94420 + Interview Questions in Chemistry in Class 12

1.	Calculate the mass percentage of benzene (C_(6)H_(6)) and carbon tetrachloride ("CCl"_(4)) if 22 g of benzene is dissolved in 122 g of carbon tetrachloride.
Answer» Solution :MASS of solution = Mass of BENZENE + Mass of barbon tetrachloride `=22g+122g=144G` `"Mass percentage of benzene"=("Mass of benzene")/("Mass of solution")xx100=(22g)/(144g)xx100=15.28%` `"Mass percentage of CCl"_(4)=("Mass of CCl"_(4))/("Mass of solution")xx100=(122)/(144)xx100=84.72%` Alternatively, mass percentage of `"CCl"_(4)=100-"Mass percentage of benzene "=100-15.28=84.72%`.

Discussion

2.	Calculate the mass percentage of benzene (C_6H_6)and carbon tetrachloride ("CCI"_4) if 22 g of benzene is dissolved in 122 g of carbon tetrachloride.
Answer» SOLUTION :Mass of solution = Mass of benzene + Mass of carbon tetrachloride = 22 g + 122 g = 144g Mass PERCENTAGE of benzene = Mass of benzene /Mass of solutionx 100 `= (22g)/(144g) XX 100 = 15.28%` Mass percentage of`"CCl"_4 = ("mass of CCl"_4)/("mass of solution") xx 100 = 122/144 xx 100 = 84.72%`

Discussion

3.	Calculate the mass percentage of aspirin (C_(9)H_(8)O_(4)) in acetonitrile (CH_(3)CN) when 6.5 g of (C_(9)H_(8)O_(4)) is dissolved in 450 g of :
Answer» Solution :6.5 g of `C_(9)H_(8)O_(4)` is dissolved in 450 g of `CH_(3)CN`. Then,total MASS of the solution `= (6.5+45)g` = 456.5 g Therefore, mass PERCENTAGE of `C_(9)H_(8)O_(4)` `=(6.5)/(456.5)xx100` `= 1.424 %`.

Discussion

4.	Calculate the mass percentage of aspirin (C_(9)H_(8)O_(4)) in acetonitrile (CH_(3)CN) when 6.5 g of C_(9)H_(8)O_(4) is dissolved in 450 g of CH_(3)CN.
Answer» Solution :`"MASS percent of ASPIRIN "=("Mass of aspirin")/("Mass of aspirin "+"Mass of acetonitrile")XX100=(6.5)/(6.5xx450)xx100=1.424%.`

Discussion

5.	Calculate the mass percentage of aspirin (C_9H_8O_4) in acetonitrile (CH_3CN) when 6.5 g of C_9H_5O_4is dissolved in 450 g of CH_3CN .
Answer» Solution :MASS per CENT of ASPIRIN = (Mass of aspirin)/(Mass of aspirin + Mass of acetonitrile)x100 ` = (6.5)/(6.5 + 450) XX 100 = 1.424%`

Discussion

6.	Calculate the mass per cent of different elements present in sodium sulphate (Na_(2)SO_(4))?
Answer» Solution :`"Mass "%" of an element"=("Mass of that element in the COMPOUND")/("Molar mass of the compound")xx100` `"Now, molar mass of "Na_(2)SO_(4)=2(23.0)+32.0+4xx16.0="142 G MOL"^(-1)` `"Mass PERCENT of sodium"=(46)/(142)xx100=32.39%` `"Mass per cent of sulphur"=(32)/(142)xx100=22.54%` `"Mass per cent of oxygen"=(64)/(142)xx100=45.07%`

Discussion

7.	Calculate the mass of zinc required to produce enough hydrogen to reduce completely 6.7 g of copper oxide of copper ?
Answer» ANSWER :(i) 0.169 G (II) 5.493 g

Discussion

8.	Calculate the mass of urea (NH_(2)CONH_(2)) required in making 2.5 kg of 0.25 molal aqueous solution.
Answer» Solution :0.25 molal aqueous solution of UREA means that Moles of urea = 0.25 mole.`""` MASS of solvent (water) = 1 kg = 1000 g `"Molar mass of urea "(NH_(2)CONH_(2))=14+2+12+16+14+2="60 g mol"^(-1)` `therefore"0.25 mole of urea "="0.25 mole "xx"60 g mol"^(-1)=15g` THUS, 1.015 kg of solution contain urea = 15 g `therefore"2.5 kg of solution will require urea "=(15 g)/(1.015 kg)xx2.5kg=37g.`

