Summer Vacation Holiday Homework 2017-18: Class XII

KENDRIYA VIDYALAYA NO.3 AF-II, JAMNAGAR

HOLIDAY HOMEWORK

SUB:ENGLISH CLASS:XII

1. Solve 2 unseen Reading compre. Passages

2. Solve 2 Note Making & summary passages

3. Write 2 Notices.

4. Write 2 Advertisements (Display)

5. Write 2 Advertisements (Classified)

6.Write 2 Posters.

7. Write 2 Invitations (one each formal & Informed)

8. Write 2 Replies (one each formal & Informal ) acceptance & Refusal.

9. Complete all your CW.

10. Read & Write the summary of the Novel ‘The Invisible Man’

ग्रीष्मकालीन अवकाश गृहकार्य कक्षा 12

प्रश्न 1 “भारत की उभरती हुई शक्ति” एवं “महिला सशक्तिकरण” विषयों पर निबंध लिखिए – (३०० शब्द)

प्रश्न 2खाद्य सामग्री में मिलावट को लेकर प्रशासन का ध्यान आकर्षित करते हुए एक सम्पादकीय लिखिए|

प्रश्न 3 किसी एक विषय पर फीचर लिखिए |

प्रश्न 4आपके द्वारा हाल ही पढ़ी गई किसी पुस्तक की समीक्षा लिखिए |

प्रश्न 5 अंग्रेजी माध्यम से शिक्षा विषय पर आलेख लिखिए |

प्रश्न 6 सचिवालय में निकले कनिष्ठ लिपिक पद हेतु आवेदन पत्र लिखिए |

प्रश्न 7कोई एक अपठित पद्यांश लिखकर सम्बन्धित प्रश्नों के उत्तर दीजिए |

STD: XII SUB: Mathematics

Extra sums: Assignment of extra sums of Matrices and Determinants

Examples from textbook (Ch.:3, 4).

CLASS XII SUBJECT- BIOLOGY

Unit One—Chapter 1 to 4 and from unit 2 only up to Inheritance of Two genes.

Define biological terms from the chapter 1 to 4 and 5.

Write the Answers of the 3 marks and 5 marks Questions which are given chapter wise separately and learn also.

Draw neat and labeled diagrams of the following:-

Microsporogenesis Megasporogenesis. Spermatogenesis. Oogenesis.Dicot Embyo and Seed

The practical 1 to 6 related with theory (unit .1) and cover 11 marks hence these are also learn with theory portions.

Prepare notes of the lesson environmental Issue.

Investigatory project work from any one lesson.

Revised prepared Answers Daily (15 to 20 minutes)

CLASS 12 PHYSICS

1. Explain the coloumbs law in vector and scalar form ?

2. Explain the guass law and its aplicaton ?

3. Write down all the numericals of electrostatistics unit, solved examples and exercise ?

Summer Holiday Home Work -2017-18

Class XII Subject: Computer Science

: Theory questions :

1. What is the difference between object oriented programming and procedural programming? And also explain how does OOP overcome the shortcomings of traditional programming approaches?

2. Reusability of classes is one of the major properties of OOP. How is it implemented in C++?

3. How is matching done in case of function overloading?

4. When should default arguments be preferred over function overloading and vice-versa?

5. When will you make a function inline and why?
6. CBSE BOARD QUESTION PAPER last 05 years solve Q.1(all sub questions) & Q 2. (a, b & c).

Programming:
1. Define a class Serial in C++ with the following specifications:
Private members of class Serial
Serialcode integer
Title 20 characters
Duration float
Noofepisodes integers
Public member function of class Serial
A constructor function to initialize Duration as 30 and noofepisodes as 10
Newserial function to accept values for serialcode and Title.
Otherenteries( ) function to assign the values of Duration and Noofepisodes with the help of corresponding values passed as parameters to this function.
Dispdata( ) function to display all the data members on the screen

2. Define a class Flight in C++ with the following specification:
Private Members:
· A Data Member Flight Number of type integer
· A Data Member Destination of type String
· A Data Member Distance of Float Type
· A Data Member Fuel of type float
A Member Function CalFuel( ) to calculate the value of fuel as per the following criteria:

Distance Fuel

<=1000 500
more than 1000 and <=2000 1100
more than 2000 2200
Public Members:

A function Feed_Info( ) to allow user to enter values for Flight Number, Destination, Distance & Call Function CalFuel( ) to calculate the quantity of fuel.
A Function Show_Fuel( ) to allow user to view the content of all the data members.

