(i) Calculate the total number of electrons present in one mole of methane. (ii) Find (a) the total number and (b) the total mass of neutrons in 7 mg of \$\displaystyle ^{ 14 }{ C }\$.(Assume that mass of a neutron =\$\displaystyle 1.675\times { 10 }^{ -27 }kg\$).(iii) Find (a) the total number and (b) the total mass of protons in 34 mg of \$\displaystyle { NH }_{ 3 }\$ at STP.Will the answer change if the temperature and pressure are changed?

(i) Total number of electrons present in 1 methane molecule \$\displaystyle =6+4=10\$.Total number of electrons resent in 1 mole of methane \$\displaystyle = 10 \times 6.023 \times 10^{23} = 6.023 \times 10^{24}\$ electrons.The number of neutrons is equal to the difference in the mass number and atomic number.It is equal to \$\displaystyle 14-6=8 \$.(ii)(a) The number of neutrons present in 7 mg of C-14 is \$\displaystyle \frac {6.023 \times 10^{23} \times 8}{14} \times 7 \times 10^{10-3} = 2.4088 \times 10^{21}\$ neutrons.(b) Total mass of neutrons in 7 mg of C-14 is \$\displaystyle 2.4088 \times 10^{21} \times 1.675 \times 10^{-27} 4.035 \times 10^{-6}\$ kg.(iii)(a)Number of protons present in 1 molecule of ammonia \$\displaystyle =7+3=10 \$.Number of molecules present in 34 mg of ammonia is \$\displaystyle \frac {6.023 \times 10^{23} \times 34 \times 10^{-3}}{17} = 1.2044 \times 10^{21}\$.Total number of protons present in 34 mg of ammonia\$\displaystyle= 10 \times 1.2044\times 10^{21} = 1.2044\times 10^{22}\$(b) Total mass of protons present in 34 mg of ammonia at STP is \$\displaystyle 1.6726\times 10^{-27} \times 1.204 \times 10^{22} = 2.0138 \times 10^{-5} \$ kg.When temperature and pressure are changed, there is no difference in the answer.

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Reason Behind Discovery of Electron

Q: Which of the following statements is false?

The e/m value of electron is approximately \$1837\$ times e/m value of proton, hence e/m value of proton is not highest. Thus the false statement is present in option C. Thus answer is option C.

Q: If \$N_{A}\$ is Avogadro's number, then number of valence electrons in 4.2 g of nitride ion \$\left ( N^{3-} \right )\$ is:

Number of moles \$=\dfrac { 4.2 }{ 14 } =0.3\$ mol\$0.3\$ mol \$=0.3\times { N }_{ A }\$ ions.\$\Rightarrow 8\times 0.3\times { N }_{ A }\$ electrons (as each ion contains \$5+3={ 8e }^{ - }\$)\$\Rightarrow 2.4{ N }_{ A }\$ electronsHence, the correct option is A.

Q: (i) Calculate the number of electrons which will together weighs one gram. (ii) Calculate the mass and charge of one mole of electrons.

The number of electrons which will weigh 1 g is \$\displaystyle \frac {10^{-3}}{9.11 \times 10^{-31}} = 1.098 \times 10^{27} \$ .Mass of one mole of electrons is \$\displaystyle 9.11 \times 10^{-31} \times 6.023 \times 10^{23}= 5.486 \times 10^{-7} \$kg.Charge of 1 mole of electrons is \$\displaystyle 1.602 \times 10^{-19} \times 6.023\times 10^{23} = 9.647\times 10^{4}\ C\$.

Q: An element with mass number \$81\$ contains \$31.7\%\$ more neutrons as compared to protons. Assign the atomic symbol.

Let \$x\$ be the number of protons. The number of neutrons is \$\displaystyle x+0.317x=1.317x\$. The mass number is \$81\$. It is equal to the sum of the number of protons and the number of neutrons.\$\displaystyle 81 = x +1.317 x =2.317x \$\$\displaystyle x = 35 \$Hence, the symbol is \$\displaystyle _{35}^{81}Br \$.

Q: Sum of number of protons, electrons, and neutron in 12g of \${}_6^{12}C\$ is:

The atomic weight of carbon = 12 g and the given weight = 12 gAs we know that, \$\text{No. of moles} = \cfrac{\text{Given weight}}{\text{Molecular weight}}\$\$\therefore\$ No. of moles of Carbon in 12 g \$= \cfrac{12}{12} = 1 \text{ mole}\$No. of atoms present in 1 mole of Carbon \$= {N}_{A} = 6.023 \times {10}^{23}\$As 1 mole of carbon contains 6 electrons, 6 protons and 6 neutrons.\$\therefore \text{ No. of electrons in }6.023 \times {10}^{23} \text{ atoms of carbon} = 6 \times 6.023 \times {10}^{23}\$Similarly,\$\text{No. of protons in }6.023 \times {10}^{23} \text{ atoms of carbon} = 6 \times 6.023 \times {10}^{23}\$\$\text{No. of neutrons in }6.023 \times {10}^{23} \text{ atoms of carbon} = 6 \times 6.023 \times {10}^{23}\$\$\therefore\$ Sum of electrons, protons and neutrons in 12g of carbon = no. of electrons in 12 g carbon + no. of protons in 12 g carbon + no. of neutrons in 12 g carbon\$ = 6 \times 6.023 \times {10}^{23} + 6 \times 6.023 \times {10}^{23} + 6 \times 6.023 \times {10}^{23}\$\$= 108.414 \times {10}^{23} = 1.084 \times {10}^{25}\$So, the correct option is \$C\$