Discussion

9.	Calculate the mass of urea (NH_2CONH_2)required in making 2.5 kg of 0.25 molal aqueous solution.
Answer» Solution :0.25 molal aqueous solution of urea means 0.25 moles of urea are present in 1000 G of water.Molar mass of urea `(NH_2 CONH_2) = 60 g "MOL"^(-1)` ` therefore ` 0.25 mole of urea CONTAINS 0.25 mole `xx 60 g "mol"^(-1) = 15 g` Total mass of solution = 1000 g + 15 g = 1015 g = 1.015 kg THUS, 1.015 kg of solution contain urea = 15 g 2.5 kg of solution will require urea = `(15)/(1.015) xx 2.5 = 36.946 g `

Discussion

10.	Calculate the mass of silver deposited from silver nitrate solution by a current of 2 amperes flowing for 30 minutes (equivalent mass of silver is 108).
Answer» ANSWER :4.029 G

Discussion

11.	Calculate the mass of silver deposited from silver nitrate solution by a current of 2 amperes flowing for 30 minutes?
Answer» SOLUTION :`M= Z l t` `M=(108 times 2 times 30 times 60)/(96,500)` `M=4.029g ` OR `Q = l t` `Q=2 times 30 times 60` =3600 C M=ZQ `=108/(96,500) times 3600` =4.029g Or `Q= l t` `Q=2 times 30 times 60=3600C` FOr 96,500 C mass of silver DEPOSITED is 108g `THEREFORE` for 3600... it is `(3600 times 108)/(96,500)=4.029g`

Discussion

12.	Calculate the mass of NaCI (molar mass =58.5 g mol^(-1)) to be dissoved in 37.2 g of water to lower the freezing point by 2^(@)C assuming that NaCI underfoes complete dissocilation in aqueous solution (K_(f) for water = 1.86 K kg mol^(-1)).
Answer» SOLUTION :`i=(DeltaT_(f))/(K_(f)xxm)=(DeltaT_(f)xxM_(B)xxM_(A))/(K_(f)xxW_(B))or W_(B)=(DeltaT_(f)xxM_(B)xxM_(A))/(ixxK_(f))` `i=2("for complete dissciation of NaCI"),K_(f)=1.86" K kg mol"^(-1)` `DeltaT_(f)=2K, M_(B)=58.5" g mol"^(-1), M_(A)=37.2 g=0.037 kg` `W_(B)=((2K)xx(58.5"gmol"^(-1))xx(0.037 kg))/(2XX(1.86" K kg mol"^(-1)))=1.16 g`

Discussion

13.	Calculate the mass of NaCl (molar mass = 58.5 g "mol"^(-1)] to be dissolved in 37.2 g of water to lower the freezing point by 2^@C, assuming that NaCl undergoes complete dissociation.
Answer» Solution :APPLY the equation, `M_2 = (K_f xx w_2 xx 1000)/(Delta T_f xx w_1)` `58.5 = (1.86 xx w_2 xx 1000)/(2 xx 37.2) ` `w_2 = (58.5 xx 2 xx 37.2)/(1.86 xx 1000) = 2.34 g ` NaCl molecule ionises to give `Na^+` and `Cl^-`ions. Thus, ONE PARTICLE gives TWO particles in solution. ` THEREFORE ` Mass of NaCl required` = 1/2 xx 2.34 = 1.17 g`

Discussion

14.	Calculate the mass of Na_(2)CO_(3) which will have the same number of molecules as contained in 12.3 g of MgSO_(4).7 H_(2)O.
Answer» Solution :`12.3" g MgSO_(4).7H_(2)O=(12.3)/((24+32+64+126))MOL=(12.3)/(246)mol=0.05mol` `0.05mol Na_(2)CO_(3)=0.05xx106g=5.3g`

Discussion

15.	Calculate the mass of mercury which can be liberated from HgO at 25^(@)C by the treatment of excess HgO with 41.84 kJ of heat at : (a) constant pressure (b) constant volume Given : Delta H_(f)^(@)(HgO, s) = -90.8 kJ mol^(-1) & M(Hg) = 200.6 g mol^(-1).
Answer» SOLUTION :(a) `HGO rarr Hg+1/2 O_(2) ""DeltaH=90.8` For `41.84 KJ` heat, MASS of `Hg=200.6/90.8 xx41.84=-92.435 g` (b) Calculate at constant VOLUME

Discussion

16.	Calculatethe mass of Mgand thevolumeof Chlorinegas at NTPproducedduringthe electrolysis ofmolten MgC1_(2)by thepassageof 2 amperesof currentfor 1 hour. MolarmassesofMg and C1_(2) are 24 g mol^(-1) and 71 g mol^(-1) resectively
Answer» Solution :MASS ofMgproduced= 0. 8953 G Volumeof `C1_(2(g))` at NTPproduced= 835.6 `cm^(3)`

