# Numericals on Mole Concept Class 11 with Answers

JEE Mains & AdvancedGet important solved numericals on Mole Concept for Class 11 and various exams. View the Important Question bank for Class 11 & 12 Chemistry complete syllabus.

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These important questions will play significant role in clearing concepts of Chemistry. This question bank is designed keeping NCERT in mind and the questions are updated with respect to upcoming Board exams. You will get here all the important questions and solved numericals for class 11 & 12 chemistry along with Mole Concepts. Learn all important tricks related to Mole Concept with these numericals and access notes too for Class 11 and 12 topics. **Click Here for Detailed Chapter-wise Notes of Chemistry for Class 11th, JEE & NEET. ** You can access free study material for all three subject’s Physics, Chemistry and Mathematics. Click Here for Detailed Notes of any chapter. eSaral provides you complete edge to prepare for Board and Competitive Exams like JEE, NEET, BITSAT, etc. We have transformed classroom in such a way that a student can study anytime anywhere. With the help of AI we have made the learning Personalized, adaptive and accessible for each and every one. Visit eSaral Website to download or view free study material for JEE & NEET. Also get to know about the strategies to Crack Exam in limited time period.

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Mass is defined as the amount of matter present in substance. The unit of mass kilogram is defined as equal to the mass of the international prototype of the kilogram.

*S.I.*base units using power of 10 notation. (example $\left.2.54 \mathrm{mm}=2.54 \times 10^{-3} \mathrm{m}\right)$ (i) $1.35 \mathrm{mm}$ (ii) 1 day (iii) $6.45 \mathrm{mL}$ (iv) $48 \mu g$ (microgram) (v) 0.0426 inches

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**Avogadro’s Law :**Equal volume of all gases under similar conditions of temperature and pressure contain equal number of molecules.

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**(i) $6.022 \times 10^{23}$**

**(ii) 5.359**

**(iii) 0.04597**

**(iv) 34.216**

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**Average atomic mass :**It is defined as average of the mass of all of the atoms of an elements, e.g., average atomic mass of is 35.5

*u*. (ii)

**Mole**is defined as amount of substance that contains as many atoms, molecules and particles as there are atoms in exactly 0.012

*kg*of Carbon-12 isotope. (iii) Molar mass : It is mass of 1 mole of substance which contains $6.022 \times 10^{23}$ particles. (iv)

**Unit factor :**The factor which is used to convert one unit into another is called Unit factor. (v)

**Molarity :**It is defined as number of moles of solute dissolved per litre of solution. (vi)

**Precision and accuracy :**Precision means how closely the experimental measurements agree with another.

**Accuracy**means how close the experimental measurements and exact values are with each other.

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*g*of this complex.

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*gm*dihydrogen gas to yield ammonia ? Also calculate the amount of ammonia formed.

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**[At. Wt. of $B a=137 u, S=3 u, O=16 u]$**

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*S.I.*units : (i) 93 million miles (this is the distance between the earth and the sun). (ii) 5 feet 2 inches (this is the average height of an Indian female). (iii) 100 miles per hour (this is the typical speed of Rajdhani Express). (v) $46^{\circ} \mathrm{C}$ (this is the peak summer temperature in Delhi). (vi) 150 pounds (this is the average weight of an Indian male).

