DPP 01 Mole Concept JH Sir-3571

TARGET

CH EMISTRY

JEE (ADVANCED) : 2014

DPP

COURSE NAME : URJAA (UR)

DAILY PRACTICE PROBLEMS

DATE : 17.06.2013 to 22.06.2013

DPP NO. 01 TO 02

* Marked Questions are having more than one correct option. DPP No. # 1 1.

A quantity of aluminium has a mass of 54.0 g. What is the mass of the same number of magnesium atoms? (A) 12.1 g

2.

(C) 48 g

(D) 97.2 g

The charge on 1 gram ions of Al3+ is : (A)

3.

(B) 24.3 g

1 N e coulomb 27 A

(B)

1 × NAe coulomb 3

(C)

1 × NAe coulomb 9

(D) 3 × NAe coulomb

The density of water at 4°C is 1.0 × 103 kg m–3. The volume occupied by one molecule of water is approximately : (A) 3.0 × 10–23 mL

4.

(D) 9.0 × 10–23 mL

(B) 9.0 × 6.02 × 1023

(C) 18.0

(D) 36.0

From 392 mg of H2SO4, 1.204 × 1021 molecules are removed. How many moles of H2SO4 are left ? (A) 2.0 × 10–3

6.

(C) 3.0 × 10–21 mL

The number of molecules of water in 333 g of Al2 (SO4)3. 18H2O is : (A) 18.0 × 6.02 × 1023

5.

(B) 6.0 × 10–22 mL

(B) 1.2 × 10–3

(C) 4.0 × 10–3

(D) 1.5 × 10–3

In which of the following pairs do 1 gm of each have an equal number of molecules ? (A) N2O and CO

(B) N2 and C3O2

(C) N2 and CO

(D) N2O and CO2

Comprehension (Q. No. 7 to 9) Equal number of atoms are contained in one gram atomic weight of each element, and that the same number of molecules is found in one gram molecular weight of any compound. The terms gram atomic weight and gram molecular weight are used to refer to a fixed number (Avogadro's number 6.023 x 1023) of particle. The term mole stands for the amount of material which contains these number of particles. 7.

If 12 g of C12 has equal number of atoms as Avogadro's number what mass of 2He4 will contain same number of atoms ? (A) 2 g

(B) 4 g

(C) 6 g

(D) 3 g

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PAGE NO. # 1

8.

What will be the gram atomic weight of Mg, if 24 g of 12Mg24 contains 6.023 × 1023 atoms ? (A) 12 g

9.

(B) 36 g

(D) 48 g

Compute gram molecular weight of H2 gas if 4 g of H2 gas contains 24.09 × 1023 atoms : (A) 4 g

10.

(C) 24 g

(B) 3 g

(C) 2 g

(D) 1 g

Fill the blanks in the table. Where NA is Avogadro number. Mass of same no. atoms of C atom as No. of removed atoms present in the original sample

Sample

Relative atomic Mass for the element

Gram Atomic mass of sample

8 gm O atom

16

___

___

For Example

16

16 gm

½ mole

230 gm Na

23

60 gm Ca

40

1 mole

20 gm He

4

3 mole

56 gm N

14

½ mole

12 gm Mg

24

NA 4

128 gm S

32

NA

93 gm P

31

3NA 2

Moles of No. of atoms Mass removed of sample from the sample sample

Mole removed

___

2 gm

___

___

___

NA 2

2 gm

1 mole 8

NA 8

6 gm

46 gm

DPP No. # 2 1.

According to common practice, if you have Avogadro number of Hydrogen then its mass is equal to : (A) 1 g

2.

(B) 2 g

(C) 4 g

(D) depends on temperature

From 64 g oxygen, 11.2 lit. Oxygen gas at S.T.P and 6.02 × 1023 oxygen atoms are removed from the oxygen container. Find the mass of the oxygen gas left. (A) zero

3.

(B) 32 g

(C) 16 g

(D) none

From the mixture of 4 moles Ca3(PO4)2 and 5 moles of P4O10 and 6 moles of H3PO3 all the phosphorus atoms are removed then moles of P4 molecule formed from all these atoms is : (A) 8.5

4.

(C) 34

(D) 10

If a sample of CuSO4.5H2O contains 3 moles of oxygen molecule the mole of H-atoms present in the sample is: (A) 5

5.

(B) 17

(B) 9

(C) 10/3

(D) 20/3

Sulphur exist in different allotropic forms like S2 , S6 and S8 etc. If equal weight of three are taken in separate container then the ratio of number of atoms is : (A) 1 : 3 : 4

(B) 1 : 1 : 1

(C) 2 : 4 : 3

(D) 1 : 1 : 4

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PAGE NO. # 2

6.

