52671 + Interview Questions in Chemistry in Class 11

1.	Calculate the molarity, molality and mole fraction of ethyl alcohol in a solution of total volume 95 mL prepared by adding 50 mL of ethyl alcohol (density = 0.789 mL^(-1)) to 50 mL water (density = 1.00 g mL^(-1)).
Answer» SOLUTION :No. of moles of ethyl ALCOHOL `= ("Vol." xx"density")/("Mol. MASS")` `= (50xx0.789)/(46)=0.8576` No. of moles of water `= ("Vol."xx"density")/("Mol. mass")=(50xx1)/(18)=2.7777` Molarity `= ("No. of moles")/("Vol. of sol. in mL")xx1000` `=(0.8576)/(95)xx1000=9.027M` Molarity `= ("No. of moles of solute")/("Mass of SOLVENT in grams")xx1000` `=(0.8576)/(50)xx1000=17.152m` Mole fraction `= (0.8576)/(0.8576+2.7777)=(0.8576)/(3.6353)=0.236`.

Discussion

2.	Calculate the molar volume of the following : (d ) 3.0115xx10^23 molecules of SO_2 gas .
Answer» Solution :`6.023xx10^(23)` MOLECULES =1 mole `3.0115 xx10^(23)` molecules `=(1)/(6.023xx CANCEL (10^(23)))xx3.0115xx 10^(cancel 23)=0.5` moles. Molar VOLUME of 1 mole of `SO_2 =2.24 xx10^(-2)m^3` Molar volume of 0.5 moles of `SO_2=2.24xx10^(-2)xx0.5` `=1.12xx10^(-2)m^3`.

Discussion

3.	Calculate the molar volume of a gas at STP.
Answer» SOLUTION :At S.T.P., P=1 ATM, T=273 K. Apply PV=nRT. Take n=1 MOL, R=0.0821 L atm `K^(-1)mol^(-1)`.

Discussion

4.	Calculate the molar volume of 146 g of HC1 gas and the number of molecules present in it.
Answer» Solution :MOLAR mass of HCl = 36.5 g The molar volume of 36.5 g (1 mole) of HCl = `2.24xx10^(-2)m^(3)` `:.` The volume of 146 g (4 moles) of HCl = `(2.24xx10^(-2))/36.5 xx 146` = `8.96xxl0^(-2)m^(3)` No. of molecules in 146 g of HCl = 4 N = 4 `xx` Avogadro Number = `4 xx 6.023 xx 10^(23)` = `24.092 xx 10^(23)` = `2.4092 xx 10^(24)` molecules.

Discussion

5.	Calculate the molar solubility of Ni(OH)_2 in 0.10 M NaOH. The ionic product of Ni(OH)_2 is 2.0xx10^(-15) .
Answer» Solution :0.1 M NaOH is STRONG base , so COMPLETE ionised Thus NaOH `to Na_((aq))^(+) + OH_((aq))^(-)` `therefore [NaOH]=[OH^-]=0.1`M , NaOH Suppose solubility of `Ni(OH)_2 ="S mol L"^(-1)` So, At equilibrium in this salt `[Ni^(2+)] = "S mol L"^(-1)` `[OH^-] = "2 S mol L"^(-1)` produce `{:(Ni(OH)_(2(s)) hArr, Ni_((aq))^(2+) + , 2OH_((aq))^(-)),("At equilibrium ", SM , 2SM):}` CONCENTRATION So, total `[OH^-]`=(0.1+2S)M (ADDITION of `HO^-`of NaOH and `Ni(OH)_2` ) but the VALUE of S is much less , So ineligible in compare to 0.1 in solution. `[OH^-]=(0.1 +2S) approx "0.1 mol L"^(-1)` `K_(sp) =[Ni^(2+)] [OH^-]^2 = 2xx10^(-15)` `therefore (S)(0.1)^2 =2xx10^(-15)` `therefore S=2.0 xx 10^(-13) "mol L"^(-1) = [Ni^(2+)]`

Discussion

6.	Calculate the molar solubility of Ni(OH)_(2)" in " 0.10mNaOH . The ionic product of Ni(OH)_(2)" is " 2.0 xx 10^(-15).
Answer» ANSWER :`2.0 XX 10^(-13)M `

