ncert solutions for class 10 maths

S.No.	Ncert solutions for class 10 maths
1	Real Numbers
2	Polynomials
3	Pair of Linear Equations in Two variables
4	Quadratic Equations
5	Arithematic Progressions
6	Triangles
7	Co-ordinate Geometry
8	Introduction to trigonometry
9	Some Applications of Trigonometry
10	Circles
11	Constructions
12	Areas Related to circles
13	Surface Areas and Volumes
14	Statistics
15	Probability

ncert solutions for class 10 science

S.No.	Ncert solutions for class 10 science
1	Chemical Reactions and Equations
2	Acids Bases and Salts
3	Metals and Non Metals
4	Carbon and its Compounds
5	Periodic Classification of Elements
6	Life Processes
7	Control and Coordination
8	How do Organisms Reproduce?
9	Heredity and Evolution
10	Light-Reflection and Refraction
11	Human Eye and Colourful World
12	Electricity
13	Magnetic Effects of Electric Currents
14	Sources of Energy
15	Our Environment
16	Management of Natural Resources

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class 10 science,
science  class 10
CBSE class 10 science solution

ncert solutions for class 12 maths

S. No.	Ncert solutions for class 12 maths
1	Relations and Functions
2	Inverse Trigonometry Functions
3	Matrices
4	Determinants
5	Continuity and Differentiability
6	Application of Derivative
7	Integrals
8	Application of Integrals
9	Differential Equations
10	Vector Algebra
11	Three Dimentianal Geometry
12	Linear Programming
13	Probability

ncert solutions for class 12 physics

S. No.	Ncert solutions for class 12 physics
1	ELECTRIC CHARGES AND FIELDS
2	ELECTROSTATIC POTENTIAL AND CAPACITANCE
3	CURRENT ELECTRICITY
4	MOVING CHARGES AND MAGNETISM
5	MAGNETISM AND MATTER
6	ELECTROMAGNETIC INDUCTION
7	ALTERNATING CURRENT
8	ELECTROMAGNETIC WAVES
9	RAY OPTICS AND OPTICAL INSTRUMENTS
10	WAVE OPTICS
11	DUAL NATURE OF RADIATION AND MATTER
12	ATOMS
13	NUCLEI
14	SEMICONDUCTOR ELECTRONICS: MATERIALS, DEVICES AND SIMPLE CIRCUITS
15	COMMUNICATION SYSTEMS

An antifreeze solution is prepared from 222.6 g of ethylene glycol (C2H6O2) and 200 g of water. Calculate the molality of the solution. If the density of the solution is 1.072 g mL–1, then what shall be the molarity of the solution?

Molecular mass of glycol C2H4(OH)2 =62

Number of moles of C2H4(OH)2 = mass of glycol / molar mass of glycol

= 222.6/62=3.59  moles

Molality of solution = number of moles of solute / mass of solvent in Kg  

Mass of solvent = 200g = 200/1000 Kg = 0.2 kg

Plug the values we get

Molality = 3.59/0.2

Molality = 17.95

Formula of molarity of solution = number of moles of solute / volume of solution in Kg

Formula of volume = mass / density

Volume = 422.6g/(1.072 g/ml)

Volume = 394.21 ml

Convert in liter   

Volume in liter = 394.21 ml /1000 liter  =0.394 liter

Molarity =9.1M

How many mL of 0.1 M HCl are required to react completely with 1 g mixture of Na2CO3 and NaHCO3 containing equimolar amounts of both?

In our problem both Na2Co3 and NaHCO3 both are equimolar that means molairy of both substance are equal

So number of moles of Na2CO3  = number of moles of NaHCO3  --------------(1)

Let find moles of both substances

Let take mass of Na2CO3 = x g

Then mass of NaHCO3  = 1-x g

Molar mass of Na2CO3 = 2*23+ 12+ 3*16 =  106 g/ mol

Molar mass of NaHCO3 = 1*23 + 1 + 12 + 3*16 = 84 g/mol

So number of moles of Na2CO3 = mass/ molar mass = x/106 moles     ……….(2)

Number of moles of NaHCO3 is = (1-x)/ 84 gram

Plug the value in equation (1)

