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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.The number of moles of oxygen obtained by the electrolytic decomposition of 90 g water is:
•  2.5
•  5
•  7.5
•  10
Solution
option a

Q2.Resultant molarity of H^+ion in a mixture of 100 mL of 0.1 M H_2 SO_4 and 200 mL of 0.1 M H_3 PO_3 is
•  0.1 M
•  0.2 M
•  0.267 M
•  0.133 M
Solution
(b) H^+ (in H_2 SO_4- a dibasic acid) =0.2 M H^+ (in H_3 PO_3- a dibasic acid) =0.2 M Total [H^+ ]=(M_1 V_1+M_2 V_2)/(V_1+V_2 )=(100 × 0.2+200 ×0.2)/300=0.2 M

Q3.An isotope of Ge_32^76 is
•  Ge_32^77
•  As_33^77
•  Se_34^77
•  Se_34^78
Solution
(a) Atoms of the same element having same atomic numbers but different mass numbers are called isotopes

Q4.An aqueous solution of a substance gives a white precipitate on treatment with dilute hydrochloric acid, which dissolves on heating. When hydrogen sulphide is passed through the hot acidic solution, a black precipitate is obtained. The substance is a
•  Hg^(2+) salt
•  Cr^(2+) salt
•  Ag^+ salt
•  Pb^(2+) salt
Solution
(d)

Q5.O_3 oxidized I^- to I_2.Equivalent weight of O_3 is
•  48
•  32
•  8
•  4
Solution
(c) O_3+6l^-+6H^+→3H_2 O+3l_2 O_3≡3l_2≡6l ∴l≡O_3/6 ∴ Equivalent mass of O_3=48/6=8

Q6.At 100℃ and 1 atm, if the density of the liquid water is 1.0 g cm^(-3) and that of water vapour is 0.0006 g cm^(-3), then the volume occupied by water molecules in 1 L of steam at this temperature is
•  6 cm^3
•  60 cm^3
• 0.6 cm^3
•  0.06 cm^3
Solution
(d) For water vapours, P = 0.0006 g cc^(-1) ∴0.0006=Mass/Volume=Mass/1000 Mass = 1000×0.0006 = 0.6 g The density of liquid water is 1 g cc^(-1) So, the volume occupied by water is Mass/density=0.6/1=0.6 cc

Q7.1.0 g of a monobasic acid when completely acted upon Mg gave 1.301 g anhydrous Mg salt. Equivalent weight of acid is
•  35.54
•  36.54
•  17.77
•  18.27
Solution
(b) (Ew of acid)/(Ew of salt of Mg)=(Weight of acid)/(Weight of salt) Mg + 2HCl (monobasic acid) →MgCl_2+H_2 Let Ew of acid = Ew of H + Ew of acid redical ∴Ew of salt of Mg=Ew of Mg+Ew of acid radical ∴(Ew of acid)/(Ew of Mg salt)=(Weight of acid)/(Weight of Mg salt) ⇒(Ew of H+Ew of acid radical (E))/(Ew of Mg+Ew of acid radical (E))=1.0/1.301 ⇒(1+E)/(12+E)=1.0/1.301 ∴E=35.54 ∴Ew of acid =Ew of H + Ew of acid radical = 1+35.54 =36.54

Q8.Cortisone is a molecular substance containing 21 atoms of carbon per molecule. The mass percentage of carbon in cortisone is 69.98%. its molar mass is
•  176.5
•  252.2
•  287.6
•  360.1
Solution
(d) If carbon content is 69.98 g then molar mass =100 g If carbon content is 21×12 g then molar mass is =100/69.98×21×12 =360.1 g mol^(-1)

Q9. 100 mL of H_2 O_2 is oxidised by 100 mL of 0.01 M KMnO_4 in acidic medium (Mn〖O_4〗^⊖reduced to Mn^(2+)). 100 mL of the same H_2 O_2 is oxidised by V mL of 0.01 M KMnO_4 in basic medium (Mn〖O_4〗^⊖ reduced to MnO_2). Hence, V is
•  500
•  100 S
•  100/3
•  500/3
Solution
(d) V=500/3 mL

Q10. A mixture contains Cu^(2+),Al^(3+) and Ni^(2+). Following steps have been adopted but written in disorder I: Filter, boil of H_2 S gas and add NH_4 Cl,heat and add NH_4 OH II: Filter, add NH_4 OH and pass H_2 S gas III: Pass H_2 S gas into acidified solution of mixture Steps will be used in the following order
•  I, II, III
•  III, I, II
•  III, II, I
• I, III, II
Solution
(b) #### Written by: AUTHORNAME

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