Discussion

7.	Calculate the molar solubility of Ni (OH)_(2) in 0.10 M NaOH. The ionic product of Ni(OH)_(2) and 2.0xx10^(-15).
Answer» Solution :SUPPOSE the solubility of `Ni(OH)_(2)` in 0.10 M NaOH = s mol `L^(-1)` `Ni(OH)_(2)` in the solution dissociates as :`Ni(OH)_(2) hArr Ni^(2+)+2OH^(-)`. Thus, s mol `L^(-1)` of `Ni(OH)_(2)` in the solution gives s mol `L^(-1) ` of `Ni^(2+)` ion and 2 s mol `L^(-1)` of `OH^(-)` IONS. But `OH^(-)` ions are also PRODUCED from NaOH . As NaOH dissociates completely , `OH^(-)` from NaOH = 0.1 M . HENCE, `[Ni^(2+)]=s "mol" L^(-1) and [OH^(-)]=2 s + 0.1 "mol " L^(-1) ~~ 0.1 "mol" L^(-1)` (`:'` 2sisvery small compared to 0.1 ) `K_(sp) ` of `Ni(OH)_(2) = [Ni^(2+)][OH^(-)]^(2)` `:. s (0.1)^(2) = 2.0 XX 10^(-15 ) or s = 2.0 xx 10^(-13) "mol" L^(-1)`

Discussion

8.	Calculate the molar mass of the following compounds. (i)Urea \|CO(NH_(2) )_(2)\| (ii) Acetone \|CH_(3)COCH_(3)\| (iii) Boric acid \|H_(3)BO_(3)\| (iv) Sulphuric acid \|H_(2) SO_(4)\|
Answer» Solution :(i) Urea`[CO(NH_(2)]` Atomic MASS of C=12 Atomic mass of O=16 Atomic mass of 2(N)=28 Atomic mass o f4(H)=4 `:.` Molar mass of Urea=60 (ii) Acetone`[CH_(3)COCH_(3)]` Atomic mass of 3( C)=36 Atomic mass of 1(O)=16 Atomic mass of 6(H)=6 `:.` Molar mass of Acetone=58 (iii) BOric acisd `[H_(3)BO_(3)]` Atomic mass of B=10 Atomic mass of 3(H)=3 Atomic mass of 3(O)=48 `:.` Molar mass of Boric acid=61 (iv) Sulphuric acid `[H_(2)SO_(4)]` Atmic mass of 2(H)=2 Atomic mass of 1(S)=32 Atomic mass of 4(O)=64 `:.` Molar mass of SUlphuric acid =98

Discussion

9.	Calculate the molar mass of the following : (i) H_(2)O (ii) CO_(2) (iii) CH_(4)
Answer» Solution :Molecular MOLAR mass is the SUM of atomic mass of all atoms in compound. Molar mass of `H=1.0079 u` Molar mass of `O=16.00u` Molar mass of `C=12.01 u` (i) Molar mass of `H_(2)O = 2 xx "atomic mass of H" + 1 xx "atomic mass of O"` `=2xx1.0079 u + 1 xx 10.00u` `=18.0158u` (ii) Molar mass of `CO_(2) = 1 xx "atomic mass of C" + 2 xx "atomic mass of O"` `= 1 xx 12.01 u + 2 xx 16.00 u` `= 44.01u` Molar mass of `CH_(4) = 1 xx "atomic mass of C" + 4 xx "atomic mass of H" ` `= 1 xx 12.01u + 4 xx 1.0079u` `=16.0416u`

Discussion

10.	Calculate the molar mass of glucose.
Answer» Answer :Molar MASS of Glucose `C_(6)H_(12)O_(6)=6xx12+12xx1+6xx16=180`

Discussion

11.	Calculate the molar mass of 20 L of gas weighing 23.2 g at STP.
Answer» Solution :Molar MASS = `(" Weight of the substance X Molar VOLUME")/("Volume of the substance at STP")` Molar volume at STP = `2.24 xx 10^(2)m^(3)` = 22.4 L (or) 22400 cc. Molar mass of the GAS at STP = `(23.2xx22.4)/20 ` = 25.984 g .

Discussion

12.	Calculate the molality of an aqueous solution containing 3.0g of urea (mol.mass=60) in 250g of water.
Answer» SOLUTION :Mass of solute= 3.0 g Moles of solute `=( " Mass of solute ")/(" Molar mass ")` ` = (3.0g )/( 60 g " mol "^(-1))`= 0.05 mol Mass of Solvent = 250 g `= (250)/(1000) = 0.25 kg ` Molality of solution `= (" Moles of solute ")/(" Mass of Solvent in kg . ")` `= (0.05 Mol )/( 0.25 kg ) =0.2 `m

Discussion

13.	Calculatedthe molality of a solution containing 7.5 g of glycine(NH_2 -CH_2 -COOH)dissolved in 500g of water
Answer» SOLUTION :Molality `= ("No. of moles of solute")/("MASS of solvent (in KG)")` No. of moles of GLYCINE `=("mass of glycine")/("Molar mass of glycine") = (7.5)/(75) =0.1` Molality `= ( 0.1)/(0.5 kg) =0.2 m`