3. Define a class REPORT with the following specification
Private:
Adno 4 digit admission number
Name 20 characters
Marks an array of floating point values
Averge average marks obtained
Getavg( ) to compute the average obtained in five subjects
Public:
Readinfo( ) function to accept values for adno, name, marks and
Invoke the function Getavg( )
Displayinfo( ) function to display all data members on the screen.
4. Define a class SHOP in C++ with the following description:
Private members

Name of the owner
Contact Number of owner
· Address of shop
· Number of employees.

Public Members

A function READ_DATA( ) to read the data.
· A function WRITE_DATA() to display the data.

Declare an array of SHOP to store the information for 100 shops. Use this array in main( ) and display the information.

5. Define a class named ADMISSION in C++ with the following descriptions:
Private members:
AD_NO integer (Ranges 10 - 2000)
NAME Array of characters (String)
CLASS Character
FEES Float
Public Members:
· Function Read_Data ( ) to read an object of ADMISSION type
· Function Display( ) to display the details of an object
· Function Draw_Nos ( ) to choose 2 students randomly and display the details. Use
random function to generate admission nos to match with AD_NO.

6. Class XI Revision Tour : 15 programs based concepts Flow of Controls , Function and Array.

" * Happy Holidays * "

SUB: CHEMISTRY UNIT (1) – SolidState Practice Paper-1

Q1 (i) Why is Frenkel defect not found in pure alkali metal halides.

(ii) What happens to the structure of CsCl when it is heated to about 760 k?

(iii) Fe₃O₄ is ferrimagnetic at room temperature and becomes paramagnetic at 850 K.Why?

(iv) When atoms are placed at the corners of all 12 edges, how many atoms are present per unit cell? 4

Q2 The composition of a sample of wustite is Fe_0.93O_1.00. What percentage of iron is present in the form of

Fe(III)? 4

Q3 Calculate the packing efficiciency in cubic close packing arrangement. 4

Q4 If the radius of octahedral void is r& radius of atoms in close packing is R, drive the relation between

r &R. 4

Q5 Niobium crystallizes in body centered cubic structure. If density is 8.55 gm cm^-3. Calculate atomic

radius of niobium . Atomic mass of niobium is 93 u. 4

Q6 Explain the following term with suitable example

(a) Schottky defect (b)Frenkel defect (c) Interstitials (d) F-centres 4

Q7 What is a semiconductor? Describe the two main types of semiconductors & contrast their conduction

mechanism. 4

Q8 (a) In terms of band theory, what is difference between

(i) Conductor and an insulator (ii) Conductor and semiconductor 2

(b) If NaCl is doped with 10^-3mol % of SrCl₂ , what is the concentration of cation vacancy? 2

Q9 (a) What type of stoichiometric defect is shown by : 2

(i) ZnS (ii) AgBr

(b) Classify the following as p-type or n-type semiconductor: 2

(i) B doped with Si (ii) Si doped with B

Q10 (i) Why is glass considered a super cooled liquid?

(ii) Name the parameters that characterize a unit cell.

(iii) What is the two dimensional coordination number of a molecule in square close-packed layer ?

(iv) Give the significance of a lattice point. 4

Answers→

Q1 Because ions can not get into interstitial sites due to their larger size.(b) CsCl structure transforms into

NaCl structure.(c) due to randomization of spins at high temp.(d) at corners 8 x 1/8 = 1 atom(ans).

