(76)atm.` `(1atm=76cmHg)` Suppose M is the MOLECULAR weight of cellulose acetate `n=(0.2)/(M)`, `V=100mL=0.1` litre, `R=0.082` lit. atm/K/mole and `T=273+27=300K` Now, `p=(n)/(V)RT` ..............(Eqn. 6) `:.(0.136)/(76)=(0.2//M)/(0.1)xx0.082xx300` `M=27500`

Discussion

16.	Calculate the molecular weight of a substance 10g of which in 1 litre of solution exerts an osmotic pressure of 81 mm Hg at 27K
Answer» ANSWER :`207.99`

Discussion

17.	Calculate the molecular mass of : (i) H_(2)O (ii)CO_(2) (iii)CH_(4)
Answer» Solution :`"(i) Molecular MASS of "H_(2)O=2(1.008amu)+16.00amu =18.016amu` `"(II) Molecular mass of "CO_(2)=12.01amu+2xx16.00amu=44.01amu` `"(III) Molecular mass of "CH_(4)=12.01amu+4(1.008amu)=16.042amu`

Discussion

18.	Calculate the molecular mass of glucose (C_(6)H_(12)O_(6)) molecule.
Answer» Solution :Molecular MASS of glucose `(C_(6)H_(12)O_(6))=6(12.011" AMU")+12(1.008" amu")+6(16.00" amu")` `=72.066" amu"+12.096" amu"+96.000" amu"=180.162" amu."`

Discussion

19.	Calculate the molecular mass of a substance, 1.0 gram of which when dissolved in 100 gram of solvent gace an elecation of 0.307 K in the boiling point. (Molar elevation constant K_(b)=1.84 K kg mol^(-1)).
Answer» SOLUTION :`W_(B)=1.0g, W_(A)=0.1 kg, DeltaT_(b)=0.307 K, K_(b)=1.84 K//m, M_(B)=?` `M_(B)=(K_(b)xxW_(B))/(DeltaT_(b)xxW_(A))=((1.84" K kg mol"^(-1))(1.0g))/((0.307 K)XX(0.1 kg))=59.93" g mol"^(-1)`.

Discussion

20.	Calculate the mole fraction of toluene in the vapour phase which is in equilibrium with a solution of benzene and toluene having a mole fraction of toluene 0.5. The vapour pressure of pure benzene is 119 torr, that of toluene is 37 torr at the same temperature.
Answer» ANSWER :0.237

Discussion

21.	Calculate the mole fraction of ethylene glycol (C_(2)H_(6)O_(2)) and water in a solution containing 20% of C_(2)H_(6)O_(2) by mass.
Answer» Solution :`20%` of `C_(2)H_(6)O_(2)` by mass MEANS 20 g of `C_(2)H_(6)O_(2)` are present in 100 g of the solution, i.e., `{:("Mass of SOLUTE "(C_(2)H_(6)O_(2))=26g,,,"Mass of solvent "(H_(2)O)=100-20=80g),("Molar mass of "C_(2)H_(6)O_(2)=62"g mol"^(-1),,,"Molar mass of"H_(2)O=18"g mol"^(-1)),(therefore" No, of moles of "C_(2)H_(6)O_(2)=(20)/(62)=0.322,,,"No. of moles of "H_(2)O=(80)/(18)=4.444):}` `"Mole FRACTION of "C_(2)H_(6)O_(2)" in the solution"=(.^(n)C_(2)H_(6)O_(2))/(.^(n)C_(2)H_(6)O_(2)+.^(n)H_(2)O)=(0.322)/(0.322+4.444)=0.068` `"Mole fraction of "H_(2)O" in the solution "=1-0.068=0.932`

