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**Q1.**In an experiment, 50 mLof 0.1 M solution of a metallic salt reacted exactly with 25 mLof 0.1 M solution of sodium sulphite. In the reaction SO_3^(2-) is oxidized to SO_4^(2-).If the original oxidation number of the metal in the salt was 3, what would be the new oxidation number of the metal?

Solution

(c) SO_3^(2-) is oxidized to SO_4^(2-) (change in O.N. = 2) 25 mL of 0.1 M SO_3^(2-)=2.5 millimol =5.0 milliequivalents of SO_3^(2-) =5.0 milliequivalents of M^(3+) 50 mL of 0.1 M M^(3+)=5 millimol (given) Thus, decrease in O.N. of M^(3+) should be 1 So that 5 millimol = 5 milliequivalents Thus, new O.N. of metal = 2

(c) SO_3^(2-) is oxidized to SO_4^(2-) (change in O.N. = 2) 25 mL of 0.1 M SO_3^(2-)=2.5 millimol =5.0 milliequivalents of SO_3^(2-) =5.0 milliequivalents of M^(3+) 50 mL of 0.1 M M^(3+)=5 millimol (given) Thus, decrease in O.N. of M^(3+) should be 1 So that 5 millimol = 5 milliequivalents Thus, new O.N. of metal = 2

**Q2.**Consider the following cases I : 60 g CH_3 COOH II : 30 g HCHO III : 60 g NH_2 CONH_2 IV : 180 g C_6 H_12 O_6 Percentage of carbon is identical in

Solution

(d) Percentage is irrespective of amount given I.CH_3 COOH≡2C 60g 24 g C_%=(24×100)/60=40 II.HCHO≡1 C 30 g 12 g C_%=(12×100)/30=40 NH_2 CONH_2≡1 C III. 60 g 12 C C%=(12×100)/60=20 IV.C_6 H_12 O_6≡6 C 180 g 72 g C_%=(72×100)/180=40 Thus, I, II and IV

**Q3.**10 g of a sample of a mixture of CaCl_2 and NaCl is treated to precipitate all the calcium as CaCO_3. This CaCO_3 is heated to convert all the Ca to CaO and the final mass of CaO is 1.62 g. The precent by mass of CaCl_2 in the original mixture is

Solution

(a) CaCl_2+ NaCl=10 g Let weight of CaCl_2=x g CaCl_2→CaCO_3→CaO 1 mol 1 mol 1 mol x/111 mol x/111 mol x/111 mol Mol of CaO =1.62/56 ∴x/111=1.62/56 x = 3.21 g % of CaCl_2=3.21/10×100=32.1 %

(a) CaCl_2+ NaCl=10 g Let weight of CaCl_2=x g CaCl_2→CaCO_3→CaO 1 mol 1 mol 1 mol x/111 mol x/111 mol x/111 mol Mol of CaO =1.62/56 ∴x/111=1.62/56 x = 3.21 g % of CaCl_2=3.21/10×100=32.1 %

**Q4.**2H_2 O_2 (l)→2H_2 O(l)+O_2 (g) 100 mL of X molar H_2 O_2 gives 3 L of O_2 gas under the condition when 1 mol occupies 24 L. The value of X is

Solution

(a) 2H_2 O_2 (l)→2H_2 O (l)+ O_2 (g) 24 L O_2=1 mol O_2 3 L O_2=1/8 mol O_2=1/4 mol H_2 O_2 in 100 mL =2.5 mol H_2 O_2 L^(-1)

(a) 2H_2 O_2 (l)→2H_2 O (l)+ O_2 (g) 24 L O_2=1 mol O_2 3 L O_2=1/8 mol O_2=1/4 mol H_2 O_2 in 100 mL =2.5 mol H_2 O_2 L^(-1)

**Q5.**100 mL of 1 M BaF_2solution is mixed with 100 mL of 2 M H_2 SO_4.Resulting mixture contains

Solution

(c) 100 mL of 1 M BaF_2=100 millimoles 100 mL of 2 M H_2 SO_4=200 millimoles BaF_2+H_2 SO_4→BaSO_4+2HF 1 mol 1 mol 1 mol 2 mol Initial 100 millimol 200 millimol Final 0 100 millimol 100 millimol 200 milliomol Thus, resulting solution has =100 millimoles H_2 SO_4+200 millimoles HF =200 millimoles of H^++200 millimoles of H^+ =400 millimoles H^+ in 200 mL solution [H^+ ]=(400/1000 moles)/(200/1000 L)=2.0 M [BaSO_4 ]=100 millimoles BaSO_4=0.1 mol Thus, (a), (b) true

