If the speed of light is 3.0 ×10^8 m s-1, calculate the distance covered by light in 2.00 ns

Q-22 If the speed of light is 3.0 ×10^8 m s-1, calculate the distance covered by light in 2.00 ns.
Answer
According to the question: 

1ns  = 10^-9 s (conversion )

Time taken to cover the distance = 2.00 ns 

                                                 = 2.00 × 10^–9 s
            Speed of light                = 3.0 × 10^8 ms–1
Distance= Speed of light × Time taken  (formula )
= (3.0 × 10^8 ms –1) (2.00 × 10^–9 s)
= 6.00 × 10^–1 m
= 0.600 m

The following data are obtained when dinitrogen and dioxygen react together to form different compounds: Mass of dinitrogen Mass of dioxygen (i) 14 g 16 g (ii) 14 g 32 g (iii) 28 g 32 g (iv) 28 g 80 g

Q- 21 The following data are obtained when dinitrogen and dioxygen react together to form
different compounds:
Mass of dinitrogen Mass of dioxygen
(i) 14 g 16 g
(ii) 14 g 32 g
(iii) 28 g 32 g
(iv) 28 g 80 g
Answer :- 
(a) Which law of chemical combination is obeyed by the above experimental data?Give
its statement.
(b) Fill in the blanks in the following conversions:
(i) 1 km = ...................... mm = ...................... pm
(ii) 1 mg = ...................... kg = ...................... ng
(iii) 1 mL = ...................... L = ...................... dm3

Answer :-

A) Fixing the mass of dinitrogen to 28g , mass of dioxygen combined with 32,64,32, and 80 g in the given four oxides . This is the ratio of 1:2:1:5 which is a simple whole number ratio . Hence the given data follow the law of multiple proportions

Law of Multiple Proportions This law was proposed by Dalton in 1803.According to this law, if two elements cancombine to form more than one compound, themasses of one element that combine with afixed mass of the other element, are in theratio of small whole numbers.For example, hydrogen combines with oxygen to form two compounds, namely, water and hydrogen peroxide.
Hydrogen + Oxygen → Water
2g                 16g           18g
Hydrogen + Oxygen → Hydrogen Peroxide
2g               32g                 34g
Here, the masses of oxygen (i.e. 16 g and 32 g)which combine with a fixed mass of hydrogen (2g) bear a simple ratio, i.e. 16:32 or 1: 2. .

B)

Q-20 Round up the following upto three significant figures: (i) 34.216 (ii) 10.4107 (iii) 0.04597 (iv) 2808

Q-20 Round up the following upto three significant figures:
(i) 34.216
(ii) 10.4107
(iii) 0.04597
(iv) 2808
Answer
(i) 34.2
(ii) 10.4
(iii) 0.0460
(iv) 2810

How many significant figures are present in the following? (i) 0.0025 (ii) 208 (iii) 5005 (iv) 126,000 (v) 500.0 (vi) 2.0034

How many significant figures are present in the following?
(i) 0.0025
(ii) 208
(iii) 5005
(iv) 126,000
(v) 500.0
(vi) 2.0034
Answer

For information of rules
green colour representing the significant figure 

(i) 0.0025          There are 2 significant figures.
(ii) 208              There are 3 significant figures.
(iii) 5005           There are 4 significant figures.
(iv) 126,000      There are 3 significant figures.

(v) 500.0           There are 4 significant figures.
(vi) 2.0034        There are 5 significant figures.

Express the following in the scientific notation: (i) 0.0048 (ii) 234,000 (iii) 8008 (iv) 500.0 (v) 6.0012

Q-18 Express the following in the scientific notation:
(i) 0.0048
(ii) 234,000
(iii) 8008
(iv) 500.0
(v) 6.0012

Answer
(i) 0.0048 = 4.8× 10^–3
(ii) 234, 000 = 2.34 ×10^5
(iii) 8008 = 8.008 ×10^3
(iv) 500.0 = 5.000 × 10^2
(v) 6.0012 = 6.0012

A sample of drinking water was found to be severely contaminated with chloroform,CHCl3, supposed to be carcinogenic in nature. The level of contamination was 15ppm (by mass). (i) Express this in percent by mass. (ii) Determine the molality of chloroform in the water sample.

A sample of drinking water was found to be severely contaminated with chloroform,CHCl3, supposed to be carcinogenic in nature. The level of contamination was 15ppm (by mass).
(i) Express this in percent by mass.
(ii) Determine the molality of chloroform in the water sample.

Answer : - 

1)   15 ppm means 15 parts in 1 millions (10^6)

                  and percentage always for 100 

           hence  15 ppm = (15/10^6 )*100 

                                   = 15*10^-4

                                   = 1.5*10^-3 

2) molar mass of chloroform  = 118.5g/ mol 

  from percentage by mass   100 gram of sample has 1.5*10^-3 g of chloroform 

then 1000 gram of sample has = 1.5*10^-3 g * 1000 gram/ 100 gram 

                                                = 1.5*10^-2 g

molality = moles of solute / mass of solvent in Kg

          

moles of solute = mass of solute / molar mass 

                         = 1.5*10^-2/ 118.5  

                         =1.266*10^-4 moles

molality  =     1.266*10^-4 moles/ 1 Kg  = 1.266*10^-4 m

    

                                   

What do you mean by significant figures ?Addition and Subtraction of SignificantFigures,Multiplication and Division of Significant Figures

Q- 16 What do you mean by significant figures and the rules of significant figures  ?

