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Multiple Choice Questions

1. Total number of atoms represented by the compound $\text{CuSO}_4 \cdot 5\text{H}_2\text{O}$ is : [BHU 2005]

(a) 27

(b) 21

(d) 8

Answer: (b)

Explanation: Total number of atoms in $\text{CuSO}_4 \cdot 5\text{H}_2\text{O} = 1 (\text{Cu}) + 1 (\text{S}) + 4 (\text{O}) + 5 \times 2 (\text{H}) + 5 \times 1 (\text{O}) = 1 + 1 + 4 + 10 + 5 = 21$.

2. The equivalent weight of phosphoric acid ($\text{H}_3\text{PO}_4$) in the reaction, $\text{NaOH} + \text{H}_3\text{PO}_4 \longrightarrow \text{NaH}_2\text{PO}_4 + \text{H}_2\text{O}$ is: [BHU 2005]

(a) 25

(b) 49

(d) 98

Answer: (d)

Explanation: $\text{NaOH} + \text{H}_3\text{PO}_4 \longrightarrow \text{NaH}_2\text{PO}_4 + \text{H}_2\text{O}$

$\text{EW} = \frac{\text{MW}}{\text{no. of ionisable } \text{H}^+} = \frac{98}{1} = 98$

3. The number of sodium atoms in 2 moles of sodium ferrocyanide is

(a) $23 \times 10^{12}$

(b) $23 \times 10^{26}$

(d) $23 \times 10^{48}$

Answer: (d)

Explanation: As we know that four sodium atom are present in sodium ferrocyanide $\text{Na}_4[\text{Fe}(\text{CN})_6]$.

Hence, number of $\text{Na}$ atoms = No. of moles $\times$ number of atom $\times$ Avogadro's number

$$2 \times 4 \times 6.023 \times 10^{23} = 48 \times 10^{23}$$

4. How many moles of electron weighs one kilogram?

(a) $6.023 \times 10^{23}$

(b) $\frac{10^{31}}{9.108}$

(d) $\frac{10^8}{9.108 \times 6.023}$

Answer: (d)

Explanation: Mass of an electron $= 9.108 \times 10^{-31}\text{ kg}$

$\because 9.108 \times 10^{-31}\text{ kg} = 1\text{ electron}$

$\therefore 1\text{ kg} = \frac{1}{9.108 \times 10^{-31}}\text{ electrons} = \frac{10^{31}}{9.108 \times 6.023 \times 10^{23}}\text{ mole of electrons} = \frac{10^8}{9.108 \times 6.023}\text{ mole of electrons.}$

5. Which has maximum number of atoms?

(a) 24 g of C (12)

(b) 56 g of Fe (56)

(d) 108 g of Ag (108)

Answer: (a)

Explanation: Number of atoms = number of moles $\times$ Avogadro's number ($N_A$)

$\Rightarrow$ Number of atoms in 24 g $\text{C} = \frac{24}{12} N_A = 2N_A$

$\Rightarrow$ Number of atoms in 56 g of $\text{Fe} = \frac{56}{56} N_A = N_A$

$\Rightarrow$ Number of atoms in 27 g of $\text{Al} = \frac{27}{27} N_A = N_A$

$\Rightarrow$ Number of atoms in 108 g of $\text{Ag} = \frac{108}{108} N_A = N_A$

Hence, 24 g of carbon has the maximum number of atoms.

6. Mixture X = 0.02 mole of $[\text{Co}(\text{NH}_3)_5\text{SO}_4]\text{Br}$ and 0.02 mole of $[\text{Co}(\text{NH}_3)_5\text{Br}]\text{SO}_4$ was prepared in 2 L solution.

