G. D. GOENKA INTERNATIONAL SCHOOL
CLASS – 11 SCIENCE
FIRST UNIT TEST
SUBJECT:CHEMISTRY
Time :1 hr
DATE: 25-7-12
Max. Marks : 30
General instructions
1. All question are compulsory.
2. Questions in section A are very short answer questions to be answered in
3.
4.
5.
6.
one sentence and carry 1 mark each.
Questions in section B are short answer questions to be answered in about
30 words and carry 2 marks each.
Questions section C are also short answer questions to be answered in
about 40 words and carry 3 marks each.
Question in section D is long answer question to be answered in about 70
words and carry 5 marks.
Molecular masses: C=12, O=16, H=1, Na=23, Cl=35.5 ,Zn=65.5 ,Mg=24,
S=32, N=14
SECTION A
1. What are the units of molarity? What is the effect of temperature on the molarity
of a solution?
2. What is water gas shift reaction.
3. How many moles of magnesium are present in 0.478 g of the element?
4. State the law of multiple proportions.
5. Give two advantages of using hydrogen as a fuel over gasoline.
SECTION B
6. If law of constant composition is true what weights of calcium and oxygen are
present in 1.5 g of calcium carbonate ,if sample of calcium carbonate from
another source contains the following percentage composition. (Ca = 40%,
C=12%, O=48%)
7. Describe the structure of hydrogen peroxide
8. How many litres of oxygen at NTP are required to burn completely 2.2 g of
propane,C3H8 ?
9. A solution is prepared by dissolving 18.25 g of NaOH in distilled water to give
200 cc of solution.Calculate the molarity of solution.
SECTION C
10. Explain inorganic ion exchange method for removing permanent hardness of water.
11. Calculate the percentage strength and strength in grams per litre of 10 volume
hydrogen peroxide solution.
12. An organic compound containing carbon,hydrogen and oxygen gave the
percentage composition as C=40.687%, H=5.085% and O= 54.228% . The vapour
density of compound is 59. Calculate the molecular formula of compound
13. What mass of zinc is required to produce hydrogen by reaction with HCl which is
enough to produce 4 mol of ammonia according to reactions
Zn + 2HCl ------------ZnCl2 +H2 , 3H2 + N2 -----2NH3
SECTION D
14. (a) What is limiting reactant?
(b) How much magnesium sulphide can be obtained from 2.0 g of magnesium and 2.0 g
of sulphur by the reaction Mg +S -- MgS Which is the limiting reagent?
Calculate the amount of one of the reactants which remain unreacted.
CLASS 11 Science
CHEMISTRY
ANSWER KEY
SECTION A
1.
Unit of molarity is moles/L
As temperature increases molarity decreases
0.5 mark
0.5 mark
FeCrO4
2.
CO + H2O +H2O ----------- CO2 + 2 H2
1 mark
673 K
3.
24 g of Mg contain 1 mole
0.478 g of Mg contains 0.478/24
0.5 mark
=0.019 mol
0.5 mark
4.
When two elements combine to form two or more chemical compounds then mass of one of
the elements which combine with fixed mass of other bear a simple ratio to one another
1 mark
5.
High heat of combustion
0.5 mark
No pollutions.
0.5 mark
SECTION B
6.
Mass of CaCO3 =100g
100 g of CaCO3 contains 40 g of Ca
1.5 g of CaCO3 contains (40/100)x1.5 =0.6g Ca
1mark
100 g of CaCO3 contains 48 g of O
1.5 g of CaCO3 contains (48/100)x1.5 =0.72g O
1mark
7.
Hydrogen peroxide is a non planar molecule . the two oxygen atoms are linked to each other by
a single covalent bond (i.e peroxide bond ) and each oxygen is further linked to a hydrogen atom by a
single covalent bond.The two O-H bonds are in different planes .The dihedral angle between the two
planes is different in gas phase and crystalline state.
1 mark
FIG
8.
44
C3H8 + 5 O2 ----
5x22.4 L
72
1 mark
4 H2O + 3 CO2
132
44 g of C3H8 require 5 x 22.4L of O2
2.2 g of C3H8 require (5 x 22.4 x 2.2 )/44 L of O2
=5.6 L of O2
9.
Molarity = no.of moles
Vol. of sol.in L
1 marks
1 marks
0.5 marks
No. of moles = 18.25/40 = 0.456
Vol. of sol. =200/1000 =0.2 L
0.5 marks
0.5 marks
Molarity=0.456 /0.2 =2.28
0.5 marks
SECTION C
10.
Complex inorganic salts like hydrated sodium aluminium silicates exchange Ca 2+ and Mg 2+ ions
present in hard water with sodium present in complex salts.zeolites are used for this purpose .The
zeolite is loosely packed over layers of graval and sand in a big tank. Hard water is introduced from the
top into the base of tank. From the bottom water rises up through the graval and sand layers and finally
percolates through the bed of permutit.During the process the Ca 2+ and Mg 2+ ions are exchanged by
sodium ions in permutit .
Na2Z +CaCl2 ----- CaZ + NaCl
Na2Z +MgCl 2 ----- MgZ + NaCl
As a result Ca2+ and Mg2+ ions get exchanged to zeolite and water which rises above the premitute
layer is soft .The softened water still contains sodium salts
2.5 marks
FIG
0.5 mark
11.
2H2O2 --- 2H2O + O2
2 X 34
22.4
22.4L of O2at NTP are obtained from 2 x 34 g = 68 g of H2O2
Thus 10 ml of O2 at NTP will be from (68/22400)x10 g of H2O2
But 10 ml of O2 at NTP are produced from 1 ml of 10 vol H2O2 sol.
Thus 1 ml of 10 vol. H2O2 sol. Contains (68/22400)x10 g of H2O2
100 ml of 10 vol. sol. Will contain(68/22400)x10x100 =3.036g
12.
1 mark
1mark
1mark
Symbol of
element
C
H
O
%
composition
40.687
5.085
54.228
Atomic
mass
12
1
16
moles
Simple molar
ratio
40.687/12=3.39 3.39/3.39=1
5.085/1=5.085
5.085/3.39=1.5
54.228/16=3.389 3.389/3.39=1
Thus, empirical formula is C2H3O2
Whole no.
ratio
1X2=2
1.5X2=3
1X2=2
1.5 marks
Emprical mass =59
Molecular mass = 2 x vapour density
= 2x 59 =118
0.5 marks
n=mol.mass /empirical mass
= 118/59 = 2
0.5 marks
Thus molecular formula is 2x C2H3O2
= C4 H6O4
0.5 marks
13.
Zn +2 HCl --------- ZnCl2 + H2
----- (a)
3 N2 + H2 -------- 2 NH3
------(b)
To produce 2 moles of ammonia 3 moles of H2 is required
To produce 4 moles of ammonia 6 moles of H2 is required
To produce 6 moles of hydrogen in reaction (a) 6 moles of Zn is required
Thus mass of zinc = 6 x 65.5 = 393 g
1 mark
1 mark
0.5 mark
0.5 mark
SECTION D
14.
Limiting reactant – The reactant which is completely consumed in a chemical reaction is called
limiting reactant.
1 mark
Mg + S --- MgS
24 g of Mg reacts with 32 g of S
2 g of Mg reacts with (32/24) x 2 g of S
= 2.6 g of S
2.6 g of S is required for complete reaction
But given is 2 g
So S is limiting reactant
1 mark
32 g of Mg gives 56 g of MgS
2 g of Mg gives (56 /32)x 2 g of MgS
=3.5 g of MgS
1 mark
Amount of Mg left unreacted = 2- 1.5 =0.5
1 mark