Q2 Let atoms of Fe⁺³ = x , atoms of Fe⁺² = 0.93—x . For electrical neutrality,

Total positive charge = Total negative charge

2(0.93—x ) + 3x = 2, then x = 0.14( Fe⁺³atoms), %Fe⁺³ = 0.14x100/0.93 = 15.05

% 0f Fe⁺² = 100—15.05 = 84.95.

Q3 Packing efficiency in ccp or fcc = 74.06 % derivation in book

Q4 r / R = 0.414 (ans).

Q5 Z = 2, M = 93, d = 8.55 gm/cm³, N₀ = 6.022 x 10²³, d = Z x M / N₀ x 10^—30 x a³

Calculate a³ = 3.61 x 10⁷ , a = 330.5 pm.

Q6 (a) Schottky—it arises if some of the atoms or ions are missing from their normal lattice sites. Equal

number of cations and anions are missing. Density is lowered. Occurs in compounds having high C.N

&Cations and anions are of similar size.eg Alkali metal halides (B) Frenkel—it arises if some oftheatoms or ions are missing from their normal positions and occupy an interstitialsite between lattice points.Density remains same. Occurs in compounds having low C.N. Cations are smaller than anions. Ag halidesAgBr contains both defects. (c) Vacant sites in the structure of crystal lattice. (d) Electrons trapped in anion vacancies arerefered as F—centres.They are responsible for colour. Nonstoichiometric KCl is violet in colour.

Q7 Those substances which are insulator at lower temperature but conduct electricity at higher temperature

are called semiconductors.Thermal energy available at room temperature is sufficient to excite electrons

from highest occupied band to the next permitted band.

Intrinsic—Electrical conductivity is due to ther effect. No extra impurity is added.(Ge& Si at room temp).

Extrinsic—Electrical conductivity is due to impurity effect. Extra substance is added. They are of two

types, n—type and p—type.

n-type—Electrical conductivity is due to electrons. eg group 13 elements doped with group 14 elements.

p-type—Electrical conductivity is due to electron deficient bond or electron vacancy or holes.eg group 14

elements doped with group 13 elements.

Q8 (a) In conductors energy gap is very-very small less than 50 kj/mole(metals).

In insulators energy is large more than 300 kj/mole(rubber).

In semiconductors energy gap is 30—300 kj/mole.

(b) One cation of Sr⁺² would create one vacancy. Conc of cation vacancy = 10^—3mole % = 10^—3/ 100 =

10^—5mole. No of Sr⁺² ions in 10^—5mole = 6.022 x 10²³x 10^—5= 6.022 x 10¹⁸ions = cation vacancies.

Q9 (a) ZnS—Frenkel or dislocation. AgBr—Frenkel and schottky both. (b) i—n-type, ii—p-type.

Q10 (a) Because it has the property to flow slowly.

(b) Three dimensions and three angles.

(d) lattice points are occupied by constituent particles.

Practice Paper- 2

Q1 (a)Why is the window glass of old buildings thick at the bottom? 3

(b)What happens when CaCl₂ is introduced to the AgCl crystal?

(c)What happens when CsCl is heated to 760 K?

Q2 (a)Give significance of a lattice point.

(b)What type of defect can arise when a solid is heated? Which physical property is affected by it

and in what way?

(c)What is meant by co-ordination number? 3

Q3 Analysis shows that Nickel Oxide has oxide Ni_0.98O_1.00. What fraction of Ni exists as Ni⁺³& Ni⁺² ions? 3

Q4 KF has NaCl structure, What is the distance between K⁺ ions and F^- ions, if density is 2.48 gm/cm³. 3

Q5 (a)Explain intrinsic and extrinsic semiconductors with suitable examples. 3

(b)Classify the following as n-type or p-type semiconductors

(i) B doped with Si (ii) Si doped with B

Q6 Explain the following with suitable examples 3

Frenkel defect, F-centres, Ferrimagnetism

Q7 Derive the relation between r and R for tetrahedral void. 3

Q8 An element has BCC structure with a edge length 288 pm. The density of element is 7.2 gm/cm³.