1 L of mixture X + excess of $\text{AgNO}_3$ solution $\longrightarrow \text{Y}$

1 L of mixture X + excess of $\text{BaCl}_2$ solution $\longrightarrow \text{Z}$

Number of moles of Y and Z are:

(a) 0.01, 0.01

(b) 0.02, 0.01

(d) 0.02, 0.02

Answer: (a)

Explanation: 1.0 L of mixture X contain 0.01 mole of each $[\text{Co}(\text{NH}_3)_5\text{SO}_4]\text{Br}$ and $[\text{Co}(\text{NH}_3)_5\text{Br}]\text{SO}_4$. Also, with $\text{AgNO}_3$, only $[\text{Co}(\text{NH}_3)_5\text{SO}_4]\text{Br}$ reacts to give $\text{AgBr}$ precipitate as

$$[\text{Co}(\text{NH}_3)_5\text{SO}_4]\text{Br} + \text{AgNO}_3 \text{ (excess)} \longrightarrow [\text{Co}(\text{NH}_3)_5\text{SO}_4]\text{NO}_3 + \text{AgBr} \downarrow \text{ (1.0 mole)}$$

With $\text{BaCl}_2$, only $[\text{Co}(\text{NH}_3)_5\text{Br}]\text{SO}_4$ reacts giving $\text{BaSO}_4$ precipitates as

$$[\text{Co}(\text{NH}_3)_5\text{Br}]\text{SO}_4 + \text{BaCl}_2 \text{ (excess)} \longrightarrow [\text{Co}(\text{NH}_3)_5\text{Br}]\text{Cl} + \text{BaSO}_4 \downarrow \text{ (1 mol)}$$

Hence, moles of Y and Z are 0.01 each.

7. Given that the abundances of isotopes $^{54}\text{Fe}$, $^{56}\text{Fe}$, and $^{57}\text{Fe}$, are 5%, 90% and 5%, respectively, the atomic mass of Fe is;

(a) 55.85

(b) 55.95

(d) 56.05

Answer: (b)

Explanation: From the given relative abundance, the average weight of $\text{Fe}$ can be calculated as:

$$A = \frac{54 \times 5 + 56 \times 90 + 57 \times 5}{100} = 55.95$$

8. Dissolving 120 g of urea (mol. wt. 60 g) in 1000 g of water gave a solution of density 1.15 g/mL. The molarity of the solution is:

(a) 1.78 M

(b) 2.00 M

(d) 2.22 M

Answer: (c)

Explanation: $\text{Molarity} = \frac{\text{Moles of solute}}{\text{Volume of solution (L)}}$

$\text{Moles of urea} = \frac{120}{60} = 2$

$\text{Weight of solution} = \text{weight of solvent} + \text{weight of solute} = 1000 + 120 = 1120\text{ g}$

$\Rightarrow \text{Volume} = \frac{1120\text{ g}}{1.15\text{ g/mL}} = 973.9\text{ mL} = 0.974\text{ L}$

$\Rightarrow \text{Molarity} = \frac{2}{0.974} = 2.05\text{ M}$

9. The molarity of a solution obtained by mixing 750 mL of 0.5 (M) $\text{HCl}$ with 250 mL of 2 (M) $\text{HCl}$ will be:

(a) 0.875 M

(b) 1.00 M

(d) 0.0975 M

Answer: (a)

Explanation: From the formula,

$$M_f = \frac{M_1V_1 + M_2V_2}{V_1 + V_2}$$

Given, $V_1 = 750\text{ mL}$, $M_1 = 0.5\text{ M}$, $V_2 = 250\text{ mL}$, $M_2 = 2\text{ M}$

$$M_f = \frac{750 \times 0.5 + 250 \times 2}{750 + 250} = \frac{375 + 500}{1000} = \frac{875}{1000} = 0.875\text{ M}$$

10. The ratio of masses of oxygen and nitrogen of a particular gaseous mixture is 1:4. The ratio of number of their molecule is:

(a) 1 : 4

(b) 7 : 32

(d) 3: 16

Answer: (b)

Explanation: $\frac{n_{\text{O}_2}}{n_{\text{N}_2}} = \frac{m_{\text{O}_2} / M_{\text{O}_2}}{m_{\text{N}_2} / M_{\text{N}_2}}$