Discussion

14.	Calculate the molality of a solution containing 20.7 g of potassium carbonate in 500 mL of solution (assume density of solution = "1g mL"^(-1))
Answer» Solution :Mass of 500 mL of the solution `= 500 XX 1 = 500 G` Mass of SOLVENT (water) `= 500 - 20.7 = 479.3 g = 0.4693 kg` Molality of solution `(m)=((20.7g))/(("138 g MOL"^(-1))xx("0.4793 kg"))=0.313 "mol kg"^(-1)=0.313` m.

Discussion

15.	Calculate the molality of a solution by dissolving 0.850g of ammonia (NH_3) in 100g of water.
Answer» ANSWER :0.5m

Discussion

16.	Calculate the molality of 1M solution of sodium nitrate. The density of solution is 1.25 "g cm"^(-3)
Answer» Solution :`1 M NaNO_(3)` solution contains 85 g of `NaNO_(3)` in 1000 ML of the solution DENSITY of solution `= 1.25 g mL^(-1)`. Therefore, mass solution `=1000 xx 1.25 = 1250 g` Mass of SOLVENT (water) `=1250 - 85 = 1165 g = 1.165 kg` Molality of solution `(m)=(("1 MOL"))/(("1.165 kg"))="0.86 mol kg"^(-1)=0.86m`.

Discussion

17.	Calculate the molality of 1 litre solution of 93% H_2SO_4 (weight/volume). The density of the solution is 1.84 g ml^(-1).
Answer» ANSWER :10.43 m

Discussion

18.	Calculate the molality of 90% H_2SO_4 (weight/volume). The density of solution is 1.80 g mL^(-1).
Answer» ANSWER :10.2 ML

Discussion

19.	Calculate the molality, if density of 8.653% (w/v) Na_(2)CO_(3) solution is 1.018g cc^(-1).
Answer» ANSWER :0.876 MOL `KG^(-1)`

Discussion

20.	Calculate the minimum uncertainty in velocity of a particle of mass 1.1 xx 10^(-27)kg if uncertainly in its position is 3xx 10^(-10) cm (h = 6.6 xx 10^(-34) kg m^(2) s^(-1))
Answer» SOLUTION :`DELTA x = 3 XX 10^(-10) CM 3 xx 10^(-12) m`

Discussion

21.	Calculate the Miller indices of crystal planes which cut through the crystal axes at (i) (2a, 3b, c) (ii) (a, b, c) (iii) (6a, 3b, 3c) and (iv) (2a, -3b, -3c).
Answer» SOLUTION : following the procedure GIVEN above, we prepare the tables as FOLLOWS:

Discussion

22.	Calculate the maximum work done (in multiple of 10^(3)) in expanding 16g of oxygen at 300K and occupying a volume of 5 dm^(3) isothermally, until the voume becomes 25 dm^(3) (Give you anwer as nearest integer)
Answer» Solution :Maximum work (or) slow process terms indicates that the process must be reversible. `w= - 2.303n TR "long" (v_(2))/(v_(1))` `= -2.303 xx (16)/(32) (300) (2) "log" (25)/(5) = -2 xx 10^(-3)J`

Discussion

23.	Calculate the maximum % efficiency of thermal engine operating between 110^(@) and 25^(@) .
Answer» SOLUTION :% Efficiency =`[(T_1-T_2)/T_1]XX100` `T_1=110^@C + 273`=383 K `T_2=25^@C` + 273 +298 K `therefore` % Efficiency =`[(383-298)/383]xx100` `=(85xx100)/383=8500/383` % Efficiency =22.2 %

Discussion

24.	Calculate the masses of (i) 2.00 gram atoms of chlorine (ii) 2.00 gram molecules of chlorine (iii) 10.50 gram molecules of ammonia.
Answer» Solution : (i) `therefore`The ATOMIC MASS of chlorine (CI) = 35.45 amu `therefore`The gram atomic mass of chlorine = 35.45 g Since, mass in grams = Number of gram atoms x gram atomic mass We have, mass of chlorine = `2.00 xx 35.45 = 70.9` g (ii) `therefore` The molecular mass of chlorine `(Cl_(2))` `therefore` The gram molecular mass of `Cl_(2) = 70.9 g` Since, mass in grams = Number of gram molecules `xx` gram molecular mass We have, mass of chlorine = `2.00 xx 70.9 = 1.42 xx 10^(2)`g (iii) The molecular mass of ammonia `(NH_(3))` `=14.01 + (3 xx 1.008) = 17.034 amu` = 14.01 + (3 x 1.008) = 17.034 amu The gram molecular mass of `NH_3= 17.034` g Hence, we have mass of `NH_(3) = 10.50 xx 17.034 = 178.8` g