Discussion

17.	Calculate the mass of mercury which can be liberated from HgO at 25^(@)C by the treatment of excess HgO with 41.84 kJ of heat at (a) constant pressure (b) constant volume conditions Given : Delta H_(f)^(@) (Hg O, s) = -90.8 kJ mol^(-1)& M (Hg) = 200.6 g mol^(-1)
Answer» SOLUTION : (a) 92.435 G(B) 93.72 g

Discussion

18.	Calculate the mass of KClO_3 required to produce 9.6 gm of oxygen.
Answer» ANSWER :24.5 G

Discussion

19.	Calculate the mass of iron which will be converted into the oxide (Fe_(3)O_(4)) by the action of 18 g of steam on it.
Answer» Solution :The chemical equation representing the reaction is : `underset("= 168 g")underset(3xx56)(3Fe)+underset("= 72 g")underset(4xx18)(4H_(2)O)RARR Fe_(3)O_(4)+4H_(2)` Thus, 72 g of STEAM react with 168g of iron `therefore"18 g of steam will react with "(168)/(72)xx18="42 g of iron"therefore"Mass of iron required = 42 g"`

Discussion

20.	Calculate the mass of iron required to be converted into its oxide (Fe_3O_4) by the action of 23.4 g of steam on it .
Answer» ANSWER :54.6 G

Discussion

21.	Calculate the mass of (i) 0.1 mole of KNO_(3) (ii) 1xx10^(23) molecules of methane and (iii) 112cm^(3) of hydrogen at STP.
Answer» Solution :(i) `"1 mole of "KNO_(3)=101g""(because" Formula mass of "KNO_(3)=1xx39+1xx14+3xx16=101u)` `therefore"0.1 mole of KNO"_(3)=101xx0.1` `=10.1" g OG "KNO_(3)` (ii) 1 mole of `CH_(4)=16g=6.022xx10^(23)` molecules i.e., `6.022xx10^(23)` molecules of methane have mass = 16 g `therefore""1xx10^(23)` molecules of methane will have meas`=(16)/(6.022xx10^(23))xx10^(23)=2.657g.` (iii) 1 mole of `H_(2)=2g=22400cm^(3)` at STP, i.e., 22400 `CM^(3)` of `H_(2)` at STP have mass = 2 g `therefore""112CM^(3)" of "H_(2)" at STP will have mass"=(2)/(22400)xx112=0.01g.`

Discussion

22.	Calculate the mass of hydrogen gas evolved by passing a current of 0.5 ampere for 40 minutes through acidified water.
Answer» SOLUTION :`m=Zit` or `m=(E l t)/(96.500)` `m=(1 times 0.5 times 40 times 60)/(96,500)` =0.0124 g

Discussion

23.	Calculate the mass of compound (molar mass = 256 g mol^(-1) ) to be dissolved in 75 g of benzene to lower its freezing point by 0.48 K(K_(f) = 5.12 "Kkg " mol^(-1)) .
Answer» SOLUTION :Given : `Delta T_(f) = (K_(f) W_(2) xx 1000)/(W_(1) xx M_(2))` 0.48 K = 5.12 k KG `mol^(-1) xx (W_(2))/(75 xx 256) xx 1000` `W_(2) = (0.48 xx 75 xx 256)/(5.12 xx 1000)` `W_(2) ` = 1.8 g

Discussion

24.	Calculate the mass of compound (molar mass = 256 g "mol"^(-1) ) to be dissolved in 75 g of benzene to lower its freezing point by 0.48 K (K_f = 5.12 K kg "mol"^(-1)).
Answer» SOLUTION :Apply the formula `M_2 = (K_f xx w_2 xx 1000)/(Delta T_f xx w_1)` substituting the values , we have `256 = (5.12 kg "MOL"^(-1) xx WG xx 1000 g kg^(-1) )/(0.48 K xx 75 g)` ` w = (256 xx 0.48 xx 75)/(5.12 xx 1000) g = 1.8 g `

Discussion

25.	Calculate the mass of compound (molar mass = "256 g mol"^(-1)) to be dissolved in 75 g of benzene to lower its freezing point by 0.48 K (K_(f)="5.12 K kg mol"^(-1))
Answer» SOLUTION :`w_(2)=(M_(2)xxDeltaT_(f)xxw_(1))/(1000xxK_(f))=(256xx0.48xx75)/(1000xx5.12)g=1.8g`

Discussion

26.	Calculate the mass of CO_(2) which contains the same number of molecules as are contained in 40 g of oxygen.
Answer» SOLUTION :40 G `O_(2)=40//32` MOLES = 1.25 moles 1.25 moles of `CO_(2)=1.25xx44g=55g` (Equal moles contain equal NUMBER of molecules)