Where, $m_{\text{O}_2}$ = given mass of $\text{O}_2$, $m_{\text{N}_2}$ = given mass of $\text{N}_2$, $M_{\text{O}_2}$ = molecular mass of $\text{O}_2 = 32$, $M_{\text{N}_2}$ = molecular mass of $\text{N}_2 = 28$

$$\frac{n_{\text{O}_2}}{n_{\text{N}_2}} = \left(\frac{m_{\text{O}_2}}{m_{\text{N}_2}}\right) \times \left(\frac{28}{32}\right) = \frac{1}{4} \times \frac{28}{32} = \frac{7}{32}$$

11. 1 gram of a carbonate ($\text{M}_2\text{CO}_3$) on treatment with excess $\text{HCl}$ produces 0.01186 mole of $\text{CO}_2$. The molar mass of $\text{M}_2\text{CO}_3$ in $\text{g mol}^{-1}$ is: [JEE Main 2017]

(a) 11.86

(b) 1186

(d) 118.6

Answer: (c)

Explanation: Given chemical equation

$$\text{M}_2\text{CO}_3 + 2\text{HCl} \longrightarrow 2\text{MCl} + \text{H}_2\text{O} + \text{CO}_2$$

From the balanced chemical equation:

$$\frac{1}{M} = 0.01186 \implies M = \frac{1}{0.01186} = 84.3\text{ g/mol}$$

12. The most abundant elements by mass in the body of a healthy human adult are: Oxygen (61.4%); Carbon (22.9%); Hydrogen (10.0%); and Nitrogen (2.6%). The weight which a 75 kg person would gain if all atoms are replaced by $^2\text{H}$ atoms is: [JEE Main 2017]

(a) 10 kg

(b) 15 kg

(d) 7.5 kg

Answer: (d)

Explanation: Mass in the body of a healthy human adult has:

Oxygen = 61.4%, Carbon = 22.9%, Hydrogen = 10.0% and Nitrogen = 2.6%

Total weight of person = 75 kg

Mass due to $^1\text{H}$ is $\frac{10}{100} \times 75 = 7.5\text{ kg}$

$^1\text{H}$ atoms are replaced by $^2\text{H}$ atoms. So, mass gain by person = 7.5 kg.

13. The ratio of mass percent of C and H of an organic compound ($\text{C}_x\text{H}_y\text{O}_z$) is 6 : 1. If one molecule of the above compound ($\text{C}_x\text{H}_y\text{O}_z$) contains half as much oxygen as required to burn one molecule of compound $\text{C}_x\text{H}_y$ completely to $\text{CO}_2$ and $\text{H}_2\text{O}$. The empirical formula of compound $\text{C}_x\text{H}_y\text{O}_z$ is: [JEE Main 2018]

(a) $\text{C}_2\text{H}_4\text{O}_3$

(b) $\text{C}_3\text{H}_6\text{O}_3$

(d) $\text{C}_3\text{H}_4\text{O}_2$

Answer: (d)

Explanation:

Element	Relative mass	Relative mole	Simplest whole number ratio
C	6	$\frac{6}{12} = 0.5$	1
H	1	$\frac{1}{1} = 1$	2

So, $x = 1, y = 2$

Equation for combustion of $\text{C}_x\text{H}_y$:

$$\text{C}_x\text{H}_y + \left(x + \frac{y}{4}\right)\text{O}_2 \longrightarrow x\text{CO}_2 + \frac{y}{2}\text{H}_2\text{O}$$

Oxygen atoms required = $2\left(x + \frac{y}{4}\right)$

As per information, $z = x + \frac{y}{4} = 1 + \frac{2}{4} = 1.5$

Molecule can be written as $\text{C}_x\text{H}_y\text{O}_z \implies \text{C}_1\text{H}_2\text{O}_{3/2} \implies \text{C}_2\text{H}_4\text{O}_3$. Wait, let's check calculations for option d vs a. Let's re-verify from given solution text:

Actually, let's use the exact derivation from the solution block provided:

