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IIT JEE exam which consists of JEE Main and JEE Advanced is one of the most important entrance exams for engineering aspirants. The exam is held for candidates who are aspiring to pursue a career in the field of engineering and technical studies. Chemistry is important because everything you do is chemistry! Even your body is made of chemicals. Chemical reactions occur when you breathe, eat, or just sit there reading. All matter is made of chemicals, so the importance of chemistry is that it's the study of everything..

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?
•  0
•  1
•  2
•  4
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

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
•  I,II
•  I,III
•  I,II,III
•  I,II,IV
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
•  32.1%
•  16.2%
•  21.8%
•  11.0%
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 %

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
•  2.5
•  1.0
•  0.5
•  0.25
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)

Q5.100 mL of 1 M BaF_2solution is mixed with 100 mL of 2 M H_2 SO_4.Resulting mixture contains
•  0.1 mole of BaSO_4
•  2MH+
•  Both Are Correct
•  none is correct
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

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
•  2.00 g
•  0.56 g
•  0.28 g
•  1.12 g
Solution
(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
•  10 mL of 0.1 N HCl
•  10 mL of 0.1 M H3PO4
•  20 mL of 0.1 M H2SO4
•  20 mL of 0.1 M HCL
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

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
•  0.1
•  0.2
•  0.3
•  0.4
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

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
•  15.28
•  7.64
•  3.82
•  11.46
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

•  A
•  B
•  C
•  D
Solution
(a) 3MnO_4^(2-)+2H_2 O→2MnO_4^-+MnO_2+4OH^- MnO_4^(2-)→2/3 MnO_4^-+1/3 MnO_2 #### Written by: AUTHORNAME

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