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: 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 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.

Question 1

Which of the following statements is false?

Accepted Answer

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.

Question 2

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

Accepted Answer

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.

Question 3

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

Accepted Answer

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 $\displaystyle 1.602 \times 10^{-19} \times 6.023\times 10^{23} = 9.647\times 10^{4}\ C$.

Question 4

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

Accepted Answer

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 $.

Question 5

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

Accepted Answer

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 $C$

Question 6

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

Accepted Answer

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 $0.7{N_A}$

Question 7

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

Accepted Answer

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.

Question 8

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:

Accepted Answer

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).

Question 9

(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 $^{14} C$ .
(Assume that mass of a neutron = $1.675 \times 10^{- 27} k g$ ).

(iii) Find (a) the total number and (b) the total mass of protons in 34 mg of $N H_{3}$ at STP.
Will the answer change if the temperature and pressure are changed?

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Accepted Answer

(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.

Question 10

The mass of $1$ mole electron is $0.55mg$.

Accepted Answer

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.

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