Element C relative mass 6, relative mole $6/12 = 0.5$, ratio 1. Element H relative mass 1, relative mole $1/1 = 1$, ratio 2. So $X = 1, Y = 2$. Combustion: $\text{C}_X\text{H}_Y + (X + Y/4)\text{O}_2 \longrightarrow X\text{CO}_2 + (Y/2)\text{H}_2\text{O}$. Oxygen atoms required = $X + Y/4$. As per info, $Z = X + Y/4 \implies Z = 1 + 1/4 = 1.25$? Wait, the text says: $\implies Z = 1 + 1/4 = 1.5$? Let's stick strictly to the solution text provided in the prompt: $\text{C}_2\text{H}_4\text{O}_3$ or $\text{C}_3\text{H}_4\text{O}_2$ - answer key gives (d).

14. 8 g of $\text{NaOH}$ is dissolved in 18 g of $\text{H}_2\text{O}$. Mole fraction of $\text{NaOH}$ in solution and molality (in $\text{mol kg}^{-1}$) of the solutions respectively are: [JEE Main 2019]

(a) 0.167, 11.11

(b) 0.2, 22.20

(d) 0.167, 22.20

Answer: (a)

Explanation: 8 g $\text{NaOH}$, mol of $\text{NaOH} = \frac{8}{40} = 0.2\text{ mol}$

18 g $\text{H}_2\text{O}$, mol of $\text{H}_2\text{O} = \frac{18}{18} = 1\text{ mol}$

$$\therefore X_{\text{NaOH}} = \frac{0.2}{0.2 + 1} = \frac{0.2}{1.2} = 0.167$$

$$\text{Molality} = \frac{0.2 \times 1000}{18} = 11.11\text{ m}$$

15. The strength of 11.2 volume solution of $\text{H}_2\text{O}_2$ is: [Given that molar mass of $\text{H} = 1\text{ g mol}^{-1}$ and $\text{O} = 16\text{ g mol}^{-1}$] [JEE Main 2019]

(a) 13.6%

(b) 3.4%

(d) 1.7%

Answer: (b)

Explanation: Volume strength = $11.2 \times \text{molarity} = 11.2$

$\implies \text{molarity} = 1\text{ M}$

$\implies \text{strength} = 34\text{ g/L}$

$$\%w/w = \frac{34}{1000} \times 100 = 3.4\%$$

16. A mixture of 100 m mol of $\text{Ca(OH)}_2$ and 2g of sodium sulphate was dissolved in water and the volume was made up to 100 mL. The mass of calcium sulphate formed and the concentration of $\text{OH}^-$ in resulting solution, respectively, are: (Molar mass of $\text{Ca(OH)}_2$, $\text{Na}_2\text{SO}_4$ and $\text{CaSO}_4$ are 74, 143 and 136 $\text{g mol}^{-1}$, respectively; $K_{sp}$ of $\text{Ca(OH)}_2$ is $5.5 \times 10^{-6}$) [JEE Main 2019]

(a) $1.9\text{ g}, 0.14\text{ mol L}^{-1}$

(b) $13.6\text{ g}, 0.14\text{ mol L}^{-1}$

(d) $13.6\text{ g}, 0.28\text{ mol L}^{-1}$

Answer: (c)

Explanation: $\text{Ca(OH)}_2 + \text{Na}_2\text{SO}_4 \longrightarrow \text{CaSO}_4 + 2\text{NaOH}$

$w_{\text{CaSO}_4} = 14 \times 10^{-3} \times 136 = 1.9\text{ gm}$

$$[\text{OH}^-] = \frac{28}{100} = 0.28\text{M}$$

17. 25 ml of the given $\text{HCl}$ solution requires 30 mL of 0.1 M sodium carbonate solution. What is the volume of this $\text{HCl}$ solution required to titrate 30 mL of 0.2 M aqueous $\text{NaOH}$ solution? [JEE Main 2019]

(a) 25 mL

(b) 50 mL

(d) 75 mL

Answer: (a)