Discussion

25.	Calculate the mass percentage of the element oxygen in hydrogen peroxide.
Answer» ANSWER :0.941

Discussion

26.	Calculate the mass percent of benzene (C_(6)H_(6)) and carbon tetrachloride (C Cl_(4)) if 22 g benzene is dissolved in 122 g of carbon tetrachloride.
Answer» Solution :Mass PERCENT (Mass %) `= ("Mass of SOLUTE")/("Mass of solution")xx100` Mass of benzene = 22 g , Mass of carbon TETRACHLORIDE = 122 g. Mass percent of benzene `= (("22 g"))/(("22g+122g"))xx100=15.28%` Mass percent of carbon tetrachloride `= 100 - 15.28 = 84.72 %`

Discussion

27.	Calculate the mass percent of calcium, phosphorus and oxygen in calcium phosphate Ca_(3)(PO_(4))_(2).
Answer» Solution :Masspercent of calcium `=(3 XX "(atomic mass of calcium)")/("molecular mass of " Ca_(3)(PO_(4))_(2))xx100` `= (120 u)/(310 u) xx 100 = 38.71%` Mass percent of PHOSPHORUS `=(2xx"(atomic mass of phosphorus)")/("molecular mass of " Ca_(3)(PO_(4))_(2))xx100` `=(2xx31 u )/(310 u) xx 100 = 20%` Mass percent of oxygen`=(8 xx "(atomic mass of oxygen)")/("molecular mass of " Ca_(3)(PO_(4))_(2))xx100` `=(8xx16u)/(310 u) xx 100 = 41.29%`

Discussion

28.	Calculate the mass per cent of different elements present in sodium sulphate (Na_2SO_4)
Answer» Solution :Gram molecular mass of `Na_2SO_4` `=(2 xx 22.99) + 32.06 + (4 xx 16.00) = 142.04 G MOL^(-1)` Mass per CENT of an element `=("Mass of the element in one MOLE of the compound")/("Gram molecular mass of the compound") xx 100` `therefore` Mass % of sodium `=(22.99 xx 2)/(142.04)xx 100 = 32.37 %` Mass % of sulphur `=32.06/(142.04) xx 100 = 22.57 %` Mass % of OXYGEN `=(16.00 xx 4)/(142.04) xx 100 = 45.06 %`

Discussion

29.	Calculate the massper cent of different elements present in sodium sulphate (Na_(2)SO_(4)).
Answer» Solution :Mass PRECENT `= ("Mass of solute")/("Mass of solution")xx100` Molecular mass of `Na_(2)SO_(4)=(2xx22.9g)+32.06+(4xx16.00)=142.04g` % of NA `=(45.98xx100)/(142.04g)=32.37%` % of S `=(32.06xx100)/(142.04)=22.57%` % of O `= (64xx100)/(142.04)=45.06%`

Discussion

30.	Calculate the mass of urea (NH_(2)CONH_(2)) required to prepare 2.5 kg of 0.25 molal aqueous solution.
Answer» Solution :Molality of solution `= ("Mass of UREA/MOLAR mass of urea")/("Mass of water in kg.")` Molality of solution `=0.25 m=("0.25 mol kg"^(-1))` Molar mass of urea `(NH_(2)CONH_(2))=2xx14+1xx12+1xx16+4xx1=60 "g mol"^(-1)` Mass of solvent (water) = 2.5 kg `("0.25 mol kg"^(-1))=("Mass of urea")/(("60 g mol"^(-1))XX("2.5 kg"))` Mass of urea `=("0.25 mol kg^(-1))xx("60 g mol"^(-1))xx(2.5 kg)` `= 38.5 g`.