7.

Which of the following contains the greatest number of atoms ? (A) 1.0 g of butane (C4H10)

(B) 1.0 g of nitrogen (N2)

(C) 1.0 g of silver (Ag)

(D) 1.0 g of water (H2O)

Under the same conditions, two gases have the same number of molecules. They must : (A) be noble gases.

(B) have equal volumes. 3

(C) have a volume of 22.4 dm each. 8.

(D) have an equal number of atoms.

A gaseous mixture contains CO2(g) and N2O(g) in a 2 : 5 ratio by mass. The ratio of the number of molecules of CO2(g) and N2O(g) is : (A) 5 :2

9.

(B) 2 : 5

(D) 5 : 4

In which of the following pairs do 1 g of each have an equal number of molecules? (A) N2O and CO

10.

(C) 1 : 2

(B) N2 and C3O2

(C) N2 and CO

(D) N2O and CO2

11.2 L of a gas at STP weights 14.0 g. The gas could be : (A) N2O

(B) NO2

(C) N2

(D) CO

ANSWER KEY DPP No. # 1 1.

(C)

2.

(D)

8.

(C)

9.

(C)

3.

(C)

4.

(B)

5.

(A)

6.

(D)

7.

(B)

5.

(A)

6.

(A)

7.

(B)

DPP No. # 2 1.

(A)

2.

(B)

3.

(A)

8.

(B)

9.

(D)

10.

(C)

4.

(D)

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PAGE NO. # 3

TARGET

CH EMISTRY

JEE (ADVANCED) : 2014

DPP

COURSE NAME : URJAA (UR)

DATE : 24.06.2013 to 29.06.2013

DAILY PRACTICE PROBLEMS

DPP NO. 03 TO 04

* Marked Questions are having more than one correct option. DPP No. # 03 1.

The chloride of a metal contains 71% chlorine by weight and the vapour density of it is 50. The atomic weight of the metal will be (A) 29 (B) 58 (C) 35.5 (D) 71

2.

The haemoglobin of most mammals contains approximately 0.33% of iron by mass. The molecular mass of haemoglobin is 67200. The number of iron atoms in each molecule of haemoglobin is(A) 3 (B) 4 (C) 2 (D) 6

3.

If law of conservation of mass was to hold true, then 20.8 gm of BaCl2 on reaction with 9.8 gm of H2SO4 will produce 7.3 gm of HCl and BaSO4 equal to (A) 11.65 gm (B) 23.3 gm (C) 25.5 gm (D) 30.6 gm

Fill in the blank : 4.

For the reaction, 2x + 3y + 4z  5w Initially if 1 mole of x, 3 mole of y and 4 mole of z is taken. If 1.25 mole of w is obtained then % yield of this reaction is _______.

5.

The vapour density of a mixture of gas A (Molecular mass = 40) and gas B (Molecular mass = 80) is 25. Then mole % of gas B in the mixture would be ________.

6.

For the reaction 2A + 3B + 5C  3D Initially if 2 mole of A, 4 mole of B and 6 mole of C is taken. With 25% yield, moles of D which can be produced are _________.

Single Correct: 7.

Mass of sucrose C 12 H 22 O 11 produced by mixing 84 gm of carbon, 12 gm of hydrogen and 56 lit. O2 at 1 atm & 273 K according to given reaction, is C(s) + H2(g) + O2 (g)  C12H22O11(s) (A) 138.5

(B) 155.5

(C) 172.5

(D) 199.5

8.

The percentage by mole of NO2 in a mixture of NO2(g) and NO(g) having average molecular mass 34 is : (A) 25% (B) 20% (C) 40% (D) 75%

9.

74 gm of a sample on complete combustion gives 132 gm CO2 and 54 gm of H2O. The molecular formula of the compound may be : (B) C4H10O (C) C3H6O2 (D) C3H7O2 (A) C5H12

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PAGE NO. # 1

DPP No. # 04 1.

Calculate percentage change in Mavg of the mixture, if PCl5 undergo 50% decomposition. PCl5  PCl3 + Cl2 (A) 50% (B) 66.66 % (C) 33.33 % (D) Zero

2.

The average atomic mass of a mixture containing 79 mole % of and 26Mg , is 24.31. % mole of 26Mg is : (A) 5 (B) 20 (C) 10

3.

and remaining 21 mole % of

25Mg

(D) 15

The density of a liquid is 1.2 g/mL. There are 35 drops in 2 mL. The number of molecules in one drop (molar mass of liquid = 70 g/mol) is : 2

 1.2  (A)   NA  35 

4.

