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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 equivalent weight of MnSO_4 is half its molecular weight when it is converted to
•  Mn2o3
•  MnO2
•  MnO4-
•  MnO4^(2-)
Solution
(b) MnSO_4 x+6-8=0,x=2 In (a) Mn_2 O_3, 2x-6=0,x=3 In (b) MnO_2, x-4=0,x=4 In (c) MnO_4^( ⊝), x-8=-1,x=7 In (d) MnO_4^( 2-),x-8=-1,x=6 In (b), the change in oxidation number is 4-2=2 Therefore, the answer is (b)

Q2.There is foul smell in presence of moisture with
•  AlCl3
•  Al2(So4)3
•  FeS
•  FeSO4
Solution

(c) FeS+2H_2 O→Fe(OH)_2+H_2 S Foul smell

Q3.  The best way to ensure complete precipitation from saturated H_2 S(aq) of a metal ion M^(2+), as its sulphide, MS(s), is to
•  increase H2S in sol.
•  Raise the PH
•  Heat the sol.
Solution
(c) H_2 S⇌2H^++S^(2-) On increasing pH, [H^+] decreases, hence [S^(2-)] increases making precipitation of M^(2+) as MS

Q4. Sucrose solution is 1 molal. Mole fraction of sucrose in the aqueous solution is
•  0.018
•  0.015
•  0.036
•  0.009
Solution
(a) Sucrose = 1 molal = 1 mol in 1000 g H_2 O Moles of water =1000/18 Mole of sucrose = 1 Total moles =1+1000/18 Mole fraction of sucrose =1/(1 +1000/18)=18/1018=0.018

Q5.[Fe(H_2 O)_5 NO]^(2+) is formed as brown ring in NO_3^- test. Fe in this complex has….. unpaired electrons
•  one
•  two
•  three
•  four
Solution
(c) Reaction and appearance of ring is due to charge transfer NO⟶NO^++e^- Fe^(2+)+e^-⟶Fe^+ ⥮ ↿ ↿ ↿ ↿ ⥮ ⥮ ↿ ↿ ↿ Three unpaired electrons [Fe^+ (H_2 O)_5 NO^+ ]≡[Fe(H_2 O)_5 NO]^(2+) Note Presence of three unpaired electrons is confirmed by magnetic moment of Fe which is √15 BM

Q6. Metal chloride A is soluble in hot water but insoluble in cold water. Select correct statement about A. Thus
•  A can give yellow ppt. with K_2 CrO_4
•  A can give white ppt. with K2SO4
•  A can give yellow ppt. with KI
•  All of the above are correct statements
Solution
(d) PbCl_2 is soluble in hot water PbCl_2 □(→┴(K_2 CrO_4 ) ) PbCrO_4↓ yellow ppt PbCl_2 □(→┴( K_2 SO_4 ) ) PbSO_4 white ppt PbCl_2 □(→┴( KI ) ) Pbl_2↓ yellow ppt

Q7.When 10 g CaCO_3 reacts with 20 g BaCl_2 BaCl_2+CaCO_3→BaCO_3+CaCl_2 then limiting reactant is
•  CaCO3
•  BaCl2
•  CaCl2
•  none of these
Solution
(b) BaCl_2+CaCO_3→BaCO_3+CaCl_2 1 mol 1 mol 208 g 100 g 10 g CaCO_3 reacts with 20.8 g BaCl_2 But BaCl_2 taken =20 g Thus, BaCl_2 is the limiting reactant

Q8.36.5% HCl has density equal to 1.20 g mL^(-1). The molarity (M) and molality (m), respectively, are
•  15.7,15.7
•  12,12
•  15.7,12
•  12,15.7
Solution
(d) M=(% by weight×10×d)/〖Mw〗_2 =(36.5×10×1.2 )/36.5=12 M m =(36.5×1000)/(36.5×(100-36.5 ))=1000/63.5=15.7 m

Q9.If 1 L of O_2 at 15°C and 750 mm pressure contains N molecules, the number of molecules in 2 liters of SO_2 under the same conditions of temperature and pressure will be
•  N/2
•  N
•  2N
•  4N
Solution
(c) Under similar conditions of temperature and pressure, equal volume of gas contains equal number of molecules ∴1 L=N molecules 2 L=2N molecules

Q10.Al and KClO_3 react together to form Al_2 O_3 according to 2KClO_3→2KCl+3O_2 4Al+3O_2→2Al_2 O_3 4 moles of KClO_3(50% pure) on reaction with excess of Al form Al_2 O_3
•  2 mol
•  4 mol
•  6 mol
•  8 mol
Solution
(a) From reaction stoichiometry 1 mol KClO_3 gives =1 mol Al_2 O_3 Thus, 4×50/100=2 mol pure KClO_3 give =2 mol Al_2 O_3 #### Written by: AUTHORNAME

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