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**Q.1.** **Calculate the molecular mass of the following:****(a) H _{2}O**

**H _{2}O Structure**

= [2(1.0084) + 1(16.00 u)]

= 2.016 u + 16.00 u

= 18.016

=

_{ CO2 Structure}

__The molecular mass of carbon dioxide, CO _{2:}__

= (1 × Atomic mass of carbon) + (2 × Atomic mass of oxygen)

= [1(12.011 u) + 2 (16.00 u)]

= 12.011 u + 32.00 u

=** 44.01 u**

**(c) ****CH**_{4}

_{}**CH _{4 }Structure**

__The molecular mass of methane, CH _{4:}__

= (1 × Atomic mass of carbon) + (4 × Atomic mass of hydrogen)

= [1(12.011 u) + 4 (1.008 u)]

= 12.011 u + 4.032 u

= **16.043 u**

**Q.2.** **Calculate the mass percent of different elements present in sodium sulfate (Na _{2}SO_{4}).**

**Ans.**

Na_{2}SO_{4}

The molecular formula of sodium sulphate is Na_{2}SO_{4}

Molar mass of Na_{2}SO_{4} = [(2 × 23.0) + (32.066) + 4 (16.00)]

= 142.066 g

► Mass percent of an element

∴ Mass percent of sodium:

= 32.379

= **32.4%**__Mass percent of sulphur:__

= 22.57

=

= 45.049

=

Given:

% of iron by mass = 69.9 %

% of oxygen by mass = 30.1 %

= 1.25

= 1.88

= 1:1.5 =

∴ The empirical formula of the iron oxide is

**(a) **As per the balanced equation, 1 mole of carbon burns in 1 mole of dioxygen (air) to produce 1 mole of carbon dioxide.**(b) **According to the question, only 16 g (0.5 mol) of dioxygen is available. Hence, it will react with 0.5 moles of carbon to give **22 g **of carbon dioxide. Hence, it is a limiting reactant.**(c) **Since oxygen is the limiting reactant here, the 16g (0.5 mol) of O_{2} will react with 6 g of carbon (0.5 mol) to form 22 g of carbon dioxide. The remaining 18 g of carbon (1.5 mol) will not undergo combustion.**Q.****5.** **Calculate the mass of sodium acetate (CH _{3}COONa) required to make 500 mL of 0.375 molars aqueous solution. The molar mass of sodium acetate is 82.0245 g mol^{–1}.**

0.375 M aqueous solution of sodium acetate ≡ 1000 mL of a solution containing 0.375 moles of sodium acetate.

∴ A number of moles of sodium acetate in 500 mL.

= 0.375/1000 x 500

= 0.1875 mole

Molar mass of sodium acetate = 82.0245 g mol

∴ Required mass of sodium acetate = (82.0245 g mol

=

The mass percent of nitric acid in the sample = 69 % (Given)

Thus, 100 g of nitric acid contains 69 g of nitric acid by mass.

= 1 +14 + 48

= 63 g mol

∴ Number of moles in 69 g of HNO

= 1.095 mol

= 70.92 mL ≡ 70.92 x 10

= 15.44 mol/L

∴ Concentration of nitric acid =

1 mole of CuSO

Molar mass of CuSO

= 63.5 +32.00 + 64.00

= 159.5 g

159.5 g of CuSO

⇒ 100 g of CuSO

of copper.

∴ Amount of copper that can be obtained from 100 g CuSO

► Mass percent of iron (Fe) = 69.9% (Given)

► Mass percent of oxygen (O) = 30.1% (Given)

► Number of moles of iron present in the oxide = 69.90/55.85 = 1.25

► Number of moles of oxygen present in the oxide = 30.1/16.0 = 1.88

► The ratio of iron to oxygen in the oxide:

The empirical formula of the oxide is Fe_{2}O_{3.}

Empirical formula mass of Fe_{2}O_{3} = [2(55.85) + 3(16.00)] g

Molar mass of Fe_{2}O_{3} = 159.69 g

The molecular formula of a compound is obtained by multiplying the empirical formula with n.

Thus, the empirical formula of the given oxide is Fe_{2}O_{3} and n is 1.

