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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.Which has maximum number of H-atoms per gram of the substance?
•  CH4
•  CuSO4.5H2O
•  H2O2
•  H2O
Solution
(a) H-atoms per gram (a) CH_4 (4N_0)/16=N_0/4=0.25 N_0 (b) CuSO_4 .5H_2 O (10N_0)/249.5=0.04N_0 (c) H_2 O_2 (2N_0)/34=0.0588N_0 (d) H_2 O (2N_0)/17=0.1176N_0

Q2.Which is not true about H_3 PO_2?
•  It is a tribasic acid
•  One mole is neutralized by 0.5 mol Ca(OH)_2
•  NaH_2 PO_2 is normal salt
•  It disproportionate to H_3 PO_3and PH_3 on heating
Solution

(a) H_3 PO_2 is monobasic acid H_3 PO_2⇌H^++H_2 PO_2^-

•  A
•  B
•  C
•  D
Solution
(b) Moles of Fe =0.0056/56=10^(-4) mol 1 mol of alum = 2 mol of Fe 2 mol of Fe = 1 mol of alum 10^(-4) mol of Fe=1/2×10^(-4) mol =0.5×10^(-4) mol

Q4. The total number of electrons in one molecular of carbon dioxide is
•  22
•  44
•  66
•  88
Solution
(a) In one molecule of CO_2, the number of electrons is 6 + 8 + 8 = 22

•  A
•  B
•  C
•  D
Solution
(d) 2Al(s)+6HCl(aq)→2Al^(3+) (aq)+6Cl^- (aq)+3H_2 (g) 6 mol 3×22.4 L 3×22.4 L H_2 (g) at STP are produced from 6 moles HCl Hence, 11.2 L H_2 (g) at STP are produced from 1 mole HCl

Q6. I_2 obtained from 0.1 mol of CuSO_4 required 100 mL of 1 M hypo solution, hence, mole percentage of pure CuSO_4is
•  100
•  50
•  25
•  40
Solution
(a) 2CuSO_4+4Kl→Cu_2 I_2+I_2+2K_2 SO_4 I_2+2Na_2 S_2 O_3→2NaL+Na_2 S_4 O_6 2CuSO_4=2Na_2 S_2 O_3 CuSO_4=Na_2 S_2 O_3 100 mL of 1 M hypo=0.1 mol hypo =0.1 mol pure CuSO_4 Hence, 100% pure

Q7.100mL of 0.01 M KMnO_4 oxidised 100 mL H_2 O_2 in acidic medium. The volume of same KMnO_4 required in strong alkaline medium to oxidise 100 mL of same H_2 O_2 will be
•  100/3 mL
•  500/3 mL
•  300/5 mL
•  None of these
Solution
(d) In acidic medium : mEq of KMnO_4≡ meq of H_2 O_2 (0.01×5)×100≡N×100 …(i) In basic medium: mEq of KMnO_4≡mEq of H_2 O_2 (0.01×1)×V=N×100 …(ii) From equations (i) and (ii), we get (0.01×1)×V≡(0.01×5)×100⇒V=500 mL

•  A
•  B
•  C
•  D
Solution
(a) Adding reactions, 2KClO_3→2KCl+3O_2 3 mol O_2 (=22.4×3 L at STP) are formed from =2 mol KClO_3 11.2 L O_2 are formed from =(2×11.2)/(3×22.4) mol KClO_3 =1/3 if yield is 100% =2/3 mol KClO_3 if yield is 50%

Q9.The oxide of an element contains 67.67% of oxygen. Equivalent weight of the element is
•  2.46
•  3.82
•  4.36
•  4.96
Solution
(b) Oxygen =67.67% Metal =32.33% Thus, 67.67 g oxygen combine with =32.33 g metal Hence, 8 g oxygen combine with =(32.33 × 8)/67.67=3.82

•  A
•  B
•  C
•  D
Solution
(a) A new Avogadro’s number =X Mass of one H-atom =1/x=1 amu #### Written by: AUTHORNAME

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SOME BASIC CONCEPTS OF CHEMISTRY Quiz-58
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