# Chemistry Basics

## NCERT Solution 2

Question 10: In three moles of ethane (C_{2}H_{6}), 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 × 10^{23} molecules

So, 3 M ethane contains 3 × 6.022 × 10^{23} molecules

= 18.066 × 10^{23}

= 1.8 × 10^{24} molecules

Question 11: What is the concentration of sugar (C_{12}H_{22}O_{11}) 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 C_{12}H_{22}O_{11}

= 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 CH_{3}OH = 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 M_{1} × V_{1} = M_{2} × V_{2}

24.78 × V_{1} = 0.25 × 2.5

Or, V_{1} = `(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 10^{34} 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 cm^{2}

For converting it to per m^{2}, we need to multiply it by 100 × 100

Pressure = 10340 × 9.8 N m^{-2}

= 1.01332 × 10^{5} 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:

Prefixes | Multiples |
---|---|

(i) micro | (a) 10^{6} |

(ii) deca | (b) 10^{9} |

(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, CHCl_{3}, 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:

- 0.0048
- 234,000
- 8008
- 500.0
- 6.0012

**Answer:** (i) 4.8 × 10^{-3} (ii) 2.34 × 10^{5} (iii) 8.008 × 10^{3} (iv) 5.0 × 10^{2} (v) 6.0012 × 10^{0}