Hence, the molecular formula of the oxide is Fe_{2}O_{3.}**Q.****9. Calculate the atomic mass (average) of chlorine using the****following data:**

**Ans:**

__The average atomic mass of chlorine:__

= 26.4959 + 8.9568

= 35.4527 u

The average atomic mass of chlorine = **35.4527 u**

**Q.****10.** **In three moles of ethane (C _{2}H_{6}), calculate the following:**

**(a) Number of moles of carbon atoms.**

**(b) Number of moles of hydrogen atoms.**

**(c) Number of molecules of ethane.**

**Ans.**

**(a) **1 mole of C_{2}H_{6} contains 2 moles of carbon atoms.

Number of moles of carbon atoms in 3 moles of C_{2}H_{6 }

= 2 × 3 = **6**

**(b) **1 mole of C_{2}H_{6 }contains 6 moles of hydrogen atoms.

Number of moles of carbon atoms in 3 moles of C_{2}H_{6 }

= 3 × 6 =** 18**

**(c) **1 mole of C_{2}H_{6 }contains 6.023 × 10^{23} molecules of ethane.

Number of molecules in 3 moles of C_{2}H_{6 }

= 3 × 6.023 × 10^{23}

= **18.069 × 10**^{2}

**Q.****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 2 L?**

**Ans.**

__Molarity (M) of a solution is given by:__

= 0.02925 mol L^{–1}

The molar concentration of sugar = 0.02925 mol L^{–1}

**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?**

**Ans.**** **

**CH _{3}OH **Molar mass of methanol (CH

= 32 g mol^{–1}

= 0.032 kg mol^{–1}

__Molarity of methanol solution:__

(Since density is mass per unit volume)

Applying,

**M _{1}V_{1} = M_{2}V_{2}**(Given solution) = (Solution to be prepared)

(24.78 mol L^{–1}) V_{1 }= (2.5 L) (0.25 mol L^{–1})

V_{1 }= 0.0252 L

V_{1 }= **25.22 mL**

**Q.****13.** **The pressure**** is determined as force per unit area of the surface.**

**an SI unit of pressure, Pascal is as shown below:**

**1 Pa = 1 N m**^{–2}

**If the mass of air at sea level is 1034 g cm ^{–2}, calculate the pressure in Pascals.**

**Ans.**

The pressure is defined as a force acting per unit area of the surface.

**P = F/A**

= 1.01332 × 10^{5} kg m^{-1} s^{-2}

We know,

► 1 N = 1 kg ms^{–2}

Then,

► 1 Pa = 1 N m^{–2} = 1 kg m^{–2}s^{–2}

► 1 Pa = 1 kg m^{1}^{–}s^{2}^{–}

∴ Pressure = **1.01332 × 10 ^{5} Pa.**

**Q.****14.** **What is the SI unit of mass? How is it defined?**

**Ans. **The SI unit of mass is the **kilogram (kg)**.

- 1 Kilogram is defined as the mass equal to the mass of the international prototype of the kilogram.

**Q.****15. Match the following prefixes with their multiples**

**Ans.****Q.****16.** **What do you mean by significant figures?****Ans. **Significant figures are those meaningful digits that are known with certainty. They indicate uncertainty in an experiment or calculated value.**Example:** If 15.6 mL is the result of an experiment, then 15 is certain while 6 is uncertain, and the total number of significant figures is 3.

Hence, significant figures are defined as the total number of digits in a number, including the last digit that represents the uncertainty of the result.**Q.****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).**

≈1.5 x 10

⇒ 1000 g of the sample contains 1.5 × 10

Molar mass of CHCl

=

=

There are

There are

There are

There are

There are

There are

**(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 = ........... dm**^{3}**Ans.****(a) **If we fix the mass of dinitrogen at 28 g, then the masses of dioxygen that will combine with the fixed mass of dinitrogen are 32 g, 64 g, 32 g, and 80 g.

The masses of dioxygen bear a whole number ratio of 1:2:2:5. Hence, the given experimental data obeys the** law of multiple proportions**.

- The law states that if two elements combine to form more than one compound, then the masses of one element that combines with the fixed mass of another element are in the ratio of small whole numbers.

**(b) ****(i)** 1 km = 1 km × 1000 m/1 km × 100 cm/1m × 10 mm/1cm

⇒ 1 km = 10^{6} mm

⇒ 1 km = 1 km × 1000 m/1 km × 1 pm/10^{-12} m

⇒ 1 km = 10^{15} pm

Hence, **1 km = 10 ^{6} mm = 10^{15} pm**

⇒ 1 mg = 10

⇒ 1 mg = 1 mg × 1 g/ 1000 mg × 1 mg/ 10

⇒ 1 mg = 10

Hence,

⇒ 1 mL = 10

⇒ 1 mL = 1 cm

**Q.****22.** **If the speed of light is 3.0 × 10 ^{8} ms^{-1}, calculate the distance covered by light in 2.00 ns.**

Time taken to cover the distance = 2.00 ns = 2.00 × 10

Speed of light = 3.0 × 10

= (3.0 × 10

= 6.00 × 10

=

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