Discussion

31.	Calculate the mass of urea (NH_(2) CONH_(2)) required in making 2.5 kgof 0.25 molar aqueous solution.
Answer» SOLUTION :`0.25` molar aqueous solution MEANS that Moles of urea `= 0.25 ` mole MASS of solvent (WATER) `=1 kg = 1000 kg` Molar mass of urea `= 14 + 2 + 12 + 16 + 14 + 2 =60 g mol ^(-1)` `therefore 0.25` mole of urea `= 60xx0.25 ` mole = 15 g Total mass of the solution `= 1000 + 15 g = 1015 g = 1. 015g` Thus, `1.015` kg of solution CONTAIN urea = 15 h `therefore 2.5 kg ` of solution will require urea `= (15)/(1.015) xx 2.5 kg =37g`

Discussion

32.	Calculate the mass of the solute in the following solutions : (i) 100 cm^(3) of N/10 KOH (ii) 150 cm^(3) of M/2 HCl
Answer» ANSWER :(i) 0.56 G, (II) 12.25 g , (III) 31.5 g , (IV) 2.74 g

Discussion

33.	Calculate the mass of a photon with wavelength 3.6Å.
Answer» ANSWER :`6.135 XX 10^(-29) KG `

Discussion

34.	Calculate the mass of the following : (i) 1 atom of sodium (ii) 10 molecules of argon (iii) 1 molecule of CO_2 (iv) 1 molecule of H_2SO_4.
Answer» Solution :(i) The gram atomic mass of sodium = 22.99 This mass contains `6.022 xx 10^(23)` sodium ATOMS. `therefore` Mass of one atom of sodium `=(22.99)/(6.022 xx 10^(23)) = 3.817 xx 10^(-23) g` (ii) Argon is monoatomic. Therefore, its one molecule contains only one atom. HENCE, gram molecular mass of argon = its gram atomic mass 39.95 g. This mass contains `6.022 xx 10^(23)` molecules Hence, mass of 10 molecules of argon `=(10 xx 39.95)/(6.022 xx 10^(23)) = 6.634 xx 10^(-22)g` (iii) Gram molecular mass of `CO_(2)` `=12.01 xx (2 xx 16.0) = 44.01 g` This mass contains `6.022 xx 10^(23)` molecules. Therefore, mass of one molecule of `CO_(2) = (44.01)/(6.022 xx 10^(23)) = 7.308 xx 10^(-23) g` (iv) Gram molecular mass of `H_(2)SO_(4)` `=(2 xx 1.008) + 2.06 + (4 xx 16.0) = 98.076` g This mass contains `6.022 xx 10^(23)` molecules. Therefore, mass of one molecule of `H_(2)SO_(4) = (98.076)/(6.022 xx 10^(23)) = 1.629 xx 10^(-22) g`

Discussion

35.	Calculate the mass of the following : (i) 1 atom of carbon (ii) 1 atom of silver (iii) 1 molecule 1 benzene (C_(6)H_(6)) (iv) 1 molecule of water (H_(2)O)
Answer» SOLUTION :(i) `6.022 xx 10^(23)` atoms of carbon have mass = 12.0 g 1.0 atom of carbon has mass `= (12.0)/(6.022xx10^(23))=1.99xx10^(-23)g` (ii) `6.022 xx 10^(23)` atoms of silver have mass = 107.8 g 1.0 atom of silver has mass `= (107.8g)/(6.022xx10^(23))=1.79xx10^(-22)g` (III) `6.022 xx 10^(23)` molecules of benzene `(C_(6)H_(6))` have mass = 78.0 g 1.0 molecule of benzene `(C_(6)H_(6))` has mass `= (78.0g)/(6.022xx10^(23)) = 1.295 xx 10^(-22)g` (IV) `6.022xx10^(23)` molecules of `H_(2)O` have mass = 18.0 g 1.0 molecule of `H_(2)O` has mass `= (18.0 g)/(6.022 xx 10^(23))=2.99 xx 10^(-23) g`.

Discussion

36.	Calculate the mass of the following atoms in amu(a) Helium (mass of He = 6.641xx10^(-24)g)(b) Silver (mass of Ag = 1.790xx10^(-22) g)
Answer» Solution :1 amu = `1.66056xx10^(-24)` (a) The MASS of Helium atom in amu =`(6.641xx10^(-24))/(1.66056xx10^(-24))`= 3.9992 amu. (b)The mass of SILVER atom in amu = `(1.790xx10^(-22))/(1.66056xx10^(-24))` = 107.79 amu

Discussion

37.	Calculate the mass of oxygenatom in amu.
Answer» SOLUTION :OXYGEN MASS of Oxygen ATOM `=2.656xx10^(-23)` 1a.m.u ( or 1 u is equal to `1.66075` multiple The mass of oxygen atom in amu `=(2.656xx10^(-23))/(1.66075xx10^(-24))=15.992` a.m.u.