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**(ii) 5 feet 2 inches $=62$ inches**\[ \begin{array}{l} {1 \text { inch }=2.54 \times 10^{-2} \mathrm{m}} \\ {\text { Unit factor }=\frac{2.54 \times 10^{-2}(\mathrm{m})}{1 \mathrm{inch}}} \\ {62 \text { inches }=\frac{62(\text { inches }) \times 2.54 \times 10^{-2} \mathrm{m}}{1 \mathrm{inch}}} \\ {=62 \times 2.54 \times 10^{-2} \mathrm{m}} \\ {=157.48 \times 10^{-2} \mathrm{m}=1.57 \mathrm{m}} \end{array} \] (iii) 1 mile $=1.60 \mathrm{km}=1.60 \times 10^{3} \mathrm{m}$ \[ \begin{array}{l} {\text { Unit factor }=\frac{1.60 \times 10^{3} \mathrm{m}}{1 \mathrm{mile}}} \\ {1 \mathrm{hr}=60 \times 60 \mathrm{s}=3.6 \times 10^{3} \mathrm{s}} \\ {\text { Unit factor }=\frac{3.6 \times 10^{3} \mathrm{s}}{1 \mathrm{hr}}} \end{array} \] $\therefore \quad$ Speed $=\frac{100 \text { miles }}{h r}$ $=\frac{100 \text { miles }}{h r} \times \frac{1.60 \times 10^{3} \mathrm{m}}{1 \mathrm{mile}} \times \frac{1 \mathrm{hr}}{3.6 \times 10^{3} \mathrm{s}}=44 \mathrm{ms}^{-1}$ (iv) $1 \dot{A}=10^{-10} \mathrm{m}$ \[ \text { Unit factor }=\frac{10^{-10} \mathrm{m}}{1(\hat{A})} \] $\therefore \quad 0.74 \mathrm{A}=\frac{0.74 \hat{A} \times 10^{-10}(m)}{1(\hat{A})}$ \[ =0.74 \times 10^{-10} \mathrm{m} \text { or }=7.4 \times 10^{-11} \mathrm{m} \] (v) $\quad 0^{\circ} \mathrm{C}=273.15 \mathrm{K}$ $\quad 46^{\circ} \mathrm{C}=273.15 \mathrm{K}+46 \mathrm{K}=319.15 \mathrm{K}$ (vi) 1 pound $=454 \times 10^{-3} \mathrm{kg}$ \[ \text { Unit factor }=\frac{454 \times 10^{-3}(k g)}{1(\text { pound })} \] $\begin{aligned} \therefore \quad 150 \text { pound }=\frac{150(\text { pound }) \times 454 \times 10^{-3}(k g)}{1(\text { pound })} & \\=& 150 \times 454 \times 10^{-3} \mathrm{kg}=68.1 \mathrm{kg} \end{aligned}$

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*g*, masses of dioxygen combined will be 32, 64, 32 and 80

*g*in the given four oxides. These are in the ratio 1 : 2 : 1 : 5 which is a simple number ratio. Hence, the given data obey the law of multiple proportion.

**(i) $11 \mathrm{g}$ of $\mathrm{CO}_{2}$**

**(ii) $3.01 \times 10^{22}$ molecules of $\mathrm{CO}_{2}$**

**(iii) 1.12 litre of $\mathrm{CO}_{2}$ at $\mathrm{S.T.P.}$**

*C*,

*H*,

*O*elements. A 4.24

*mg*sample of butyric acid is completely burnt in oxygen. It gives 8.45

*mg*of carbon dioxide and 3.46

*mg*of water. What is the mass percentage of each element ? Determine the empirical and molecular formula of butyric acid if molecular mass of butyric acid is determined to be 88

*u*.

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*g*of carbon dioxide and 0.690

*g*of water and no other products. 10.0

*L*of this welding gas at

*S.T.P.*is found to weigh 11.6

*g*. Calculate : (i) empirical formula, (ii) molar mass, and (iii) molecular formula of welding gas.

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**(i) Calculation of empirical formula:**Mass of carbon in 3.38 g carbon dioxide $=\frac{12 \times 3.38}{44}=0.922 g$ Mass of hydrogen in 0.690

*g*of water $=\frac{2 \times 0.690}{18}=0.077 g$ The ratio by mass of $C$ and $H$ in the sample $=0.922: 0.077$ The mole ratio of $C$ and $H$ in the sample $=\frac{0.922}{12}: \frac{0.077}{1}=0.077: 0.077=1: 1$ Thus empirical formula of the gas is $C H$ (ii) Calculation of molar mass: $10.0 \mathrm{L}$ of gas at S.T.P. weighs $=11.6 \mathrm{g}$ $22.4 \mathrm{L}$ of gas at S.T.P. weighs $=\frac{11.6 \times 22.4}{10}=25.98 g$ Thus, molar mass of gas $=25.98 \mathrm{g} \mathrm{mol}^{-1}$ (iii) Calculation of molecular formula: $n=\frac{\text { Molecular mass }}{\text { Empirical formula mass }}=\frac{25.98}{13}=1.998 \approx 2$ Thus, molecular formula is $(C H)_{2}=C_{2} H_{2}$

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