Discussion

22.	Calculate the mole fraction of ethylene glycol (C_(2)H_(6)O_(2)) in a solution containing 20% of C_(2)H_(6)O_(2) by mass.
Answer» Solution :Assume that we have 100 g of solution (one can start with any amount of solution because the results obtained will be the same). Solution will CONTAIN 20g of ethylene glycol and 80 g of water. Molar mass of `C_(2)H_(6)O_(2)=12xx2+1xx6+16xx2` `= 62 g mol^(-1)` Moles of `C_(2)H_(6)O_(2)=(20g)/(62g mol^(-1))=0.322` mol Moles of water `= (80g)/(18g g mol^(-1))=4.444` mol `X_("glycol")=("Moles of "C_(2)H_(6)O_(2))/("Moles of " C_(2)H_(6)O_(2)+" moles of " H_(2)O` `=(0.322 mol)/(0.322 mol+4.444 mol)=0.068` Similarly,`X_("water")=(4.444 mol)/(0.322 mol + 4.444 mol)=0.932` Mole fraction of water can also be CALCULATED as :`1-0.068=0.932`

Discussion

23.	Calculate the mole fraction of ethanol and water in a sample of rectifiedspirit which contains 95 per cent of ethanol by weight .
Answer» SOLUTION :`x_(C_(2)H_(5)OH)=0.88, x_(H_(2(O)))=0.12`

Discussion

24.	Calculate the mole fraction of CH_3 OH in a solution containing 30% by mass in water.
Answer» Solution :Mass of `CH_3 OH = 30 g.` Mass of WATER = 70 g No. of MOLES of `CH_3 OH =(30)/(32) = 0.938` No. of moles of `H_2O ==(70)/(18) = 3.89` `X_(CH_3OH) = ( 0.938 )/(0.938 +3.89 _=(0.938 )/(4.83 )`

Discussion

25.	Calculate the mole fraction of Co, in one litre of soda water sealed under a pressure of "3.5 bar at 298 K". K_(H)=1.67xx10^(3)" bar"
Answer» <P> Solution :`"W.K.T"P=K_(H)X` Given `"Pressure = 3.5 BAR"P_(CO_(2))=K_(H)X_(CO_(2))` `K_(H)=1.67xx10^(3)" bar"X_(CO_(2))=(P_(CO_(2)))/(K_(H))` `=(3.5)/(1.67xx10^(3))=2.095xx10^(-3)`

Discussion

26.	Calculate the mole fraction of benzene in solution containing 30% by mass in CCl_4
Answer» Solution :Mass of solution =100G Mass of BENZENE = 30G Number of moles of benzene = (Mass)/(Molar mass) = 30/78 = 0.385 Mass of `CCl_4` = 100-30 = 70G Number of moles of `CCl_4`= (Mass)/(Molar mass) = 70/154 = 0.458 `therefore` MOLE fraction of benzene =(Number of moles of `C_6H_6`)/(Total number of moles)= `0.385/(0.385+0.455) = 0.458`

Discussion

27.	Calculate the mole fraction of benzene in solution containing 30% by mass in carbontetrachloride.
Answer» Solution : Let the mass of solution = 100 g Then, mass of benzene in the solution = 30 g ` therefore ` Mass of CARBON tetrachloride = 100 - 30 = 70 g MOLAR mass of benzene `(C_6H_6) = (6 XX 12) + (6 xx 1) = 78 g "mol"^(-1)` Molar mass of `"CCl"_4 = 12 + 4 xx 35.5 = 154 g "mol"^(-1)` Number of moles of benzene =`(30g)/(78 g"mol"^(-1) ) = 0.385 `mol Number of moles of `"CCl"_4 = (70g)/(154 g "mol"^(-1) ) = 0.455 ` mol MOLE FRACTION of benzene = `(0.385)/(0.385 + 0.455) = (0.385)/(0.84) =0.458 ` Mole fraction of `"CCl"_4 = 1 - 0.458 = 0.542` .