Discussion

38.	Calculate the mass of sodium which contains the same number of atoms as are present in 10 g of magnesium.
Answer» ANSWER :9.456 G

Discussion

39.	Calculate the mass of sodium (in kg) present in 95 kg of a crude sample of sodium nitrate whose percentage purity is 70%.
Answer» Solution :Sodium Nitrate = `NaNO_(3)` Molecular MASS of Sodium Nitrate = NaNO3 = 23 + 14 + 48 = 85 100% pure 85 g of `NaNO_(3)` CONTAINS 23 g of Sodium. 100% pure `95xx10^(3)` g of `NaNO_(3)` will contains` 23/85 xx95 xx 10^(3)` = `25.70xx10^(3)` g of Sodium. 100% pure NaNO3 contains `25.70xx10^(3)` g of Sodium. `:. `100% pure `NaNO_(3)` contains `25.70xx10^(3)`g of sodium `:. `70% pure `NaNO_(3)` will contains `(25.70xx10^(3))/100 xx 70` 17999 (g) or 17.99 kg of Na

Discussion

40.	Calculate the mass of sodium chloride that is to be treated with 19.6g of sulphuric acid in order to produce 7.3g of hydrogen chloride and 24g of sodium bisulphate. Hint: NaCl +H_(2)SO_(4) to NaHSO_(4)+HCl
Answer» ANSWER :11.7g

Discussion

41.	Calculate the mass of sodium acetate (CH3COONa) required to make 500 mL of 0.375 molar aqueous solution. Molar mass of sodium acetate is 82.0245 g mol^(–1).
Answer» Solution :Molarity of an aqueous solution is given by `W=(MM.V)/1000` In the present case, `M=0.375, M. 82.0245, V=500 mL, w`=? SUBSTITUTING the VALUES, we get `w=(0.375 XX 82.0245 xx 500)/1000 = 15.38 g` Here, mass of sodium ACETATE required = 15.38 g.

Discussion

42.	Calculate the mass of sodium acetate (CH_3COONa)required to make 500 mL of 0.375 M aqueous solution, (Molar mass of CH_3COONa = 82,0245g mol^(-1)
Answer» SOLUTION :Number of moles of `CH_3COONa` in 1000 mL=0.375 Number of moles of`CH_3COONa` in 500 Ml=0.375/2 MASS of `CH_3COONa` REQUIRED = `0.375/2xx82.0245 = 15.38g`

Discussion

43.	Calculate the mass of sodium acetate (CH_(3)COONa) required to make 500 mL of 0.375 molar aqueous solution. Molar mass of sodium acetate is 82.0245 g "mol"^(-1).
Answer» SOLUTION :Molarity `= (w xx 1000)/(m xx "volume of solution ( in ML)")` Where, w `=` weight of solute ( in gm) m `=` MOLECULAR MASS of solute `w=(0.375xx82.0245xx500)/(1000)` `= 15.379~=15.38g`

Discussion

44.	Calculate the mass of oxalic (H_(2)C_(2)O_(4).2H_(2)O) which can be oxidised to CO_(2) by 100 mL of MnO_(4)^(-)(acidic) solution, 10 mL of whichare capable of oxidising 50.0 mL of 1.0 m I^(-) to I_(2). Also calculate the weight of FeC_(2)O_(4) oxidised by same amount of MnO_(4)^(-).
Answer» ANSWER :`H_(2)C_(2) O_(4) =31.5g, FeC_(2)O_(4) = 24g`

Discussion

45.	Calculate the mass of oxygen obtained by complete decomposition of 10kg of pure potassium chlorate (Atomic mass K=39, O=16 and Cl = 35.5) 2KClO_3 to 2KCl+ 3O_2
Answer» SOLUTION :MOLECULAR mass of `KClO_3 = 39 + 35.5 + 48 = 122. 5` Molecular mass of `O_2 = 16 + 16 = 32` According to the Stoichiometric equation written above ` ( 2 xx 122.5) xx 10^(-3)`KG of `KClO_3` on heating gives `(3 xx 32) xx 10^(-3)` kg of oxygen . 10 kg of `KClO_3` gives =`(3 xx 32 xx 10^(-3))/( 2 xx 122.5 xx 10^(-3)) xx 10` `=3 .92 kg ` of O_2`

Discussion

46.	The mass of a non-volatile solute (molar mass 80 g mol^(-1) ) which should be dissolved in 92g of toluene to reduce its vapour pressure to 90%
Answer» Solution :`(Delta P)/(P^(@))=X_(2)` `(100-90)/(100)=(n_(2))/(n_(1)+n_(2))` `(1)/(10)=(n_(2))/(n_(1)+n_(2))` `(n_(2))/(n_(1)+n_(2))=10` `(n_(1))/(n_(2))+1 =10 "" [n_(1)=(92)/(92)=1]` `(1)/(n_(2))=9` `n_(2)=(1)/(9)` `(W_(2))/(M_(2))=(1)/(9)` `W_(2)=(M_(2))/(9)=(80)/(9)=8.89 g`