How many atoms are present in 208 gm of this element? 3

Q9 Calculate the efficiency of packing in case of a metal crystal for face centred cubic. 3

Q10 A metal crystallises into two cubic phases FCC & BCC whose edge lengths are 3.5 & 3.0 A⁰ respectively.

Calculate the ratio of densities of FCC & BCC. 3

ANSWERS

Q1 (a) Glass has the property to flow slowly (b) vacancies are created(schottky defect).

Q2 (a) particles are present on lattice points (b) schottky defects, density decreases (c) definition

Q3 NCERT page No-31, Q -1.16 Ni⁺² = 96 %, Ni⁺³ = 4 %

Q4 Z = 4, M = 39 + 19 = 58, d = 2.48 , N_A = 6.022 x 10²³, calculate a (5.375 x 10^—10cm )

distance between K⁺ and F^—ions = 5.375 x 10^—10cm / 2 = 2.688 x 10^—10 cm.

Q5 extrinsic contains impurities & intrinsic no impurities (i) n-type (ii) p-type

Q6 definitions in book

Q7 derivation r / R = 2.25 in book

Q8 Z = 2, M = ?, N_A = 6.022 x 10²³, d = 7.2, calculate M(51.8),

No of atoms = 6.022 x 10²³ x 208/51.8 = 24.16 x 10²³ANS

Q9 efficiency FCC = 74 % derivation in book

Q10 d(fcc) = 4 x M / N_A x (3.5)³, d(bcc) = 2 x M / N_A(3)³

d(fcc) / d(bcc) = 4 x 3³ / 2 x (3.5)³ = 1.26 ans

Practice paper-3

Q1 (a) Give significance of a lattice point

(b) What type of defect can arise when a solid is heated? Which physical property is

affected by it & in what way?

Q2 Analysis shows that Nickel oxide has the formula Ni_0.98O_1.00. What fraction of Nickel

exists as Ni⁺³& Ni⁺²ions? 3

Q3 (a) Explain extrinsic and intrinsic semiconductors with examples.

(b) Classify the followings as p-type or n-type of semiconductors

(i) B doped with Si (b) Si doped with B

Q4 Derive the relation between r & R in tetrahedral void. 3

Q5 An element has BCC structure with a edge length of 288 pm. The density of element is

7.2gm/cm³. How many atoms are present in 208 gm of this element? 3

Q6 State Henry^,S law for solubility of a gas in liquid. Explain the significance of Henry^,S

Constant(K_H) at the same temperature, H₂ is more soluble than He in water.Which of

them will have higher value of K_H and why? 1+1+1/2+1/2

Q7 Define the followings;

(i) Raoult^,S law (ii) Azeotropes (iii) Colligative properties 3

Q8 What do you understand by relative lowering in V.P? Show that it is a colligative

Property. How will determine the molar mass by relative lowering in V.P? 3

Q9 How many ml of 0.1 M HCl are required to react completely with 1gm mixture of

Na₂CO₃ and NaHCO₃ containing equimolar amounts of both. 3

Q10 1.00 g of a non-electrolye solute dissolved in 50 g of benzene lowered the freezing point of benzene by 0.40 K. Find the molar mass of the solute.( K_f for benzene = 5.12 K kg mol^-1)

ANSWERS

Q1 (a) particles are presents on lattice points (b) schottky defect, density decreases (c) definition

Q2 NCERT page 31 Q 1.16 Ni⁺² =96% Ni⁺³= 4%

Q3 (a) extrinsic contains impurities , intrinsic no impurities (b) (i) n-type (ii) p-type

Q4 derive r/R = 0.225

Q5 Z = 2, M =?, Na = 6.022 X 10²³, No of atoms = Na x 208 / M , answer = 24.16 x 10²³.

Q6 Statement, solubility α 1/ K_H, H₂ has lower value of K_H

Q7 definitions in book

Q8 Derivation in book

Q9 NCERT page 60 Q-2.6, ans = 158.7 ml

Q10

UNIT(2) —SOLUTION Practice paper-1

Q 1 (i) Which is more concentrated, 1Molar aqueous solution or 1 molal aqueous solution & why?