Discussion

27.	Calculate the mass of calcium oxide formed when 25 g of 80% pure lime stone is completely decomposed on heating.
Answer» ANSWER :11.2 G of CAO FORMED

Discussion

28.	Calculate the mass of CaO and CO_2 formed by heating 10 gms of CaCo_3.
Answer» ANSWER :5.6 G ,4.4 g

Discussion

29.	Calculate the massof ascorbic acid (Vitamin C, C_(6)H_(8)O_(6)) to be dissolved in 75g of acetic acid to lower its melting point by 1.5^(@)C. K_("f")=3.9" K kg mol"^(-1). Since, lowering of melting point is given apply the formula for lowering of melting point, i.e., DeltaT_("f")=K_("f").m DeltaT_("f")=(K_("f").W_(B))/(M_(B)xxW_(A))" or "W_(B)=(DeltaT_("f").M_(B).W_(A))/(K_("f"))
Answer» SOLUTION :Mass of ascorbic acid `(W_(A))=75" g"=0.075" kg"` Depression in melting point `(DeltaT_("F"))=1.5^(@)C=1.5" K"` Molar mass of ascorbic acid `(M_(B))=(12xx6)+(8xx1)+(16xx6)=176" g mol"^(-1)` Molal depression constant `(K_("f"))=3.9" K kg mol"^(-1)` `W_(B)=((176" g mol"^(-1))xx(1.5" K")xx(0.075" kg"))/((3.9" K kg mol"^(-1)))=5.08" g"`

Discussion

30.	Calculate the mass of ascorbic acid (Vitamin C, C_6H_8O_6) to be dissolved in 75 g of acetic acid to lower its melting point by 1.5^@C. K_f = 3.9 K kg "mol"^(-1) .
Answer» Solution : Lowering in melting point `(DELTA T_f) = 1.5^@` Mass of solvent `(CH_3COOH), w_1 = 75 g` Molar mass of solvent `(CH_3COOH), M_1= 60 g "mol"^(-1)` Molar mass of solute `(C_6H_8O_6), M_2 = 72 + 8 + 96 = 176 g "mol"^(-1)` For ACETIC acid, `K_f = 3.9 K kg "mol"^(-1)` Applying the formula, ` M_2 = (1000 K_f w_2)/(w_1 Delta T_f) " or " w_2 = (M_2 xx w_1 xx Delta T_f)/(1000 xx K_f)` Substituting the values, we get ` w_2 = ((176 g "mol"^(-1) )(75G)(1.5K))/((1000 g kg^(-1) )(3.9 K kg "mol"^(-1) )) = 5.077 g `

Discussion

31.	Calculate the mass of ascorbic acid (Vitamin C, C_(6)H_(8)O_(6)) to be dissolved in 75 g g of acetic acid to lower its melting point by 1.5^(@)C, [K_(f)=3.9" K kg mol"^(-1)].
Answer» Solution :Mass of ACETIC acid, `w_(1)=75 G` Molar mass of ascoric acid `(C_(6)H_(8)O_(6))`, `M_(2)=6xx12+8xx1+6xx16=176g mol^(-1)` Lowering of melting point, `Delta T_(f)=1.5 K` We know that : `K_(f)=3.9" K kg mol"^(-1)` `Delta T_(f)=(K_(f)XX w_(2)xx1000)/(M_(2)xx w_(1))` `w_(2)=(Delta t_(f)xx M_(2)xx w_(1))/(K_(f)xx1000)=(1.5xx176xx75)/(3.9xx1000)` = 5.08 gram = 5.08 g (approx) Hence, 5.08 g of ascorbic acid is needed to be dissolved.

Discussion

32.	Calculate the mass of ascorbic acid (Molar mass = 176 g mol^(-1)) to be dissolved in 75 g of acetic acid, to lower its freezing point by 1.5^(@)C. [K_(f)=3.9"K kg mol"^(-1)]
Answer» Solution :Apply the relation `M_(2)=(K_(f)XX w_(2) xx 1000)/(DeltaT_(f) xx w_(1))` Given: `K_(f)=3.9"K kg MOL"^(-1), M_(2)=176" g mol"^(-1), DeltaT_(f)=1.5""^(@)C, w_(1)=75g` Substituting the values in the equation above, we have `176" g mol"^(-1)=(3.9"K kg mol"^(-1) xx w_(2) xx 1000)/(1.5K xx 75g)` or `w_(2)=(176"g mol"^(-1)xx1.5K xx 75g)/(3.9"K kg mol"^(-1) xx 1000"g kg"^(-1))` `=(176xx1.5xx75)/(3.9xx1000)=5.077g`

Discussion

33.	Calculate the mass of aluminium deposited at cathode when 193 C ofcurrent is passed through molten electrolyte containing dissolved alumina.
Answer» SOLUTION :`3xx96500`Ccan DEPOSIT 1 mole of Al atoms i.e 27 G. `therefore193C......(193xx27)/(3xx96.500)=0.018g` of Al