Discussion

28.	Calculate the mole fraction of benzene in solution containing 30% by mass in carbon tetrachloride.
Answer» Solution :`30%` of benzene in carbon tetrachloride by mass means that Mass of benzene in the solution = 30 g , `""` Mass of solution = 100 g `therefore""` Mass of carbon tetrachloride `=100-30g=70g,""` MOLAR mass of benzene `(C_(6)H_(6))="78 g mol"^(-1)` `"Molar mass of CCl"_(4)=12+4xx35.5="154 g mol"^(-1)` `therefore"No. of moles of benzene"=("Mass")/("Molar mass")=("30 g")/("78 g mol"^(-1))=0.385` `"No. of moles of CCl"_(4)=("Mass")/("Molar mass")=("70 g")/("154 g mol"^(-1))=0.455` `"Mole fraction of benzene "=("Moles of benzene")/("TOTAL moles in the solution")=(0.385)/(0.385+0.455)=(0.385)/(0.84)=0.458` `"Mole fraction of CCl"_(4)=1-0.458=0.542`

Discussion

29.	Calculate the molarityes and normalities of the solution obtained on mixing (i) 100 mL of 0.2 M H_(2)SO_(4) with 50 mL of 0.1 M HCl (ii) 100 mL of 0.2 N H_(2)SO_(4) with 50 mL of 0.1 N HCl (iii) 100 mL of 0.1 M H_(2)SO_(4) with 50 mL of 0.1 M NaOH (iv) 50 mL of 0.1 N H_(2)SO_(4) with 100 mL of 0.1 N NaOH.
Answer» Solution :(i) `"100 mL of 0.2 M "H_(2)SO_(4)" contain "H_(2)SO_(4)=(0.2)/(1000)xx100=0.02mol="0.02 g eq."` `"50 mL of 0.1 M HCl contain HC"=(0.1)/(1000)xx50="0.005 mol = 0.005 g eq."` `"Total no. of MOLES present "=0.02+0.005mol=0.025 mol` `"Total VOLUME after mixing "=100+50mL=150mL=0.150L` `"Molarity"=(0.025mol)/(0.150 L)=0.167M` `"Total no. of g eq."=0.04+0.005="0.045g eq."` `"Normality"=("0.045 g eq")/("0.150 L")=0.3N` Alternatively, `M_(1)V_(1)+M_(2)V_(2)=M_(3)(V_(1)+V_(2))` `therefore""M_(3)=(0.2xx100+0.1xx50)/(150)=0.167M` Similarly, `N_(1)V_(1)+N_(2)V_(2)=N_(3)(V_(1)+V_(2))` `"As0.2 M "H_(2)SO_(4)="0.4 N "H_(2)SO_(4) and " 0.1 M HCl = 0.1 N HCl"` `N_(3)=(0.4xx100+0.1xx50)/(150)=0.3N` (ii) `"0.2 N "H_(2)SO_(4)="0.1 M "H_(2)SO_(4)"and0.1 N HCl = 0.1 M HCl"` `"Applying"M_(1)V_(1)+M_(2)V_(2)=M_(3)(V_(1)+V_(2))` `M_(3)=(0.1xx100+0.1xx50)/(150)=(15)/(150)=0.1M` `"Applying"N_(1)V_(1)+N_(2)V_(2)=N_(3)(V_(1)+V_(2))` `N_(3)=(0.2xx100+0.1xx50)/(150)=(25)/(150)=0.167N` (III) `"100 mL of 0.1 M "H_(2)SO_(4)" contain "H_(2)SO_(4)=(0.1)/(1000)xx100="0.01 mol = 0.02 g eq"` `"50 mL of 0.1 M NAOH contain NaOH"=(0.1)/(1000)xx50="0.005 mol = 0.005 g eq"` `"0.005 g eq. of NaOH neutralize 0.005 g eq. of "H_(2)SO_(4)," therefore, "H_(2)SO_(4)" left unneutralized in the solution"` `=0.02-0.005="0.015 g eq."` `"Total volume of solution after mixing = 150 mL = 0.150 L"` `therefore"Normality of "H_(2)SO_(4)" in the solution"=("0.015g eq")/(0.150L)="0.1 N. Hence, molairy"=("Normality")/("Basicity")=(0.1)/(2)="0.05 M"` (iv) `"50 mL of 0.1 N "H_(2)SO_(4)" contain "H_(2)SO_(4)=(0.1)/(1000)xx"50 g eq. = 0.005 g eq."` `"100 mL of 0.1 N NaOH contain NaOH"=(0.1)/(1000)xx"100 g eq. = 0.01 g eq."` `"0.005 g eq. of "H_(2)SO_(4)" neutralize 0.005 g eq. of NaOH, therefore, NaOHleft in the solution"` `=0.01-0.005="0.005 geq."` `"Total volume of solution after mixing = 150 mL = 0.150 L"` `therefore"Normality of NaOH in the solution"=("0.005 g eq")/("0.150 L")="0.033 N. Hence,"` `"molarity x "=("Mormality")/("Acidity")=(0.033)/(1)=0.033M`