(c) 100 mL of 1 M BaF_2=100 millimoles 100 mL of 2 M H_2 SO_4=200 millimoles BaF_2+H_2 SO_4→BaSO_4+2HF 1 mol 1 mol 1 mol 2 mol Initial 100 millimol 200 millimol Final 0 100 millimol 100 millimol 200 milliomol Thus, resulting solution has =100 millimoles H_2 SO_4+200 millimoles HF =200 millimoles of H^++200 millimoles of H^+ =400 millimoles H^+ in 200 mL solution [H^+ ]=(400/1000 moles)/(200/1000 L)=2.0 M [BaSO_4 ]=100 millimoles BaSO_4=0.1 mol Thus, (a), (b) true

**Q6.**Temporary hardness is due to HCO_3^- of Mg^(2+) andCa^(2+). It is removed by addition of CaO Ca(HCO_3 )_2+CaO→2CaCO_3+H_2 O Mass of CaO required to precipitate 2 g CaCO_3 is

Solution

(b) 200 g CaCO_3=56 g CaO

(b) 200 g CaCO_3=56 g CaO

**Q7.**1.06 g Na_2 CO_3 is dissolved in 100 mL solution. 10 mL of this solution can be neutralized by

Solution

(d) 1.06 g Na_2 CO_3=1.06/106 mol=0.01 mol in 100 mL ∴[Na_2 CO_3 ]=0.01/(100/(1000 L))=0.1 M =0.2 N (being diacid base) 10 mL of 0.2 N Na_2 CO_3=2 milliequivalent (a) 10×0.1 N HCl=1 milliequivalent (b) 10×0.1 MH_3 PO_4=10×0.3 N H_3 PO_4=3milliequivalent (c) 20 mL of 0.1 M H_2 SO_4=20 mL of 0.2 N H_2 SO_4=4milliequivalent (d) 20 mL of 0.1 M HCl=20×0.1 N HCl=2milliequivalent

(d) 1.06 g Na_2 CO_3=1.06/106 mol=0.01 mol in 100 mL ∴[Na_2 CO_3 ]=0.01/(100/(1000 L))=0.1 M =0.2 N (being diacid base) 10 mL of 0.2 N Na_2 CO_3=2 milliequivalent (a) 10×0.1 N HCl=1 milliequivalent (b) 10×0.1 MH_3 PO_4=10×0.3 N H_3 PO_4=3milliequivalent (c) 20 mL of 0.1 M H_2 SO_4=20 mL of 0.2 N H_2 SO_4=4milliequivalent (d) 20 mL of 0.1 M HCl=20×0.1 N HCl=2milliequivalent

**Q8.**Mg_2 C_3 (X) is decomposed by H_2 O forming a gaseous hydrocarbon (Y). 8.4 g ofX gives ……… mol of Y

Solution

(a) Mg_2 C_3+H_2 O→Mg(OH)_2+CH_3 C≡CH (X)(Y) 1mol propyne 8.4 g 1 mol 8.4/84=0.1 mol 0.1 mol

(a) Mg_2 C_3+H_2 O→Mg(OH)_2+CH_3 C≡CH (X)(Y) 1mol propyne 8.4 g 1 mol 8.4/84=0.1 mol 0.1 mol

**Q9.**The vapour density of a chloride of an element is 39.5. The Ew of the elements is 3.82. The atomic weight of the element is

Solution

(b) Mw of metal chloride (MCl_x )=M+x×35.5 =2×VD=2×39.5 = 79.0 Mw=Ew×x Valency of metal = x Atomic weight of element = Ew×x MCl_x=3.82×x+x×35.5=79 x=79/(3.82+35.5)≈2 Atomic weight =Ew×x=3.82×2=7.64

(b) Mw of metal chloride (MCl_x )=M+x×35.5 =2×VD=2×39.5 = 79.0 Mw=Ew×x Valency of metal = x Atomic weight of element = Ew×x MCl_x=3.82×x+x×35.5=79 x=79/(3.82+35.5)≈2 Atomic weight =Ew×x=3.82×2=7.64

Solution

(a) 3MnO_4^(2-)+2H_2 O→2MnO_4^-+MnO_2+4OH^- MnO_4^(2-)→2/3 MnO_4^-+1/3 MnO_2

(a) 3MnO_4^(2-)+2H_2 O→2MnO_4^-+MnO_2+4OH^- MnO_4^(2-)→2/3 MnO_4^-+1/3 MnO_2