Chapter 1 Some Basic Concepts of Chemistry Solutions
Question - 11 : - What is the concentration of sugar (C12H22O11)in mol L–1 if its 20 g are dissolved in enough water to make afinal volume up to 2 L?
Answer - 11 : -
Molarity (M) of a solution is given by,
= 0.02925 mol L–1
Molarconcentration of sugar = 0.02925 mol L–1
Question - 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 - 12 : -
Molar mass of methanol (CH3OH) =(1 × 12) + (4 × 1) + (1 × 16)
= 32 g mol–1
= 0.032 kg mol–1
Molarity of methanol solution 
= 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 = 0.0252 L
V1 = 25.22 mL
Pressure is determined as force per unit area of thesurface. 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 - 13 : -
Pressure is defined as force acting per unit area of thesurface.
= 1.01332 × 105 kg m–1s–2
We know,
1 N = 1 kg ms–2
Then,
1 Pa = 1 Nm–2 = 1 kg m–2s–2
1 Pa = 1 kg m–1s–2
Pressure= 1.01332 × 105 Pa
Question - 14 : - What is the SI unit of mass? How is it defined?
Answer - 14 : -
The SI unit of mass is kilogram (kg). 1 Kilogram isdefined as the mass equal to the mass of the international prototype ofkilogram.
Question - 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 - 15 : -
| Prefix | Multiples |
| (i) | micro | 10–6 |
| (ii) | deca | 10 |
| (iii) | mega | 106 |
| (iv) | giga | 109 |
| (v) | femto | 10–15 |
Question - 16 : - What do you mean by significant figures?
Answer - 16 : -
Significant figures are those meaningful digits that areknown with certainty.
They indicate uncertainty in an experiment or calculatedvalue. For example, if 15.6 mL is the result of an experiment, then 15 iscertain while 6 is uncertain, and the total number of significant figures are3.
Hence, significant figures are defined as the total numberof digits in a number including the last digit that represents the uncertaintyof the result.
Question - 17 : - A sample of drinking water was found to beseverely contaminated with chloroform, CHCl3, supposed to becarcinogenic in nature. The level of contamination was 15 ppm (by mass).
(i) Express this in percent by mass.
(ii) Determine the molality of chloroform in the watersample.
Answer - 17 : -
(i) 1 ppm is equivalent to 1 part out of 1 million (106)parts.
Mass percent of 15 ppm chloroform in water
(ii) 100 g of the sample contains 1.5 × 10–3 gof CHCl3.
⇒ 1000 g of the sample contains 1.5 × 10–2 gof CHCl3.
Molality ofchloroform in water
Molar mass of CHCl3 = 12.00+ 1.00 + 3(35.5)
= 119.5 g mol–1
Molality ofchloroform in water = 0.0125 × 10–2 m
= 1.25 × 10–4 m
Question - 18 : - Express the following in the scientific notation:
(i) 0.0048
(ii) 234,000
(iii) 8008
(iv) 500.0
(v) 6.0012
Answer - 18 : -
(i) 0.0048 = 4.8× 10–3
(ii) 234, 000 = 2.34 ×105
(iii) 8008 = 8.008 ×103
(iv) 500.0 = 5.000 × 102
(v) 6.0012 = 6.0012
Question - 19 : - 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 - 19 : -
(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.
Question - 20 : - Round up the following upto three significant figures:
(i) 34.216
(ii) 10.4107
(iii) 0.04597
(iv) 2808
Answer - 20 : -
(i) 34.2
(ii) 10.4
(iii) 0.0460
(iv) 2810