Discussion

34.	Calculate the mass of Ag deposited at cathode when a current of 2 amperes was passed through a solution AgNO_(3) for 15 minutes.
Answer» ANSWER :2.014

Discussion

35.	Calculate the mass of a single atom of sulphur and a single molecule of carbon dioxide.
Answer» Solution :Gram -atomic mass of SULPHUR =32g Mass of one sulphur atom `=("gram atomic mass")/(6.02xx10^(23))` `=(32)/(6.02xx10^(23))=5.33xx10^(-23)g` Formula of carbon dioxide `=CO_(2)` Molecular mass of `CO_(2)=12+2xx16=44` Gram-molecular mass of `CO_(2)=44g` Mass of one molecular of `CO_(2)=("Gram molecular mass")/(6.022xx10^(23))` `=(44)/(6.02xx10^(23))=7.308xx10^(-23)g`.

Discussion

36.	Calculate the mass of a non - volatile solute (molar mass 40 g) which should be dissolved in 114 g octane to reduce its vapour pressure to 80%.
Answer» <P> Solution :Let the vapour pressure of pure OCTANE be `p_(1)^(0)` `(80)/(100)p_(1)^(0)=0.8 p_(1)^(0)` Molar mass of solute, `M_(2)=40 g mol^(-1)` Mass of octane, `w_(1)=114 g` Molar mass of octane, `(C_(3)H_(18))`, `= M_(1)=8xx12+18xx1=114 g mol^(-1)` Applying the reaction, `(p_(1)^(0)-p_(1))/(p_(1)^(0))=(w_(2)xxM_(1))/(M_(2)xx w_(1))` `therefore (p_(1)^(0)-0.8 p_(1)^(0))/(p_(1)^(0))=(w_(2)xx114)/(40xx114)` `therefore (0.2 p_(1)^(0))/(p_(1)^(0))=(w_(2))/(40)` `therefore w_(2)=8 g`

Discussion

37.	Calculate the mass of a non-volatile solute (molar mass "40 g mol"^(-1)) which should be dissolved in 114 g octane to reduce its vapour pressure to 80%.
Answer» Solution :Reduction of vapour pressure to `80` means that if `p^(@)=" 100 mm, then "p_(s)="80 mm."` Applying complete formula `(p^(@)-p_(s))/(p^(@))=(n_(2))/(n_(1)+n_(2))=(w_(2)//M_(2))/(w_(1)//M_(1)+w_(2)//M_(2))` `=(100-80)/(100)=(w_(2)//40)/(114//114+w_(2)//40)""("Mol. mass of octane "C_(8)H_(18)="114 g mol"^(-1))` `"or"(20)/(100)=(w_(2)//40)/(1+w_(2)//40)"or"(1)/(5)(1+(w_(2))/(40))=(w_(2))/(40)"or"w_(2)=10g` Note that complete formula is required because concentration of solution is greater than `5%`. Complete formula can also be applied in the form `(p^(@)-p_(s))/(p_(s))=(w_(2)//M_(2))/(w_(1)//M_(1))"or"(100-80)/(80)=(w_(2)//40)/(114//114)"or"(1)/(4)=(w_(2))/(40)"or"w_(1)=10g` Alternatively, SUPPOSE mass of SOLUTE dissolved = 2 g Moles of solute `=(w)/(40 )g` Moles of solvent (octane) `=(114)/(114)="1 mole"("Mol. mass of "C_(8)H_(18)="114 g mol"^(-1))` `therefore"Mole fraction of solvent "=(1)/(1+w//40)` For a non-volatile solution, Vapour pressure of solution = Mole fraction of solvent in the solution `xx` Vapour pressure of pure solvent `p_(s)=x_(1)xxp^(@)` `therefore""80=(1)/(1+w//40)xx100"or"1+(w)/(40)=(100)/(8)"or"(w)/(40)=(10)/(8)-1=(2)/(8)=(1)/(4)"or"w=10 g`

Discussion

38.	Calculate the mass of a non-volatile solute (molar mass 40 g "mol"^(-1) ] which should be dissolvedin 114 g octane to reduce its vapour pressure to 80%.
Answer» Solution :Let the vapour PRESSURE of solvent be `p^0`. Then, vapour pressure of solution = `0.80p^0` Let the MASS of solute = w G Number of moles of the solute ` = (w)/(40)` Number of moles of solvent (octane) =` (114 g)/(114 g "mol"^(-1))` = 1 mol [ molar mass of `C_8H_18=114 g "mol"^(-1)` ] APPLYING Raoult.s law and substituting the values in the equation, we have `(p^0 -p_s)/(p^0) = x_2` `(p^0 - 0.80 (p^0)/(p^0)) = (w/40)/(w // 40 +1)` `0.2 ((w)/(40) + 1) = (w)/(40) rArr (0.8 w)/(40) = 0.2` ` w = 10 g `