24Mg

1.2  1  NA (B) 35  35 

1.2 (C) (35)3 NA

(D) 1.2 NA

Sulphur trixoide is prepared by the following two reactions S8(s) + 8 O2(g)  8 SO2(g) ; 2SO2(g) + O2(g)  2SO3(g) How many grams of SO3 are produced from 1 mol of S8? (A) 1280.0 (B) 640.0 (C) 960.0

(D) 320.0

5.

A compound of Mg contains 6% of Mg then find the minimum molar mass of compound.

6.

A sample contains 9.81 g Zn, 1.8 × 1023 atoms of Cr and 0.6 g-atoms of O. What is empirical formula of compound ? (Zn = 65 g/mol)

7.

If 240 g of carbon is taken in a container to convet it completely to CO2 but in industry it has been found that 280 g of CO was also formed along with CO2. Find the percentage yield of CO2. The reaction occurring are C + O2  CO2 (A) 25%

;

C+

(B) 50%

1 O CO 2 2 (C) 75%

(D) 100%

8.

100 g impure CaCO3 on heating gives 5.6 lit. CO2 gas at STP. Find the percentage of calcium in the lime stone sample. [At wt. : Ca = 40 ; C = 12 ; O = 16] (A) 10 (B) 20 (C) 1 (D) 30

9.

The mass of Mg3N2 produced if 48 gm of Mg metal is reacted with 34 gm NH3 gas is : Mg + NH3  Mg3N2 + H2 (A)

200 3

(B)

100 3

400 3 ANSWER KEY DPP NO. # 3 (C)

1. 8.

(A) (A)

2. 9.

(B) (C)

3.

(B)

1. 7.

(C) (B)

2. 8.

(C) (A)

3. 9.

(C) (A)

4.

50%

(D)

5.

25%

6.

5.

400 g/mole

150 3

0.75

7.

(B)

6.

ZnCr2O4

DPP NO. # 4 4.

(B)

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PAGE NO. # 2

TARGET

CH EMISTRY

JEE (ADVANCED) : 2014

DPP

COURSE NAME : URJAA (UR)

DAILY PRACTICE PROBLEMS

DATE : 01.07.2013 to 06.07.2013

DPP NO. 05 TO 06

* Marked Questions are having more than one correct option. DPP No. # 05 1.

84 g of iron (Fe) is reacted with sufficient amount of steam to produced 44.8 lit., H2 gas at S.T.P. according the following reaction, a Fe + b H2O c Fe3O4 + d H2. The stoichiometric coefficients of the reaction is (Atomic wt. : Fe = 56, O = 16, H = 1) (A) 4, 3, 1, 4

2.

(B) 3, 4, 1, 4

(D) None of these

If 240 g of carbon is taken in a container to convert it completely to CO2 but in industry it has been found that 280 g of CO was also formed along with CO2. Find the percentage yield of CO2. The reactions occuring are : C + O2  CO2 ; C + (A) 25%

3.

(C) 1, 4, 2, 3

1 O  CO 2 2

(B) 50%

(C) 75%

(D) 100%

120 g Mg was burnt in air to give a mixture of MgO and Mg3N2. The mixture is now dissolved in HCl to form MgCl2 and NH4Cl, if 107 grams NH4Cl is produced. Then the moles of MgCl2 formed is: (Atomic wt. : Mg = 24, N = 14, Cl = 35.5)] (A) 3 moles

4.

(B) 20

(D) 10 moles

(C) 1

(D) 30

75 mL of H2SO4 (specific gravity = 1.18) containing 49% H2SO4 by mass is diluted to 590 mL. Calculate molarity of the diluted solution. [S = 32] (A) 0.7 M

6.

(C) 5 moles

100 g impure CaCO3 on heating gives 5.6 litre CO2 gas at STP. Find the percentage of calcium in the lime stone sample. [Atomic weight : Ca = 40; C = 12 : O = 16] (A) 10

5.

(B) 6 moles

(B) 7.5 M

(C) 0.75 M

(D) 0.25 M

Calculte the molarity when : (a) 4.9 g H2SO4 acid dissolved in water to result 500 mL solution (b) 2 gram molecule of KOH dissolved in water to result 500 mL solution

7.

How would you prepare exactly 3 litre of 1.0 M NaOH by mixing proportions of stock solution of 2.5 M NaOH and 0.4 M NaOH if no water is to be used. Find the ratio of the volume (v1/v2).

8.

