SOME BASIC CONCEPTS OF CHEMISTRY Quiz-50
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Q1.A metal oxide has the formula Z_2 O_3. It can be reduced by hydrogen to give free metal and water 0.2 g of the metal oxide requires 12 mg of hydrogen for complete reduction. The atomic weight of the metal is
Solution
Q2.1 mol of MnO_4^( 2-) in neutral aqueous medium disproportionates to
:
Solution
Q3. A dark violet colour mixture in presence of dil. HCl changes to pale yellow solution. Mixture may contain
Solution
Q4. Of the following solutions the one that is acidic is
Solution
((a) ZnSO_4+2H_2 O⟶Zn(OH)_2↓+2H_2 SO_4 acidic
((a) ZnSO_4+2H_2 O⟶Zn(OH)_2↓+2H_2 SO_4 acidic
Q5.H_2 S contains 94.11% sulphur; SO_2 contains 50% oxygen and H_2 O contains 11.11% hydrogen. Thus,
Solution
(b) H_2 S 5.89 g H combines with=94.11 gS Hence, 1 g H combines with = 16g S SO_2 50 g O combine with=50 g S Hence, 1 g O combines with = 1 g S H_2 O 11.11 g H combines with=88.89 g O 1 g H combine with = 8 g O Thus, law of reciprocal proportion is followed
(b) H_2 S 5.89 g H combines with=94.11 gS Hence, 1 g H combines with = 16g S SO_2 50 g O combine with=50 g S Hence, 1 g O combines with = 1 g S H_2 O 11.11 g H combines with=88.89 g O 1 g H combine with = 8 g O Thus, law of reciprocal proportion is followed
Q6. 100 mLof ozone at STP were passed through 100 mL of 10 volume H_2 O_2 solution. What is the volume strength of H_2 O_2 after the reaction?
Solution
(a) O_3 ⟶ O_2+O…(i)
H_2 O_2⟶H_2 O+O …(ii)
O + O⟶O_2…(iii)
1/2 vol 1/2 vol 1 vol
From equations (i) and (ii), we inter that 100 mL of O_3 at STP will produce 100 mL of molecular O_2 as such and 100 mL of oxygen molecule after reaction with H_2 O_2
This new volume of 100 mL of molecular oxygen after reaction with H_2 O_2 is contributed equally by O_3 and H_2 O_2. Thus, 50 mL of oxygen have been contributed by H_2 O_2
Again, we know
Volume of H_2 O_2× Volume strength of H_2 O_2 =Volume of O_2 at STP
∴100 mL of ’10 V’ H_2 O_2≡1000 mL of O_2 at STP After utilization of 50 mL of O_2, according to equation(iii), the balance (1000-50)=950 mL of O_2at STP are still retainable by 100 mL of H_2 O_2
Hence volume strength of H_2 O_2 after reaction (Volume of O_2 at STP)/(Volume of H_2 O_2 )=950/100=9.5 V
∴ Volume strength =9.5
(a) O_3 ⟶ O_2+O…(i)
H_2 O_2⟶H_2 O+O …(ii)
O + O⟶O_2…(iii)
1/2 vol 1/2 vol 1 vol
From equations (i) and (ii), we inter that 100 mL of O_3 at STP will produce 100 mL of molecular O_2 as such and 100 mL of oxygen molecule after reaction with H_2 O_2
This new volume of 100 mL of molecular oxygen after reaction with H_2 O_2 is contributed equally by O_3 and H_2 O_2. Thus, 50 mL of oxygen have been contributed by H_2 O_2
Again, we know
Volume of H_2 O_2× Volume strength of H_2 O_2 =Volume of O_2 at STP
∴100 mL of ’10 V’ H_2 O_2≡1000 mL of O_2 at STP After utilization of 50 mL of O_2, according to equation(iii), the balance (1000-50)=950 mL of O_2at STP are still retainable by 100 mL of H_2 O_2
Hence volume strength of H_2 O_2 after reaction (Volume of O_2 at STP)/(Volume of H_2 O_2 )=950/100=9.5 V
∴ Volume strength =9.5
Q7.The pH of 10^(-5) M HCl solution if 1 mL of it is diluted to 1000mL is
Solution
(d) Concentration of H^⊕ in 1 mL =10^(-5) M Concentration of H^⊕ in 1000 mL=10^(-8) M (pH=6.98)
(d) Concentration of H^⊕ in 1 mL =10^(-5) M Concentration of H^⊕ in 1000 mL=10^(-8) M (pH=6.98)
Q8.Number of moles of NH_3 formed when 0.535 g of NH_4 Cl is completely decomposed by NaOH, is
NH_4 Cl+NaOH→NH_3+NaCl+H_2 O
Solution
(d) NH_4 Cl+NaOH→NH_3+NaCl+H_2 O 53.2 g =1 mol=17 g NH_3 0.535 g=0.17 g NH_3
(d) NH_4 Cl+NaOH→NH_3+NaCl+H_2 O 53.2 g =1 mol=17 g NH_3 0.535 g=0.17 g NH_3
Q9.A mixture of ethylene and excess of H_2 had a pressure of 600 mm Hg. The mixture was passed over nickel catalyst to convert ethylene to ethane. The pressure of the resultant mixture at the similar conditions of temperature and volume dropped to 400 mm Hg. The fraction of C_2 H_4 by volume in the original mixture is
Solution
Q10.AgNO_3 gives white ppt with hypo changing to black after sometime. Black ppt is of
a)
Solution