Discussion

47.	Calculate the mass of one mole of electrons?
Answer» Solution :Mass of one electron `=9.1095 XX 10^(-31)kg` Mass of one MOLE electrons `=9.1095 xx 10^(-31) xx 6.022xx 10^(23)=5.5 xx 10^(-7)Kg` =0.55mg

Discussion

48.	Calculate the mass of lime that can be prepared by heating 200 kg of limestone that is 90% pure CaCO_3 CaCO_3 to CaO + CO_2 100 kg xx 10^(-3)"" 56 kg xx 10^(-3)
Answer» Solution :`200kg ` of `90%` PURE `CaCO_3 = 200 xx (90)/(100)` `=180 KG ` pure ` CaCO_3` ` 100 xx 10^(-3)` kg of pure `CaCO_3 ` on heating gives ` 56 xx 10^(-3)` kg of CaO 180 kg of `CaCO_3` gives on heating `=(56 xx 10^(-3) xx 180)/( 100 xx 10^(-3))` `= 100.8 kg ` CaO

Discussion

49.	Calculate the actual mass of : the atom of silver. Given mass (Ag = 108, N = 14, H = 1)
Answer» ANSWER :(i) `1.792 XX 10^(-22)G`, (ii) `1.297 xx 10^(-22)g`

Discussion

50.	Calculate the mass of : (i) 1.2 gram atom of oxygen (ii) 5.2 gram atom of iodine (iii) 5.6 gram atom of chlorine.
Answer» SOLUTION :(i) 1.0 gram atom of oxygen = 16.0 g 1.2 gram atoms of oxygen `= 16.0 XX 1.2 = 19.2 g` (ii) 1.0 gram atom of iodine = 127.0 g 5.2 gram atom of iodine `= 5.2 xx 127.0 = 660.4 g` (III) 1.0 gram atom of CHLORINE `= 35.5 g` 5.6 gram atom of chlorine `= 5.6 xx 35.5 = 198.8 g`

Discussion

Explore topic-wise InterviewSolutions in Current Affairs.

Calculate the molarity, molality and mole fraction of ethyl alcohol in a solution of total volume 95 mL prepared by adding 50 mL of ethyl alcohol (density = 0.789 mL^(-1)) to 50 mL water (density = 1.00 g mL^(-1)).

Calculate the molar volume of the following : (d ) 3.0115xx10^23 molecules of SO_2 gas .

Calculate the molar volume of a gas at STP.

Calculate the molar volume of 146 g of HC1 gas and the number of molecules present in it.

Calculate the molar solubility of Ni(OH)_2 in 0.10 M NaOH. The ionic product of Ni(OH)_2 is 2.0xx10^(-15) .

Calculate the molar solubility of Ni(OH)_(2)" in " 0.10mNaOH . The ionic product of Ni(OH)_(2)" is " 2.0 xx 10^(-15).

Calculate the molar solubility of Ni (OH)_(2) in 0.10 M NaOH. The ionic product of Ni(OH)_(2) and 2.0xx10^(-15).

Calculate the molar mass of the following compounds. (i)Urea |CO(NH_(2) )_(2)| (ii) Acetone |CH_(3)COCH_(3)| (iii) Boric acid |H_(3)BO_(3)| (iv) Sulphuric acid |H_(2) SO_(4)|

Calculate the molar mass of the following : (i) H_(2)O (ii) CO_(2) (iii) CH_(4)

Calculate the molar mass of glucose.

Calculate the molar mass of 20 L of gas weighing 23.2 g at STP.

Calculate the molality of an aqueous solution containing 3.0g of urea (mol.mass=60) in 250g of water.

Calculatedthe molality of a solution containing 7.5 g of glycine(NH_2 -CH_2 -COOH)dissolved in 500g of water

Calculate the molality of a solution containing 20.7 g of potassium carbonate in 500 mL of solution (assume density of solution = "1g mL"^(-1))

Calculate the molality of a solution by dissolving 0.850g of ammonia (NH_3) in 100g of water.

Calculate the molality of 1M solution of sodium nitrate. The density of solution is 1.25 "g cm"^(-3)

Calculate the molality of 1 litre solution of 93% H_2SO_4 (weight/volume). The density of the solution is 1.84 g ml^(-1).

Calculate the molality of 90% H_2SO_4 (weight/volume). The density of solution is 1.80 g mL^(-1).

Calculate the molality, if density of 8.653% (w/v) Na_(2)CO_(3) solution is 1.018g cc^(-1).

