SOME BASIC CONCEPTS OF CHEMISTRY Quiz-57

SOME BASIC CONCEPTS OF CHEMISTRY Quiz-57

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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.

•  A
•  B
•  C
•  D

Solution
(c) Cl_2+H_2 O⟶HCl+HClO (X) Mg+2HCl⟶MgCl_2+H_2 (Y)

Q2.0.5 g of an organic substance containing phosphorus was heated with concHNO_3 in the carius tube. The phosphoric acid thus formed was precipitated with magnesia mixture (MgNH_4 PO_4) which on ignition gave a residue of 1.0 g of magnesium pyrophosphate (Mg_2 P_2 O_7). The percentage of phosphorous in the organic compound is

•  55.85%
•  29.72%
•  19.81%
•  20.5%

Solution

 

(a) Mw of Mg_2 P_2 O_7=(24×21+31×2+16×7)=222 ∴ % of P =62/222×(Weight of Mg_2 P_2 O_7)/(Weight of compound)×100 =62/222×1.0/0.5×100=55.85%

Q3.  

•  A
•  B
•  C
•  D

Solution
(a) In basic medium,Cr_2 O_7^( 2-) changes to CrO_4^( 2-)(no change in oxidation number)

Q4. 1 mol of ferric oxalate is oxidized by x mol of MnO_4^( ⊖) in acidic medium. Hence, the value of x is

•  1.2
•  1.6
•  1.8
•  1.5

Solution
(a) Fe_2 (C_2 O_4 )_3≡3C_2 O_4^( 2-),Fe^(3+) is not affected

Q5.The mass of 70% H_2 SO_4 required for neutralization of 1 mole of NaOH is

•  70 g
•  35 g
•  30 g
•  95 g

Solution
 (a) 2NaOH+H_2 SO_4→Na_2 SO_4+2H_2 O 2 mol 1 mol 1 mol 0.5 mol =49 g H_2 SO_4 (pure) Thus, 70% H_2 SO_4 required =49×100/70=70 g

Q6. Aluminium sulphate (X) is slightly insoluble in water. It is converted into soluble sodium sulphate by using Na_2 CO_3 in the preparation of sodium carbonate extract. Moles of (Y), required for complete conversion of 1 mole of (X) into soluble sulphate, is

•  1
•  2
•  3
•  4

Solution
(c) Al_2 (SO_4 )_3+3Na_2 CO_3→3Na_2 SO_4+Al_2 (CO_3 )_3 Insoluble 1 : 3 soluble

Q7.What volume of 0.1 M NaOH will be required to neutralize 100 mL of 0.1 M H_3 PO_4 using methyl red indicator to change the colour form pink (acidic medium) to yellow (basic medium?)

•  300 mL
•  200 mL
•  100 mL
•  30 mL

Solution
(c) Methyl red indicates first step ionization of H_3 PO_4 (n-factor=1) (Ew=Mw/1)(N=M×1) NaOH≡H_3 PO_4 N_1×V_1=N_2×V_2 0.1×1×V_1=0.1×1×100 V_1=100 mL

Q8.A molal solution is one that contains 1 mol of a solute in

•  1000 g of solvent
•  1 L of solvent
•  1 L of sol.
•  22.4 L of sol.

Solution
(a) A molal solution is one that contains 1 mol of a solute in 1000 g or kg of the solvent. Molality (m)=(Moles of solute)/(Mass of solvent in kg)

Q9.Number of millimoles of Cl^- in 100 mL of 1 M BaCl_2 solution (assume 100% ionization) is

•  100
•  200
•  300
•  50

Solution
(b) BACl_2⇌Ba^(2+)+2Cl^- Millimoles of Cl^-=2×millimoles of BaCl_2 =2×100×1 =200 millimoles

Q10.

•  A
•  B
•  C
•  D

Solution
(c) Number of electrons in CO_2=6+16=22 Electrons Electrons in neutral species = proton X^+ 22 23 Y^(2+) 22 24 Z^- 22 21 Thus, increasing order of proton is Z^-