(ii) Will the elevation in boiling point be same if 0.1 mole NaCl or 0.1 mole of sugar is

dissolved in 1 L of water?

Equimolar solutions of NaCl& glucose are not isotonic . Why?

(iv) Distinguish between boiling point & normal boiling point of a liquid . 1+1+1+1

Q 2 State Raoult,s law for solution containing nonvolatile solute in volatile solvent.Derive a

Mathematical expression for the law. 1+3

Q 3 What are ideal and nonideal solutions? Give reasons for their formation . Give one example

in each case. 1+1+1+1

Q 4 What isVant^,Hoff factor ? What possible values can it have if the solute molecules undergo

(i) Association (ii) dissociation in the solution .Prove that osmotic pressure is a colligative

Property. 1+1+1+1

Q 5 Define the following terms : 1+1+1+1

(i) Mole fraction (ii) Azeotropes (iii) Henry^,s law (iv) Molal depression constant

Q 6 How many mL of 0.1 M HCl are required to react completely with 1 gm mixture of

Na₂CO₃ and NaHCO₃ containing equimolar amount of both ? 4

Q 7 Calculate the depression in freezing point of water when 10 gm of CH₃CH₂CHClCOOH

is added to 250 gm of water .K_a= 1.4 x 10^-3, K_{f =}1.86 K Kg mole^-1. 4

Q8 Henry ^,s law constant for CO₂ in water is 1.67 x 10⁸ Pa at 298 K. Calculate the quantity of

CO₂ in 500 ml of soda water when packed under 2.5 Atm CO₂ pressure at 298 K. 4

Q9 18 gm of glucose, is dissolved in 1 kg of water in a saucepan. At what temperature will

Water boil at 1.013 bar? K_b for water is 0.52 K kg mol^-14

Q10 What volume of 95 wt % sulphuric acid (density 1.85 gm/cm³ ) and what mass of water

must be taken to pre pare 100 cm³0f 15 wt % solution sulphuric acid (density 1.10 gm/cm³)

2+2

Answers→

Q1 (i) 1 Molar because it contains both solvent & solute.(ii) No. More in case of 0.1M NaCl because it

dissociates into ions.(iii) NaCl dissociates to give ions & osmotic pressure is directly proportional to

number of particles.(iv) b.p is when v.p of liquid becomes equal to surrounding pressure.Normalb.p

is the temp at which v.p is 1 atm.

Q6 mass of mixture = 1gm , let Na₂CO₃ = x gm, mass of NaHCO₃ = 1—x , moles of Na₂CO₃= x / 106.

Moles of NaHCO₃ = 1—x / 84. equivimolar mixture means x/106 = 1—x / 84. x = 0.557

Mass of Na₂CO₃ = 0.557gm, mass of NaHCO₃ = 1—0.557 = 0.443gm.

Na₂CO₃ + 2 HCl → 2 NaCl+ H₂O + CO₂

106gm 2x36.5=73 106 gm Na₂CO₃need = 73gm HCl

0.557gm will need = 73 x 0.557 / 106 =0.384gm HCl

Similarly—

NaHCO₃ + HCl → NaCl+ CO₂ + H₂O

84gm 36.5gm 84gm NaHCO₃ need = 36.5gm HCl

0.443 will need = 36.5x 0.443 / 84 = 0.192gm HCl

Total HCl used = 0.384 + 0.192 = 0.576gm

M_HCl= moles of HCl / vol in L

0.576/ 36.5

0.1 = — -------- , V = 0.157 Litre(ans)

Q7 . Molar mass of CH₃CH₂CHClCOOH = 122.5,

moles of CH₃CH₂CHClCOOH= 10/122.5 = 0.0816, m = 0.0816/ 0.250kg = 0.3264

α = √ K_a/C = √ 1.4 x 10^—3/ 0.3264 = 0.065, i—1 i—1 i—1

α = --------, 0.065 = -------- = -------, i = 1.065

m—1 2—1 1

∆T_f = i x K_f x m = 1.065 x 1.86 x 0.3264 = 0.65⁰ (ANS)