Discussion

39.	Calculate the mass of a non-volatile solute (molar mass = 40 g/mol) which should be dissolvedin 114 g octane to reduce its vapour pressure to 80%.
Answer» SOLUTION :Let the vapour pressure of PURE octane = `p_0` Vapour pressure of the solution = `0.8p_0` Lowering of vapour pressure = `0.2p_0` Relative lowering `= (0.2p_0)/(p_0)` According to Raoult.s law, Relative lowering of vapour pressure of a solution is equal to the mole fraction of the solute. Let the mass of solute be = w g Its molecular mass is GIVEN as = 40 u No. of moles of the solute =`(w)/(40)` No. of moles of the solvent ` = 114/114 = 1`[ mol. mass of octane = 114] Mole fraction of the solute ` = (w//40)/(1 + w//40)` According to Raoult.s law, `(0.2p_0)/(p_0) = (w/40)/(1 + w//40) ` on simplification `0.2 = (w//40)/((40 + w)/(40)) ` `0.2 = (w)/(40 + w)` or 0.2(40 + w) = WOR 8 + 0.2w - w = 0 or - 0.8w = -8or w = 10 g

Discussion

40.	Calculatate the mass of compound (molar mass = 256 g mol^(-1) be the dissolved in 75 g of benzene to lower its freezing point by0 .48 K(k_(f) = 5. 12 K kg mol ^(-1) .
Answer» Solution :`W_(A)=75 G=0.075 KG, T_(f)=0.48K, M_(B)=256" g mol"^(-1)` `K_(f)=5.12" K kg mol"^(-1)` `W_(B)=(M_(B)xxDeltaT_(f)xxW_(A))/K_(f)=((256"g mol"^(-1))XX(0.48K)xx(0.075 kg))/((5.12" K kg mol"^(-1)))=1.8 g`

Discussion

41.	Calculate the mass of 60%H_(2)SO_(4) required to decompose 50 g of chalk (calcium carbonate).
Answer» SOLUTION :`underset(100g)(CaCO_(3))+underset(98g)(H_(2)SO_(4))rarrCaSO_(4)+H_(2)O+CO_(2)` `"50 G chalk will REQUIRED PURE "H_(2)SO_(4)=49g` `therefore""60%H_(2)SO_(4)" required "=(100)/(60)xx49=81.67g`

Discussion

42.	Calculate the mass of 6.02xx10^(21) molecules of nitrogen.
Answer» ANSWER :0.28g

Discussion

43.	Calculate the mass of 1.5 gram molecule of sulphuric acid.
Answer» Solution :MOLECULAR mass of `H_(2)SO_(4)` `=2xx1+32+4xx16=98.0` amu Gram molecular mass of `H_(2)SO_(4)=98.0` G Mass of 1.5 gram molecule of `H_(2)SO_(4)=98.0xx1.5=147.0g`

Discussion

44.	Calculatethe massin gramsof animpurityof molarmass100 g "mol"^(-1)whichwouldberequiredto raisethe boilingpointof 50 gof chloroformby 0.30^(@)C (k_(B) for chloroform = 3.63 K kg "mol"^(-1)
Answer» SOLUTION :0.4132 G

Discussion

45.	Calculate the mass ascorbic acid (Vitamin C, C_6H_8O_6) to be dissolved in 75g of acetic acid to lower its melting point by 1.5^@c.Kf =3.9 K kg/ mol
Answer» Solution :Molar MASS of `C_6H_8O_6` (absorbic ACID) = `176g mol^(-1)` `Delta T_f =1.5^@ W_2` (mass of solute) = `(Delta T_bM_2W_1)/(1000K_f) (1.5xx176xx75)/(1000xx3.9)=5.08`

Discussion

46.	Calculate the magnetic moments of the following complexes: (i) [Fe(CN)_(6)]^(-4) (ii)[FeF_(6)]^(-3)
Answer» SOLUTION :(i) `Fe^(+2) rightarrow 3d^(6)``CN^(-)` is a strong LIGAND so `E^(-)` pair 4p, no unpaired `e^(-)` then magnetic moment is zero. (ii) `mu_(BH)=sqrt(n(n+2)) = 0` n=5,so `MU=sqrt(n(n+2))` `mu_=sqrt(5(5+2))` `mu=sqrt(35) = 5.96 BM`