Q: Number of neutrons present in \$1.7\$ g of ammonia is:

Let us consider the problem.For, moles of \$N{H_3} = \frac{{1.7}}{{17}} = 0.1\$Let the Neutrons in 1 molecules of \$N{H_3} = 7 + 3 \times 0 = 7\$Hence,Number of Neurons \$= 0.1 \times {N_A} \times 7\$The correct answer is \$0.7{N_A}\$

Q: The order (lowest to highest) of e/m (charge/mass) ratio for electron (e), proton (p), neutron (n) and alpha particle \$(\alpha)\$ is:

The increasing order (lowest to highest) of e/m (charge/mass) ratio is \$ n < \alpha < p < e\$.Neutron being neutral has zero charges and lowest e/m ratio. An electron has a unit charge but negligible mass. Hence, the electron has a maximum e/m ratio.For an alpha particle, the e/m ratio is 2/4 or 0.5.For proton, e/m ratio is 1/1 or 1.

Q: The ion of an element has 3 positive charges. The mass number of the atom is 27 and the number of neutrons is 14. The number of electrons in the ion is:

From mass number and number of neutrons, we can calculate number of protons or electrons which is 13 (mass number - number of neutrons). Hence, the electronic configuration of this element would be 2, 8, 3. In order to become stable it would lose the last 3 electrons and gain 3 positive charges as given in the question. So, the resultant ion is left with 10 electrons. The element here is Al (aluminium).Hence correct option is (B).

Q: The mass of \$1\$ mole electron is \$0.55mg\$.

Mass of one electron is \$9.1 \times 10^{-28}gm\$So mass of one mole of electronis \$ ( 6.02 \times 10^{23} )\times ( 9.1 \times 10^{-28} )\$= \$0.55 \times 10^{-3}\$ gm \$=0.55mg\$ So, the given statement is true.

Chemistry

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e/m ratio by J.J Thompson

In 1897, the cathode ray tube or CRT, which is now a part of most TV sets, was the last word in advanced laboratory instrumentation and TV was still 40 to 50 years in the future. At Cambridge University in England, J.J Thomson studied the rays emanating from the cathode of the CRT. Thomson measured the ratio of charge to mass (e/m) of these corpuscles of which the rays were composed. We now call them electrons. Well do his experiment, with a somewhat modified apparatus. Main parts of the apparatus electron gun consists of filament, cathode and anode. He setup the experiment in the same way as William Crookes. When the rays got emitted he measured the e/m ratio by use of magnetic and electric fields.

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J.J. Thomson experiment for determination of charge to mass ratio

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e/m ratio by J.J Thompson

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Subatomic Particles - Electron

Subatomic Particles - Protons

Discovery of Neutrons

Subatomic Particles - Comparison

Which of the following statements is false?

If NA​ is Avogadro's number, then number of valence electrons in 4.2 g of nitride ion (N3−) is:

(i) Calculate the number of electrons which will together weighs one gram. (ii) Calculate the mass and charge of one mole of electrons.

An element with mass number 81 contains 31.7% more neutrons as compared to protons. Assign the atomic symbol.

Sum of number of protons, electrons, and neutron in 12g of 612​C is:

Number of neutrons present in 1.7 g of ammonia is:

The order (lowest to highest) of e/m (charge/mass) ratio for electron (e), proton (p), neutron (n) and alpha particle (α) is:

The ion of an element has 3 positive charges. The mass number of the atom is 27 and the number of neutrons is 14. The number of electrons in the ion is:

The mass of 1 mole electron is 0.55mg.

If $N_{A}$ is Avogadro's number, then number of valence electrons in 4.2 g of nitride ion $(N^{3 -})$ is:

(i) Calculate the number of electrons which will together weighs one gram.
(ii) Calculate the mass and charge of one mole of electrons.

An element with mass number $81$ contains $31.7 %$ more neutrons as compared to protons. Assign the atomic symbol.

Sum of number of protons, electrons, and neutron in 12g of $_{6}^{12} C$ is:

Number of neutrons present in $1.7$ g of ammonia is:

The order (lowest to highest) of e/m (charge/mass) ratio for electron (e), proton (p), neutron (n) and alpha particle $(α)$ is:

The mass of $1$ mole electron is $0.55 m g$ .