Q9 Moles of glucose = 18 g/ 180 g mol–1 = 0.1 mol

Number of kilograms of solvent = 1 kg

Thus molality of glucose solution = 0.1 mol kg-1

For water, change in boiling point

ÄTb = Kb × m = 0.52 K kg mol–1 × 0.1 mol kg–1 = 0.052 K

Since water boils at 373.15 K at 1.013 bar pressure,therefore, the boiling point of solution will be 373.15 + 0.052 = 373.202K.

Q10 M of 95 % H₂SO₄ = moles x 1000 / V of solution in L

Moles of H₂SO₄ = 95/98, volume of solution = m/d = 100/1.85 = 54.05ml.

M = 95 x 1000/ 98 x 54.05 = 17.93 M

Similarly Molarity of 15% H₂SO₄ can be calculated as— moles = 15/98, V = 100/1.10 = 90.91ml

M = 15 x 1000 / 98 x 90.91 = 1.68 M

Apply molarity equation — M₁V₁ = M₂V₂

95 % 15 %

17.93 x V₁ = 1.68 x 100 , V₁ = 9.37 ml (ans)

Mass of 100 ml 15 % H₂SO₄to prepared = 1.10 x 100 = 110 gm

Mass of 9.4 ml H₂SO₄of 95 % = 9.4 x 1.85 = 17.4 gm, Mass of H₂O = 110—17.4 = 92.6 gm(ans)

Practice paper-2

Q1 (a)Why is elevation in boiling point of water different in the following cases; 3

(i) 0.1 M NaCl solution (ii) 0.1 M sugar solution

(b)Which is more concentrated 1 M or 1 m solution & why?

(c)Two liquids A & B boil at 145⁰C 190⁰C respectively. Which of them will have higher V.P at 80⁰C?

Q2 State Henry^,S law for solubility of a gas in liquid.Explain the significance of Henry^,S constant(K_H).

At the same temperature, H₂ is more soluble than He in water. Which of them will have higher value of

K_H& Why? 3

Q3 Define the terms: (a) Raoult^,Slaw (b) Azeotropes (c) Colligative properties 3

Q4 What do you understand by relative lowering in V.P? Show that it is a colligative property. How will you

determine the molar mass of by relative lowering in V.P ? 3

Q5 What are ideal and non-ideal solutions? Discuss the positive and negative deviations from ideal

behavior with the help of graphs. 3

Q6 Calculate (a) Molality (b) Molarity and (c) Mol fraction of KI if density of 20 % (mass/mass) aqueous

KI is 1.202 gm /ml. (K = 39, I = 127). 3

Q7 Vapour pressure of pure liquids A & B are 450 and 700 mm Hg respectively at 350 K. Find the

composition of liquid mixture if total VP is 600 mm Hg. Also find the composition in vapour phase. 3

Q8 How many mL of 0.1 M HCl are required to react completely with 1 gm mixture of

Na₂CO₃ and NaHCO₃ containing equimolar amount of both ? 3

Q9 Two elements A & B form compounds having formula AB₂ and AB₄. When dissolved in 20 gm benzene

1 gm of AB₂ lowers the freezing point by 2.3 K whereas 1 gm of AB₄ lowers it by 1.3 K. K_f for benzene

is 5.1 K kg mol^—1. Calculate atomic masses of A & B. 3

Q10 Calculate the depression in freezing point of water when 10 gm of CH₃CH₂CHClCOOH is added to

250 gm of water.(K_a = 1.4 x 10^—3, K_f = 1.86 = K kg mol^—1 . C = 12, H = 1, Cl =35.5, O =16) 3