  We get

x/106 = 1-x/84

cross multiply them we get

84x = 106  - 106 x  solve it we get

X = 106/190 = 0.558 gram

Plug the value back in equation (2) we get = 0.558/106    = 0.00526 moles

Our reactions will

Na2CO3 + 2HCl ----> 2NaCl  +  H2CO3

NaHCO3+ HCl ------> NaCl +  H2CO3

Here in our reaction 2 moles of  HCl is required each for of Na2CO3 so that

 Moles of HCl required to react with Na2CO3 = 2* 0.00526   = 0.01052

And moles of HCl required for NaHCO3 will also 0.00526

Total number of HCl moles required = 0.01052 + 0.00525 = 0.01578 moles 

Molarity of HCl  = 0.1M   given in our problem

Molarity = number of moles of solute / volume of solution in liter

0.1  = 0.01578/ volume in liter  

Cross multiply now we get

0.1 * volume in liter  = 0.01578

Divide by 0.1 we get

Volume in liter  = 0.01578/0.1   = 0.1578 liter

Volume in ml = 0.01578 liter * 1000 ml / 1liter  = 157.8 ml 

 

A solution is obtained by mixing 300 g of 25% solution and 400 g of 40% solution by mass. Calculate the mass percentage of the resulting solution.

Mass of component in first solution = 25% of 300 gram  = 25*300/100 = 75 gram

Mass of component in second solution   = 40 % of 400 gram = 40*400/100 = 160 gram

Total mass of solution = 300+400 = 700 gram

Total mass of component  = 75+ 160  = 235 grams

Mass % of final solution     = mass of component * 100 / total mass

                                         =235*100/700  = 33.5%

Total of % is always 100

So % of water in solution = 100 – 33.5  = 66.5 % 

A solution of glucose in water is labelled as 10% w/w, what would be the molality and mole fraction of each component in the solution? If the density of solution is 1.2 g mL–1, then what shall be the molarity of the solution?

Step-1

What is given in problem?

In this problem mass of glucose is 10%w/w

So take mass of total solution = 100 g

Then mass of glucose will = 10% of 100 g = 10 grams

Mass of water will = 100 – 10 = 90 grams

And density = 1.2gmL-1

Step-2

find molality

Molality = number of moles of solute / mass of solvent in Kg   ……(1)

Here have to find number of moles and mass of solvent in Kg both

mass of solute  = mass of glucose / molar mass of glucose  …..(2)

Molar mass of glucose (C6H12O6)  = 6*C +  12*H + 6*O

= 6*12 + 12*1 +  6*16

=180 g /mol

Plug the value back in equation (2) we get

Number of moles =  10/180  = 0.0556 moles

We have mass of solvent (H2O)  =  90 g

Mass in Kg  = 90 g* 1 Kg / 1000 g    = 0.09 Kg

Plug the value in equation (1) we get

Molality  =  0.0556/0.09   = 16.7M

Step – 3 

Find molar fraction

Molar fraction of glucose  = number of moles of glucose / total number of moles

Molar mass of water (H2O) = 2*H+ O  = 2*1 + 16  = 18 g/ mol

Number of moles of water  = 90/18 = 5 moles

Plug the value in formula we get

 Mole fraction  = 0.0556/(0.0556+ 5)  = 0.01

It is binary solution because it has 2 component only so molar fraction of

Water  =1 – mole fraction of glucose

            = 1-0.01  =0.99

Step – 4

Find the molarity ,

molarity = number of moles of solute / volume of solution in liter        ……….(1)

Volume = mass / density

            = 100 g / 1.2 g/ ml  = 83.33 ml

Mass in liter  = 83.33 * 1 liter /1000 ml   = 0.0833liter

Plug the value back equation (1) we get

Molarity = 0.0556/ 0.0833

               = 0.67 M

Concentrated nitric acid used in laboratory work is 68% nitric acid by mass in aqueous solution. What should be the molarity of such a sample of the acid if the density of the solution is 1.504 g mL–1?

Answer :-

Step-1  find the given values

Mass % of nitric acid  = 68%

Density of nitric acid  = 1.504 g mL-1

Step -2

Let take total mass of solutions = 100    (why ?)

