SOME BASIC CONCEPTS OF CHEMISTRY Quiz-20

 

SOME BASIC CONCEPTS OF CHEMISTRY Quiz-20

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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.Solution of (X) in dil. HCl+H_2 O→ white turbidity. (X) □(→┴(H_2 S/HCl ) ) black ppt (Y).(Y) is soluble in  :

•  H_2 SO_4 -
•  YAS
•  HNO_3
•  HCl

Solution

Q2.11.2 L of CO_2 is absorbed in 1 mole NaOH,Na_2 CO_3 formed is 2NaOH+CO_2→Na_2 CO_3+H_2 O

•  5.6 mol
•  11.2 mol
•  0.5 mol
•  1.0 mol

Solution

Q3.X^-,Y^(2-) and Z^(3-) are isotonic and isoelectronic. Thus, increasing order of atomic number of X,Y and Z is  

•  X
•  Z
•  X=Y=Z
•  Z

Solution

Q4.The ion most difficult to remove as a precipitate is

•  Ag^+
•  NH_4^+
•  Fe^(3+)
•  Cu^(2+)

Solution

Q5.HNO_3 is 0.001 M. Hence, concentration in ppm is

•  

  1×10^(-3)

•  100
•  212
•  63

Solution
 

Q6.100 mLof 1 M KMnO_4oxidized 100 mLof H_2 O_2 in acidic medium (when MnO_4^- is reduced to Mn^(2+)); volume of same KMnO_4required to oxidize 100 mLof H_2 O_2in basic medium (when MnO_4^-is reduced to MnO_2) will be

•  100/3 mL
•  500/3 mL
• 300/5 mL
•  100 mL

Solution

Q7.If 100 mL of H_2 SO_4 and 100 mL of H_2 O are mixed, the mass percent of H_2 SO_4 in the resulting solution is (d_(H_2 SO_4 )=0.09 g mL^(-1),d_(H_2 O)=1.0 g mL^(-1))

•  90
•  47.36
•  50
•  60

Solution

Q8.A salt contains cation A^(2+) and again B^(2-). Both decolourise MnO_4^- in acidic medium. Salt is

•  Ferrous oxalate
•  Ferric oxalate
•  Both (a) and (b)
•  None of these

Solution

Q9.K_2 Cr_2 O_7 is obtained in the following steps: 2FeCrO_4+2Na_2 CO_3+O→Fe_2 O_3+2Na_2 CrO_4+2CO_2 2Na_2 CrO_4+H_2 SO_4→Na_2 Cr_2 O_7+H_2 O+Na_2 SO_4 Na_2 Cr_2 O_7+2KCl→K_2 Cr_2 O_7+2NaCl To get 0.25 mol of K_2 Cr_2 O_7 , mol of 50% pure FeCrO_4 required

•  1 mol
•  0.50 mol
•  0.25 mol
•  0.125 mol

Solution

Q10.The Ew of H_3 PO_4 in the reaction is Ca(OH)_2+H_3 PO_4→CaHPO_4+2H_2 O (Ca=40,P=31,O=16)  

•  49
•  98
•  32.66
• 147

Solution

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