Discussion

30.	Calculate the mole fraction of a benzene-toluene liquid solution that is inequilibrium with a vapour phase that contains 62 mole % C_6H_6 . Vapour pressures of benzene and toluene at 25^@Care 95.1 mm and 28.4 mm respectively.
Answer» SOLUTION :0.328 for `C_6H_6`

Discussion

31.	Calculate the mole fraction in 2.5 m aqueous solution .
Answer» Solution :2.5 m AQUEOUS solution MEANS that 2.5 MOLES of the solute are dissolved in 1000 g of water. `"No. of moles of water in 1000 g" =("Mass of water")/("Gram molar mass")=((1000g))/((18"g mol"^(-1)))=55.5 mol` `"Mole fraction of solute"=("No. of moles of solute")/("No. of moles of solute + No. of moles of water")` `((2.5 mol))/((2.5 mol+55.5 mol))=((2.5 mol))/((58 mol))=0.043`.

Discussion

32.	Calculate the molarity when 73 grams of HCl is dissolved in water to make 1500 ml solution.
Answer» Solution :`MOLARITY (M) = W/"mol.wt" XX 1000/(V "in" ml)` =`73/36.5 xx 1000/1500 = 2/1.5= 1.33`

Discussion

33.	Calculate the molarity the following solution 30g Co(NO_3)_2.6H_2O in 4.3L of solution
Answer» Solution :MOLAR mass of `CO(NO_3)_2.6H_2O`= `59+ 2(14+48)+ 6xx18=291` `therefore` Molarity=`W_1/ (M_1xxV)`(in L) = `30/(291xx4.3)= 0.024`

Discussion

34.	Calculate the molarity of water.
Answer» Solution :Supposing the density to be 1, ONE litre of water WEIGHS `1000G` or `(1000)/(18)` moles `(H_(2)O=18)` `:.` molarity `=(1000)/(18)=55.55M` (moles/litre)

Discussion

35.	Calculate the molarity of water if its density is 1000 kg//m^(3).
Answer» SOLUTION :Molarity of water MENAS NUMBER of moles of water in 1 litre of water ` "L of water"= 100 cm^(3)= 1000 G""(because d=1000kg//m^(2)=1g//cm^(3))` LTBRGT `1000gH_(2)O=(1000)/(18)"moles"="55.56 moles."` `therefore"Molarity of water"=55.56"mol L"^(-1)`.