In this problem mass % is given whenever % is given take total quantity as 100 g ,

So  nitric acid =  68% of 100 g   = 68 g

Water will = 100 – 68  =  32 g

Step-3

We have to find molarity

Molarity(m) =  number of moles of solute / volume of solution in liter           ………..(1)

Here we have to find number of moles of solute  and volume of solution in liter                    

 Number of moles of solute = mass of solute /molar mass of solute             …………(2)

So molar mass of nitric acid HNO3  = 1*H + 1*N + 3*O  = 1*1 + 1*14 + 3*16  = 1+ 14 + 48 = 63 g

Plug the value in equation (2)  we get

Number of moles  = 68/63 = 1.079 moles

Volume = mass / density

              = 100g/1.504 g mL–1    =  66.5 ml

Now convert it in liter we get 

                                    66.5 ml * 1 liter / 1000ml =  0.0665 mL

Plug the value in equation first now we get

Molarity  = 1.079/ 0.0665  =16.23 M

Suppose a solid solution is formed between two substances, one whose particles are very large and the other whose particles are very small. What kind of solid solution is this likely to be?

Interstitial solid solution
In this solution, particles of solid solute fit into voids or interstices of solvent.

Define the following terms: (i) Mole fraction (ii) Molality (iii) Molarity (iv) Mass percentage

 

(i) Mole fraction (XA):-

It is the ratio of number of moles of one component to the total number of moles present in that solution

Let take a binary solution in which A is solute and B is solvent then

Molar fraction (XA) = number of moles of A/(number of moles of A + Number of moles of B)

If there are there are n number of component  in solution then

Molar fraction of component A =   number of moles of A / total number of moles

Sum of molar fraction of all components will always = 1

ie  NA+ NB+NC …………….= 1  (NA, NB  and NC .. are number of moles of components)

(ii) Molality(m) :-

Molality of any solution is represent the number of moles of solute present in per kg of solvent.

Molality(m) =  number of moles of solute / mass of solvent in Kg

Let take mass of solvent is = 1 kg then

Molality is numerically equal to number of moles of solute

So we can write the definition of molality that  “ molality  is the number of moles of solute present in 1 kg of solvent  “

Example 1.   Find the molality of solution containing 3 moles of  NaCl in 2 liter water ?

Density of water is 1kg/ liter so that    2 liter water = 2 kg of water

 Use above formula

    Molarity = 3/ 2 = 1.5 m

Calculate the molality of a solution containing 164 g of HCl in 753g of H20(water)

Number of moles of HCl = given mass of HCl / molar mass of HCl

Molar mass of HCl =  35.5 + 1  = 36.5 g /mol

Number of moles of HCl   = 164/36.5

                                           =4.5 moles

Mass of solvent in Kg  = 753 g* 1Kg/100 g  =  0.753 Kg

Molality = 4.5/0.753

            = 5.97m

Molarity(M):-

Molarity of any solution  is number of moles of solute per liter of solution

Molarity = number of moles of solute / volume of solution in liter

When volume of solution is 1 liter

Then molarity is numerically equal to number of moles of solute

So we can write a another definition of molarity that  “molarity is equal to number of moles of solute present in 1 liter of solution”

Mass percentage :-

(i) Mass percentage (w/w): The mass percentage of a component A of a solution is defined as mass of component A present in 100 unit mass of the solution. 

Mass % of a component A = mass of component A *100/Total mass of the solution

For example  let consider  a solution which has 15% salt in water by mass,  it has 15 g of glucose is dissolved in 85 g of water  in a 100 g solution.

(ii) Volume percentage (v/v): The volume percentage of a component A is defined as volume of component A present in 100 unit volume of the solution.

 Volume % of a component A = volume of component A *100/Total volume of the solution

For example, 15% methanol solution in water means that 15 mL of

Methanol is present in 100  ml of solution , and water will 100 – 15 = 85 ml

(iii) Mass by volume percentage (w/v): ): The mass by volume  percentage of a component A is defined as mass of  component A in gram present  100 ml volume of the solution.

Mass  by volume % of a component A = mass of component A in gram  *100/Total total volume  of the solution in ml .