What volume of 0.25 M HNO3 (nitric acid) reacts with 50 mL of 0.15 M Na2CO3 (sodium carbonate) in the following reaction? 2HNO3 (aq) + Na2CO3(aq)  2NaNO3 (aq) + H2O() + CO2(g)

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PAGE NO. # 1

9.

Zinc and hydrochloric acid react according to the reaction : Zn(s) + 2HCl (aq)  ZnCl2(aq) + H2(g) If 0.30 mole of Zn are added to hydrochloric acid containing 0.52 mole HCl, how many moles of H2 are produced :

10.

Match the following : Column-I

Column-II

For 1 mole of reactant placed in

Product

open container in each reaction (A)

 PCl5 (g)   PCl3 (g) + Cl2(g)

(p)

2NA molecules

(B)

 CaCO3(s)   CaO(s) + CO2(g)

(q)

67.2 litre gaseous product at STP

(C)

 2HCl (g)   H2(g) + Cl2(g)

(q)

22.4 litre gaseous product at STP

(D)

 NH4COONH2(s)   2NH3(g) + CO2(g)

(s)

44.8 litre gaseous product at STP

DPP No. # 06 1.

What volume of a 0.8 M solution contains 100 millimoles of the solute? (A) 100 mL

(B) 125 mL

(C) 500 mL

(D) 62.5 mL

2.

Calculate the volume in litre of 1 M solution of HCl which contain 36.5 g HCl?

3.

The molarity of the KOH solution which is containing 2.8% mass-volume solution of KOH is : (A) M/10

4.

(B) M/2

6*.

(D) 1 M

75 mL of H2SO4 (specific gravity = 1.18) containing 49% H2SO4 by mass is diluted to 590 mL. Calculate molarity of the diluted solution. [S = 32] (A) 0.7 M

5*.

(C) M/5

(B) 7.5 M

(C) 0.75 M

(D) 0.25 M

If 100 mL of 1 M H2SO4 solution is mixed with 100 mL of 9.8% (w/w/) H2SO4 solution (d = 1 gm/ mL) then : (A) concentration of solution remains same

(B) volume of solution become 200 mL

(C) mass of H2SO4 in the solution is 98 gm

(D) mass of H2SO4 in the solution is 19.6 gm

For 100 mL of 0.2 M AlCl3 solution + 400 mL of 0.1 M HCl solution correct datas are : (A) Total concentration of cation (s) = 0.12 M (B) Total concentration of cation(s) = 0.06 M (C) [Cl–] = 0.1 M (D) [Cl–] = 0.2 M

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PAGE NO. # 2

Comprehension : Read the following comprehension carefully and answer the questions (Q. 7 to Q. 10) The concentrations of solutions can be expressed in number of ways ; viz : mass fraction of solute (or mass percent), Molar concentration (Molarity) and Molal concentration (molality). These terms are known as concentration terms and also they are related with each other i.e. knowing one concentration term for the solution, we can find other concentration terms also. The definition of different concentration terms are given below : Molarity : It is number of moles of solute present in one litre of the solution. Molality : It is the number of moles of solute present in one kg of the solvent Mole fraction =

moles of solute moles of solute + moles of solvent

If molality of the solution is given as 'a' then mole fraction of the solute can be calculated by

a  Msolvent a  1000 (a  Msolvent  1000) where a = molality and M Mole fraction = a+ solvent = Molar mass of Msolvent solvent We can change : Mole fraction  Molality  Molarity. 7.

120 g of solution containing 40% by mass of NaCl are mixed with 200 g of a solution containing 15% by mass NaCl. Determine the mass percent of sodium chloride in the final solution : (A) 24.4%

8.

(B) 78%

(B) 5.5 m

(C) 24.4 m

(D) None

(C) 0.09

(D) 0.25

What is the mole fraction of the solute : (A) 0.18

10.

(D) 19.68%

What is the molality of the above solution ? (A) 4.4 m

9.

(C) 48.8%

(B) 0.75

What is the molarity of solution if density of solution in 1.6 g/mL ? (A) 5.5 M

(B) 6.6 M

(C) 2.59 M

(D) None

ANSWER KEY DPP No. # 05 1.

(B)

2.

(B)

3.

6.

(a) 0.1 M

10.

(A) p,s (B) p, r (C) r (D) q

(b) 4 M

(C)

4.

(A)

5.

(C)

7.

2 : 5

8.

60 mL 9.

0.26

DPP No. # 06 1.

(B)

2.

1

3.

(B)

4.

(C)

5*.

(ABD)

6*.

(AD)

7.

(A)

8.

(B)

9.

(C)

10.

(B)

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PAGE NO. # 3