Discussion

36.	Calculate the molarity of water if its density is "1000 kg/m"^(3).
Answer» Solution :Molarity of water means number of moles of water in 1 litre of water `"1 L of water = 1000 cm"^(3)"= 1000 g"(because d = 1000"kg/m"^(3)="1 g/cm"^(3))` `"1000 g "H_(2)O=(1000)/(18)"MLES = 55.56 moles"` Hence, `"molarity = 55.56 M."`

Discussion

37.	Calculate the molarity og hudrogen chloride in a solution when 0.365 g of it has been dissolved in 100 mL of the solution . 3 g of a salt of molecular weight 30 is dissolved in 250 g of water . The molarity of the solution is …
Answer» SOLUTION :(a) `0.1`M (B) `0.4` m

Discussion

38.	Calculate the molarity of pure water using its denstiy to be 1000 kg m^(-3)
Answer» ANSWER :`55.5 MOL L^(-1)`

Discussion

39.	Calculate the molarity of pure water. Given that density of pure wateris 1 g/mol.
Answer» Solution :`"Mass of 1000 mL of pure water "= Vxxd=(1000 mL)XX(1g//mL)=1000 g` `"Moles of water in 1000 g"=("Mass")/("Molar mass")=((1000g))/((18" g mol"^(-01)))=55.55 mol` `"Molarity (M) "=("Moles of water")/("Volume of WASTER in litres")=((55.55 mol))/(1L)` =55.55 mol/L = 55.55 M.

Discussion

40.	Calculate the molarity of the following solution 30mL 0.5m H_2SO_4 diluted to 500mL
Answer» Solution :For DILUTION `V_1M_1=V_2M_2` `30xx0.5=500xxM_2` `THEREFORE` Molarity of the DILUTED solution `(M_2)`= `(30xx0.5)/500= 0.03`

Discussion

41.	Calculate the molarity of pure water (d = 1 g mL^(-1))
Answer» SOLUTION :DENSITY of water = `1 g mL^(-1)` Mass of 1000 ml of water = `V xx d` `=1000 m L xx 1 gm^(-1)` `=1000 g` Moles of water = `(1000)/(18) = 55.55` mol Now mole of `H_(2) O ` present in 1000 mL or 1 L of water . So , molarity = 55.55 M

Discussion

42.	Calculate the molarity of H^(+) ions obtained by mixing 200 ml of 0.5 M H_(2)SO_(4) solution (d_("solution"=xgm//ml), 100 ml of 0.7 M HNO_(3)) solution (d_("solution")=1.2xxgm//ml) and 100 ml of 0.3 M HCl solution (d_("solution")=1.3xxgm//ml) such that density of final solution is 1.5xx gm//ml.
Answer» Solution :MMOLE of `H^(+)` ions `=200xx0.5xx2+10xx0.7+100xx0.3=300` `V_("FINAL")` of solution `=(200x+120x+130x)//` `d_("final solution")=(450x)/(1.5x)` `V_("final")=300ml` `[H^(+)]=(300)/(100)=1`

Discussion

43.	Calculate the molarity of each of the following solutions : (a) 30 g of Co(NO_3)_2. 6H_2O in 4.3 L of solution (b) 30 mL of 0.5 M H_2SO_4 diluted to 500 mL.
Answer» SOLUTION :(a) Molar mass of` Co(NO_3)_2.6H_2 O = 58.7 + 2(14 + 48) + 6 xx 18 = 310.7 g "mol"^(-1)` Number of moles of `Co(NO_3)_(2.6)H_2O = (30G)/(310 g "mol"^(-1) ) = 0.0966 `mol Volume of solution = 4.3 L MOLARITY of solution = 0.0966/4.3 =0.022 M. (b)1000 ML of 0.5 M `H_2SO_4` contain `H_2SO_4` = 0.5 mole ` therefore `30 mL of 0.5 M `H_2SO_4` contain `H_2SO_4`= 0.5/1000x30 mole = 0.015 mole Volume of solution = 500 mL = 0.500 L Molarity of solution =Number of moles of solute/Volume of solution in L `= (0.015)/(0.500) = 0.03 M`