Answer :-

Significant figures are meaningful digits which are known with certainty. The uncertainty is indicated bywriting the certain digits and the last uncertain digit.  It is the maximum precision (reliability of a measurement) for a measurement. Thus, if we write a result as 11.2 mL, wesay the 11 is certain and 2 is uncertain and the uncertainty would be +1 in the last digit

There are certain rules for determining the number of significant figures. These are stated
below:

(1) All non-zero digits are significant. For example in 285 cm, there are three  significant figures and in 0.25 mL, there are two  significant figures.
 

(2) Zeros preceding to first non-zero digit are not significant. Such zero indicates the position of decimal point. Thus, 0.03 has one significant figure and  0.0052 has two significant figures.
 

(3) Zeros between two non-zero digits are significant. Thus, 2.005 has four significant figures.
 

(4) Zeros at the end or right of a number are significant provided they are on the right side of the decimal point. For example, 0.200 g has three significant figures. But, if otherwise, the zeros are not significant. For example, 100 has only one significant figure.

(5) Exact numbers have an infinite number of significant figures. For example, in 2 balls or 20 eggs, there are infinite significant figures as these are exact numbers and can be represented by writing infinite number of zeros after placing a decimal i.e., 2 = 2.000000 or 20 = 20.000000 When numbers are written in scientificnotation, the number of digits between 1 and10 gives the number of significant figures.Thus, 4.01×102 has three significant figures, and 8.256 × 10–3 has four significant figures.

Addition and Subtraction of SignificantFigures
The result cannot have more digits to the right of the decimal point than either of the original numbers.
                    12.11    +   18.0   + 1.012  =  31.122
Here, 18.0 has only one digit after the decimal point and the result should be reported only up to one digit after the decimal point which is 31.1.

Multiplication and Division of Significant Figures
In these operations, the result must be reported with no more significant figures as are there in the measurement with the few significant figures  2.5×1.25 = 3.125 Since 2.5 has two significant figures, theresult should not have more than two significant figures, thus, it is 3.1.

Match the following prefixes with their multiples: Prefixes Multiples (i) micro 10^-6 (ii) deca 10 (iii) mega 10^6 (iv) giga 10^9 (v) femto 10^-15

Q-15 Match the following prefixes with their multiples:
Prefixes Multiples
(i) micro 106
(ii) deca 109
(iii) mega 10–6
(iv) giga 10–15
(v) femto 10

Answer 

(i) micro 10^-6
(ii) deca 10
(iii) mega 10^6
(iv) giga 10^9
(v) femto 10^-15

What is the SI unit of mass? How is it defined?

Question 14  What is the SI unit of mass? How is it defined?
Answer
The SI unit of mass is kilogram (kg). 1 Kilogram is defined as the mass equal to the mass of the international prototype of kilogram.

Pressure is determined as force per unit area of the surface. The SI unit of pressure, pascal is as shown below : 1Pa = 1N m–2 If mass of air at sea level is 1034 g cm–2, calculate the pressure in pascal.

Q- 13 Pressure is determined as force per unit area of the surface. The SI unit of  pressure, pascal is as shown below :
1Pa = 1N m–2
If mass of air at sea level is 1034 g cm–2, calculate the pressure in pascal.

Answer
Pressure is defined as force acting per unit area of the surface 

                P =  F/ A  

                

                                  

                                  = 1.01332 × 105 kg m–1s–2
                                                           
                   We know,
                            1 N = 1 kg ms–2
                 Then  1 Pa = 1 Nm–2 = 1 kg m s–2*m-2
                          1 Pa = 1 kg m–1s–2
                                        Pressure = 1.01332 × 10^5 Pa

If the density of methanol is 0.793 kg L–1, what is its volume needed for making 2.5 L of its 0.25 M solution?

Q- 12 If the density of methanol is 0.793 kg L–1, what is its volume needed for making 2.5 L of its 0.25 M solution?

Answer
Molar mass of methanol (CH3OH) = (1 × 12) + (4 × 1) + (1 × 16)
                                                  = 32 g mol–1
                                                  = 0.032 kg mol–1   (divide by 1000 to convert in Kg )
Molarity of methanol solution=0.793 Kg L-1/ 0.032 Kg mol -1
                                            = 24.78 mol L–1
 

(Since density is mass per unit volume)
Applying,
                     M1V1                      =                    M2V2
               (Given solution)                 (Solution to be prepared)
(24.78 mol L–1) V1= (2.5 L) (0.25 mol L–1)V1
                     V1 = 0.0252 L   (multiply by 1000 to convert in ml )
                          = 25.22 mL