Answers-

1	(a) 0.1M NaCl due to more number of ions (b) 1Molar because it contains solvent & solute both(c) liq A
2	statement, Solubility α 1 / K_H, H₂ has lower value of K_H
3	definitions, Q4- derivation, Q5- book,
4	book
5	book
6	NCERT page 37( 2.5) m =1.5, M = 1.44, X_KI = 0.0263 (a) mass of solvent = 100—20 = 80gm = 0.08 kg, molar mass of KI = 166, moles of KI= 20/166= 0.120 m = 0.120/0.08 = 1.5 (b) vol of solution = m/d = 100/1.202 = 83.2ml= 0.0832 L, M = 0.120/ 0.0832 = 1.44, (c) moles of H₂O = 80/18= 4.44, X_KI = 0.120/0.120 +4.44 = 0.0263
7	NCERT PAGE 47(2.8) X_A = 0.40, X_B = 0.60, vapour pressure X_A = 0.30, X_B = 0.70 P_A⁰ = 450, P_B⁰ = 700 , Total = 600mm of Hg, Raoults law(P_TOTAL) = P_A⁰ + P_B⁰ = P_A⁰ X_A + P_B⁰X_B (P_TOTAL) = P_A⁰ X_A + P_B⁰(1—X_A) , 600 = 450 x X_A + 700(1—X_A), X_A = 0.40 and X_B = 0.60 P_A = P_A⁰ X_A = 450 x 0.40 = 180 mm of Hg, P_B = P_B⁰X_B = 700 X 0.60 = 420 mm of Hg X_A in vapour phase = partial pressure of A / Total vapourpressure(P_A + P_B) = 180/ 180 + 420 = 0.30 X_B in vapour phase = 1—0.30 = 0.70
8	NCERT page 60 (2.6) ans—158.7ml , mass of mixture = 1gm , let Na₂CO₃ = x gm, mass of NaHCO₃ = 1—x , moles of Na₂CO₃= x / 106. Moles of NaHCO₃ = 1—x / 84. equivimolar mixture means x/106 = 1—x / 84. x = 0.557 Mass of Na₂CO₃ = 0.557gm, mass of NaHCO₃ = 1—0.557 = 0.443gm. Na₂CO₃ + 2 HCl → 2 NaCl+ H₂O + CO₂ 106gm 2x36.5=73 106 gm Na₂CO₃need = 73gm HCl 0.557gm will need = 73 x 0.557 / 106 =0.384gm HCl Similarly— NaHCO₃ + HCl → NaCl+ CO₂ + H₂O 84gm 36.5gm 84gm NaHCO₃ need = 36.5gm HCl 0.443 will need = 36.5x 0.443 / 84 = 0.192gm HCl Total HCl used = 0.384 + 0.192 = 0.576gm M_HCl= moles of HCl / vol in L 0.576/ 36.5 0.1 = — -------- , v = 0.157 Litre(answer) V in L
9	K_f x W_B x 1000 K_f x W_B x 1000 NCERT page 60(2.21), For AB₂, ∆T_f = ---------------------- , For AB₄ , ∆T_f = ------------------- M_B x W_AM_B x W_A 5.1 x 1 x 1000 5.1 x 1 x 1000 2.3 = --------------------- , 1.3 = ------------------- M_B x 20 M_B x 20 M_B = 110.86 M_B = 196.15 AB₂ = A + 2B = 110.86, AB₄ = A + 4B = 196.15, solve equations and find A & B values A = 25.58, B = 42.64 (ans)
10	NCERT page 61(2.32) Molar mass of CH₃CH₂CHClCOOH = 122.5, moles of CH₃CH₂CHClCOOH= 10/122.5 = 0.0816, m = 0.0816/ 0.250kg = 0.3264 α = √ K_a/C = √ 1.4 x 10^—3/ 0.3264 = 0.065, i—1 i—1 i—1 α = --------, 0.065 = -------- = -------, i = 1.065 m—1 2—1 1 ∆T_f = i x K_f x m = 1.065 x 1.86 x 0.3264 = 0.65⁰ (ANS)

Wednesday 26 April 2017

Class XII