Calculate the minimum uncertainty in velocity of a particle of mass 1.1 xx 10^(-27)kg if uncertainly in its position is 3xx 10^(-10) cm (h = 6.6 xx 10^(-34) kg m^(2) s^(-1))

Calculate the Miller indices of crystal planes which cut through the crystal axes at (i) (2a, 3b, c) (ii) (a, b, c) (iii) (6a, 3b, 3c) and (iv) (2a, -3b, -3c).

Calculate the maximum work done (in multiple of 10^(3)) in expanding 16g of oxygen at 300K and occupying a volume of 5 dm^(3) isothermally, until the voume becomes 25 dm^(3) (Give you anwer as nearest integer)

Calculate the maximum % efficiency of thermal engine operating between 110^(@) and 25^(@) .

Calculate the masses of (i) 2.00 gram atoms of chlorine (ii) 2.00 gram molecules of chlorine (iii) 10.50 gram molecules of ammonia.

Calculate the mass percentage of the element oxygen in hydrogen peroxide.

Calculate the mass percent of benzene (C_(6)H_(6)) and carbon tetrachloride (C Cl_(4)) if 22 g benzene is dissolved in 122 g of carbon tetrachloride.

Calculate the mass percent of calcium, phosphorus and oxygen in calcium phosphate Ca_(3)(PO_(4))_(2).

Calculate the mass per cent of different elements present in sodium sulphate (Na_2SO_4)

Calculate the massper cent of different elements present in sodium sulphate (Na_(2)SO_(4)).

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

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

Calculate the mass of the solute in the following solutions : (i) 100 cm^(3) of N/10 KOH (ii) 150 cm^(3) of M/2 HCl

Calculate the mass of a photon with wavelength 3.6Å.

Calculate the mass of the following : (i) 1 atom of sodium (ii) 10 molecules of argon (iii) 1 molecule of CO_2 (iv) 1 molecule of H_2SO_4.

Calculate the mass of the following : (i) 1 atom of carbon (ii) 1 atom of silver (iii) 1 molecule 1 benzene (C_(6)H_(6)) (iv) 1 molecule of water (H_(2)O)

Calculate the mass of the following atoms in amu(a) Helium (mass of He = 6.641xx10^(-24)g)(b) Silver (mass of Ag = 1.790xx10^(-22) g)

Calculate the mass of oxygenatom in amu.

Calculate the mass of sodium which contains the same number of atoms as are present in 10 g of magnesium.

Calculate the mass of sodium (in kg) present in 95 kg of a crude sample of sodium nitrate whose percentage purity is 70%.

Calculate the mass of sodium chloride that is to be treated with 19.6g of sulphuric acid in order to produce 7.3g of hydrogen chloride and 24g of sodium bisulphate. Hint: NaCl +H_(2)SO_(4) to NaHSO_(4)+HCl

Calculate the mass of sodium acetate (CH3COONa) required to make 500 mL of 0.375 molar aqueous solution. Molar mass of sodium acetate is 82.0245 g mol^(–1).

Calculate the mass of sodium acetate (CH_3COONa)required to make 500 mL of 0.375 M aqueous solution, (Molar mass of CH_3COONa = 82,0245g mol^(-1)

Calculate the mass of sodium acetate (CH_(3)COONa) required to make 500 mL of 0.375 molar aqueous solution. Molar mass of sodium acetate is 82.0245 g "mol"^(-1).

Calculate the mass of oxalic (H_(2)C_(2)O_(4).2H_(2)O) which can be oxidised to CO_(2) by 100 mL of MnO_(4)^(-)(acidic) solution, 10 mL of whichare capable of oxidising 50.0 mL of 1.0 m I^(-) to I_(2). Also calculate the weight of FeC_(2)O_(4) oxidised by same amount of MnO_(4)^(-).

Calculate the mass of oxygen obtained by complete decomposition of 10kg of pure potassium chlorate (Atomic mass K=39, O=16 and Cl = 35.5) 2KClO_3 to 2KCl+ 3O_2

The mass of a non-volatile solute (molar mass 80 g mol^(-1) ) which should be dissolved in 92g of toluene to reduce its vapour pressure to 90%

Calculate the mass of one mole of electrons?

Calculate the mass of lime that can be prepared by heating 200 kg of limestone that is 90% pure CaCO_3 CaCO_3 to CaO + CO_2 100 kg xx 10^(-3)"" 56 kg xx 10^(-3)

Calculate the actual mass of : the atom of silver. Given mass (Ag = 108, N = 14, H = 1)

Calculate the mass of : (i) 1.2 gram atom of oxygen (ii) 5.2 gram atom of iodine (iii) 5.6 gram atom of chlorine.