Discussion

44.	Calculate the molarity of each of the following solutions :30 g of Co(NO_(3))_(2).6H_(2)O in 4.3 L of solution
Answer» SOLUTION :MOLARITY is given by : Molarity `= ("Moles of solute")/("Volume of solution in litre")` Moles of `Co(NO_(3))_(2).6H_(2)O` `=59+2(14+3xx16)+6xx18` `=291 g mol^(-1)` `therefore` Moles of `Co(NO_(3))_(2).6H_(2)O` `=(30)/(291)=0.103` mol Therefore, molarity `= (0.103 mol)/(4.3 L)`

Discussion

45.	Calculate the molarity of each of the following solutions :30 mL of 0.5 MH_(2)SO_(4) diluted to 500 mL.
Answer» Solution :NUMBER of MOLES PRESENT in 1000 mL 0.5 M `H_(2)SO_(4)=0.5` mol `therefore` Number of moles present in 30 mL of 0.5 M `H_(2)SO_(4)=(0.5xx30)/(1000)` mol = 0.015 mol Therefore, MOLARITY `= (0.015)/(0.5L)` mol = 0.03 mol.

Discussion

46.	Calculate the molarity of each of the following solution : (a) "30 g of Co "(NO_(3))_(2).6H_(2)Oin 4.3 L of solution (Atomic mass of cobalt = 58.7)
Answer» Solution :(a) Molar MASS of `Co(NO_(3))_(2).6H_(2)O=58.7+2(14+48)+6xx"18 g mol"^(-1)` `=58.7+124+"108 g mol"^(-1)="290.7 g mol"^(-1)` `"No. of moles of "Co(NO_(3))_(2).6H_(2)O=("Mass")/("Molar mass")=("30 g")/("290.7 g mol"^(-1))=0.1032` `"Volume of solution = 4.3 L"` `"Molarity of solution "=("No. of moles of solute")/("Volume of solution in L")=("0.1032 mole")/(4.3 L)=0.024 M`

Discussion

47.	Calculate the molarity of each of the following solution : "30 mL of 0.5 M "H_(2)SO_(4) diluted to 500 mL. (Atomic mass of cobalt = 58.7)
Answer» Solution :(b) 1000 ML of 0.5 M `H_(2)SO_(4)` contain `H_(2)SO_(4)="0.5 mole"` `therefore"30 mL of 0.5 M "H_(2)SO_(4)" contain "H_(2)SO_(4)=(0.5)/(1000)xx"30 mole = 0.015 mole"` `"VOLUME of solution = 500 mL = 0.500 L"` `therefore"Molarity of solution"=("No. of moles of SOLUTE")/("Volume of solution in L")=("0.015 mole")/("0.500 L")="0.03 M"`

Discussion

48.	Calculate the molarity of a solution of CaCl_(2) if on chemical analysis it is found that 200 ml of CaCl_(2) solution contains 3.01xx10^(22) chloride ions.
Answer» Solution :`underset("1 mole")(CaCl_(2))rarrCa^(2+)+underset(2xx6.02xx10^(23)" IONS")(2Cl^(-))` `therefore 3.01xx10^(22)Cl^(-1)"ions will be present in CaCl"_(2)=(1)/(2xx6.02xx10^(23))xx3.01xx10^(22)" mole" = "0.025 mole"` `therefore"MOLARITY of solution "=("0.025 mole")/("0.200 L")=0.125M.`

Discussion

49.	Calculate the molarity of an aqueous solution of ammonia of pH 9.3. K_bfor ammonia is 1.8 xx 10^(-5) and K_w = 1 xx 10^(-14)
Answer» SOLUTION :`1.95 XX 10^(-5) M`

Discussion

50.	Calculate the molarity of a solution containing 5g of NaOH in 450 mL solution.
Answer» SOLUTION :Moles of `NaOH = (5G)/(40 mol^(-1))=0.125 mol` Volume of the solution in liters `= 450 mL//1000 mL L^(-1)` Molarity `= (0.125 mol xx 1000 mL L^(-1))/(450 mL)=0.278 M` `= 0.278 mol L^(-1)` `= 0.278 mol DM^(-3)`

Discussion

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