Discussion

47.	Calculate the magnitude of work done (in atm lit.) for the given cyclicprocessinvoling 1moleof an ideal gas .[Given: R =0.08 atmlit./mole/k]
Answer» Solution :`W _(AB) = W_(CD)=0` `W_(BC)=- nRDeltaT =- 1 xxR XX 300 = 300 R` `W_(DA)=- nRDeltaT =- 1 xxR xx -200 = 200 R` `W_("total")=- 100 R` `RARR \|W\|=8 "lit atm "`

Discussion

48.	Calculate the magneticmomentof Ni^(2+).
Answer» Solution :For `Ni^(2+)= 3d^(8) = `` (##PR_CHE_V01_XII_C08_E09_026_S01.png" width="80%"> . Hence,`N=2 ,=SQRT(n(n+2))= sqrt(2(2+2))= sqrt(8)=2.84BM.`

Discussion

49.	Calculate the magnetic moment ofV^(3+)
Answer» SOLUTION : The electronic CONFIGURATION of `V^(3+)` is `[Ar] 45^(@)` 3d^(2)` In the d-orbitals, there are 2 unpaired electrons. `thereforemu= sqrt(N(n+2)) B.M.` = `sqrt(2(2 + 2)) = sqrt8 B.M. = 2.73 B.M.`

Discussion

50.	Calculate the magnetic moment of Ti^(2+)
Answer» Solution :`mu = SQRT(N(n + 2))` `n = 2` ANSWER : `sqrt(2(2 + 2)) or sqrt(8) or 2.825 B.M.` • Unit is not a .value point.

Discussion

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Calculate the mass percentage of benzene (C_(6)H_(6)) and carbon tetrachloride ("CCl"_(4)) if 22 g of benzene is dissolved in 122 g of carbon tetrachloride.

Calculate the mass percentage of benzene (C_6H_6)and carbon tetrachloride ("CCI"_4) if 22 g of benzene is dissolved in 122 g of carbon tetrachloride.

Calculate the mass percentage of aspirin (C_(9)H_(8)O_(4)) in acetonitrile (CH_(3)CN) when 6.5 g of (C_(9)H_(8)O_(4)) is dissolved in 450 g of :

Calculate the mass percentage of aspirin (C_(9)H_(8)O_(4)) in acetonitrile (CH_(3)CN) when 6.5 g of C_(9)H_(8)O_(4) is dissolved in 450 g of CH_(3)CN.

Calculate the mass percentage of aspirin (C_9H_8O_4) in acetonitrile (CH_3CN) when 6.5 g of C_9H_5O_4is dissolved in 450 g of CH_3CN .

Calculate the mass per cent of different elements present in sodium sulphate (Na_(2)SO_(4))?

Calculate the mass of zinc required to produce enough hydrogen to reduce completely 6.7 g of copper oxide of copper ?

Calculate the mass of urea (NH_(2)CONH_(2)) required in making 2.5 kg of 0.25 molal aqueous solution.

Calculate the mass of urea (NH_2CONH_2)required in making 2.5 kg of 0.25 molal aqueous solution.

Calculate the mass of silver deposited from silver nitrate solution by a current of 2 amperes flowing for 30 minutes (equivalent mass of silver is 108).

Calculate the mass of silver deposited from silver nitrate solution by a current of 2 amperes flowing for 30 minutes?

Calculate the mass of NaCI (molar mass =58.5 g mol^(-1)) to be dissoved in 37.2 g of water to lower the freezing point by 2^(@)C assuming that NaCI underfoes complete dissocilation in aqueous solution (K_(f) for water = 1.86 K kg mol^(-1)).

Calculate the mass of NaCl (molar mass = 58.5 g "mol"^(-1)] to be dissolved in 37.2 g of water to lower the freezing point by 2^@C, assuming that NaCl undergoes complete dissociation.

Calculate the mass of Na_(2)CO_(3) which will have the same number of molecules as contained in 12.3 g of MgSO_(4).7 H_(2)O.

Calculate the mass of mercury which can be liberated from HgO at 25^(@)C by the treatment of excess HgO with 41.84 kJ of heat at : (a) constant pressure (b) constant volume Given : Delta H_(f)^(@)(HgO, s) = -90.8 kJ mol^(-1) & M(Hg) = 200.6 g mol^(-1).

Calculatethe mass of Mgand thevolumeof Chlorinegas at NTPproducedduringthe electrolysis ofmolten MgC1_(2)by thepassageof 2 amperesof currentfor 1 hour. MolarmassesofMg and C1_(2) are 24 g mol^(-1) and 71 g mol^(-1) resectively

Calculate the mass of mercury which can be liberated from HgO at 25^(@)C by the treatment of excess HgO with 41.84 kJ of heat at (a) constant pressure (b) constant volume conditions Given : Delta H_(f)^(@) (Hg O, s) = -90.8 kJ mol^(-1)& M (Hg) = 200.6 g mol^(-1)

Calculate the mass of KClO_3 required to produce 9.6 gm of oxygen.

