# Chemistry Basics

## NCERT Solution 2

Question 10: In three moles of ethane (C2H6), calculate the following:

(i) Number of moles of carbon atoms.

Answer: 1 M of ethane contains 2 M of carbon atoms

So, 3 M of ethane contains 6 M of carbon atoms

(ii) Number of moles of hydrogen atoms.

Answer: 1 M of ethane contains 6 M of hydrogen atoms

So, 3 M of ethane contains 18 M of hydrogen atoms

(iii) Number of molecules of ethane.

Answer: 1 M ethane contains 6.022 × 1023 molecules

So, 3 M ethane contains 3 × 6.022 × 1023 molecules

= 18.066 × 1023

= 1.8 × 1024 molecules

Question 11: What is the concentration of sugar (C12H22O11) in mol L–1 if its 20 g are dissolved in enough water to make a final volume up to 2L?

Answer: Molar mass of C12H22O11

= 12 × 12 + 22 × 1 + 11 × 16

= 144 + 22 + 176 = 342 g

No. of moles in 20 g sugar = (20)/(342)=0.585

So, concentration in given solution =(0.585)/2=0.293 M 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: Molar mass of CH3OH = 12 + 4 × 1 + 16 = 32 g per mol = 0.032 kg per mol

Molarity of given solution = 0.793 kg per L ÷ 0.032 kg per mol = 24.78 mol per L

Applying M1 × V1 = M2 × V2

24.78 × V1 = 0.25 × 2.5

Or, V1 = (0.25×2.5)/(24.78)=0.02522 L

= 25.22 mL

Question 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: Let us first calculate the weight of given mass of air

Weight = m × g

= 1034 g × 9.8 m s-2

= 1.034 kg × 9.8 m s-2

= 1.034 × 9.8 N

This is the weight per cm2

For converting it to per m2, we need to multiply it by 100 × 100

Pressure = 10340 × 9.8 N m-2

= 1.01332 × 105 N m-2 Pa

Question 14: What is the SI unit of mass? How is it defined?

Answer: The SI unit of mass is kilogram. Kilogram is equal to the mass of the international prototype of the kilogram (a platinum-iridium alloy cylinder) kept at international Bureau of Weights and Measures, at Sevres, near Paris, France.

Question 15: Match the following prefixes with their multiples:

PrefixesMultiples
(i) micro(a) 106
(ii) deca(b) 109
(iii) mega(c) 10–6
(iv) giga(d) 10–15
(v) femto(e) 10

Answer: (i) c, (ii) e, (iii) a, (iv) b, (v) d

Question 16: What do you mean by significant figures?

Answer: The significant figures of a number are digits that carry meaningful contributions to its measurement resolution.

• All non-zero digits are significant: 1, 2, 3, 4, 5, 6, 7, 8, 9.
• Zeros between non-zero digits are significant: 102, 2005, 50009.
• Leading zeros are never significant: 0.02, 001.887, 0.000515.
• In a number with or without a decimal point, trailing zeros (those to the right of the last non-zero digit) are significant provided they are justified by the precision of their derivation: 389,000; 2.02000; 5.400; 57.5400.

Question 17: 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 15 ppm (by mass).

(i) Express this in per cent by mass.

Answer: 15 ppm means 15 parts per million

So, percentage =(15)/(10^6)× 100=0.0015%

(ii) Determine the molality of chloroform in the water sample.

Answer: Molar mass of chloroform = 12 + 1 + 3 × 35.5

= 13 + 106.5 = 119.5 g per mole

Now, 100 g of sample contains 0.0015 g chloroform

So, 1000 g of sample contains 0.0015 × 10 = 0.015 g = 1.5 × 10-2 g chloroform

No. of moles =(1.5×10^(-2))/(119.5)=1.266×10^(-4) m

Question 18: Express the following in the scientific notation:

1. 0.0048
2. 234,000
3. 8008
4. 500.0
5. 6.0012

Answer: (i) 4.8 × 10-3 (ii) 2.34 × 105 (iii) 8.008 × 103 (iv) 5.0 × 102 (v) 6.0012 × 100