Calculate the mass of iron which will be converted into the oxide (Fe_(3)O_(4)) by the action of 18 g of steam on it.

Calculate the mass of iron required to be converted into its oxide (Fe_3O_4) by the action of 23.4 g of steam on it .

Calculate the mass of (i) 0.1 mole of KNO_(3) (ii) 1xx10^(23) molecules of methane and (iii) 112cm^(3) of hydrogen at STP.

Calculate the mass of hydrogen gas evolved by passing a current of 0.5 ampere for 40 minutes through acidified water.

Calculate the mass of compound (molar mass = 256 g mol^(-1) ) to be dissolved in 75 g of benzene to lower its freezing point by 0.48 K(K_(f) = 5.12 "Kkg " mol^(-1)) .

Calculate the mass of compound (molar mass = 256 g "mol"^(-1) ) to be dissolved in 75 g of benzene to lower its freezing point by 0.48 K (K_f = 5.12 K kg "mol"^(-1)).

Calculate the mass of compound (molar mass = "256 g mol"^(-1)) to be dissolved in 75 g of benzene to lower its freezing point by 0.48 K (K_(f)="5.12 K kg mol"^(-1))

Calculate the mass of CO_(2) which contains the same number of molecules as are contained in 40 g of oxygen.

Calculate the mass of calcium oxide formed when 25 g of 80% pure lime stone is completely decomposed on heating.

Calculate the mass of CaO and CO_2 formed by heating 10 gms of CaCo_3.

Calculate the mass of ascorbic acid (Vitamin C, C_6H_8O_6) to be dissolved in 75 g of acetic acid to lower its melting point by 1.5^@C. K_f = 3.9 K kg "mol"^(-1) .

Calculate the mass of ascorbic acid (Vitamin C, C_(6)H_(8)O_(6)) to be dissolved in 75 g g of acetic acid to lower its melting point by 1.5^(@)C, [K_(f)=3.9" K kg mol"^(-1)].

Calculate the mass of ascorbic acid (Molar mass = 176 g mol^(-1)) to be dissolved in 75 g of acetic acid, to lower its freezing point by 1.5^(@)C. [K_(f)=3.9"K kg mol"^(-1)]

Calculate the mass of aluminium deposited at cathode when 193 C ofcurrent is passed through molten electrolyte containing dissolved alumina.

Calculate the mass of Ag deposited at cathode when a current of 2 amperes was passed through a solution AgNO_(3) for 15 minutes.

Calculate the mass of a single atom of sulphur and a single molecule of carbon dioxide.

Calculate the mass of a non - volatile solute (molar mass 40 g) which should be dissolved in 114 g octane to reduce its vapour pressure to 80%.

Calculate the mass of a non-volatile solute (molar mass "40 g mol"^(-1)) which should be dissolved in 114 g octane to reduce its vapour pressure to 80%.

Calculate the mass of a non-volatile solute (molar mass 40 g "mol"^(-1) ] which should be dissolvedin 114 g octane to reduce its vapour pressure to 80%.

Calculate the mass of a non-volatile solute (molar mass = 40 g/mol) which should be dissolvedin 114 g octane to reduce its vapour pressure to 80%.

Calculatate the mass of compound (molar mass = 256 g mol^(-1) be the dissolved in 75 g of benzene to lower its freezing point by0 .48 K(k_(f) = 5. 12 K kg mol ^(-1) .

Calculate the mass of 60%H_(2)SO_(4) required to decompose 50 g of chalk (calcium carbonate).

Calculate the mass of 6.02xx10^(21) molecules of nitrogen.

Calculate the mass of 1.5 gram molecule of sulphuric acid.

Calculatethe massin gramsof animpurityof molarmass100 g "mol"^(-1)whichwouldberequiredto raisethe boilingpointof 50 gof chloroformby 0.30^(@)C (k_(B) for chloroform = 3.63 K kg "mol"^(-1)

Calculate the mass ascorbic acid (Vitamin C, C_6H_8O_6) to be dissolved in 75g of acetic acid to lower its melting point by 1.5^@c.Kf =3.9 K kg/ mol

Calculate the magnetic moments of the following complexes: (i) [Fe(CN)_(6)]^(-4) (ii)[FeF_(6)]^(-3)

Calculate the magnitude of work done (in atm lit.) for the given cyclicprocessinvoling 1moleof an ideal gas .[Given: R =0.08 atmlit./mole/k]

Calculate the magneticmomentof Ni^(2+).

Calculate the magnetic moment ofV^(3+)

Calculate the magnetic moment of Ti^(2+)