Explanation: $\text{HCl}$ with $\text{Na}_2\text{CO}_3$

$\text{Eq. of HCl} = \text{Eq. of Na}_2\text{CO}_3$

$$\frac{25 \times M_1}{1000} = \frac{30 \times 0.1 \times 2}{1000}$$

$$M_1 = \frac{6}{25}$$

$\text{Eq. of HCl} = \text{Eq. of NaOH}$

$$\frac{6}{25} \times \frac{V}{1000} = \frac{30 \times 1 \times 0.2}{1000}$$

$$V = 25\text{ mL}$$

18. Freezing point of a 4% aqueous solution of X is equal to freezing point of 12% aqueous solution of Y. If molecular weight of X is A, then molecular weight of Y is: [JEE Main 2019]

(a) A

(b) 3A

(d) 2A

Answer: (b)

Explanation: For same freezing point, molality of both solution should be same.

$$m_x = m_y$$

$$\frac{4 \times 100}{M_x \times 96} = \frac{12 \times 100}{M_y \times 88}$$

$$M_y = \frac{12 \times 96}{4 \times 88} M_x = 3.27 A \approx 3A$$

19. The volume strength of $1\text{M H}_2\text{O}_2$ is: (Molar mass of $\text{H}_2\text{O}_2 = 34\text{ g mol}^{-1}$) [JEE Main 2019]

(a) 16.8

(b) 11.35

(d) 5.6

Answer: (b)

Explanation: $1\text{L} - 1\text{M H}_2\text{O}_2$ solution will produce $11.35\text{ L O}_2$ gas at STP.

Integer and Subjective

20. Find the molarity of water. Given: $\rho = 1000\text{ kg/m}^3$

Answer: 55.56

Explanation: Mass of $1.0\text{ L water} = 1000\text{ g}$

$$\text{Molarity} = \frac{1000}{18} \times \frac{1}{1} \approx 55.56\text{ mol L}^{-1}$$

21. In a solution 100 mL, 0.5 M acetic acid, one gram of active charcoal is added, which adsorbs acetic acid. It is found that the concentration of acetic acid becomes 0.49 M. If surface area of charcoal is $3.01 \times 10^2\text{ m}^2$, calculate the area in $10^{-19}\text{ m}^2$ occupied by single acetic acid molecule on surface of charcoal.

Answer: 5

Explanation: Initial millimole of $\text{CH}_3\text{COOH} = 100 \times 0.5 = 50$

Millimole of $\text{CH}_3\text{COOH}$ remaining after adsorption $= 100 \times 0.49 = 49$

$\Rightarrow$ millimole of $\text{CH}_3\text{COOH}$ adsorbed $= 50 - 49 = 1$

$\Rightarrow$ Number of molecules of $\text{CH}_3\text{COOH}$ adsorbed $= \frac{1}{1000} \times 6.023 \times 10^{23} = 6.023 \times 10^{20}$

$\Rightarrow$ Area covered up by one molecule $= \frac{3.01 \times 10^2}{6.02 \times 10^{20}} = 5 \times 10^{-19}\text{ m}^2$

22. 20% surface sites have adsorbed $\text{N}_2$. On heating $\text{N}_2$ gas evolved from sites and were collected at 0.001 atm and 298 K in a container of volume is $2.46\text{ cm}^3$. Density of surface sites is $6.023 \times 10^{14}/\text{cm}^2$ and surface area is $1000\text{ cm}^2$, find out the number of surface sites occupied per molecule of $\text{N}_2$.

Answer: 2

Explanation: Partial pressure of $\text{N}_2 = 0.001\text{ atm}$, $T = 298\text{ K}$, $V = 2.46\text{ dm}^3$. From ideal gas law: $pV = nRT$

$$n(\text{N}_2) = \frac{pV}{RT} = \frac{0.001 \times 2.46}{0.082 \times 298} = 10^{-7}$$

$\Rightarrow$ No. of molecules of $\text{N}_2 = 10^{-7} \times 6.023 \times 10^{23} = 6.023 \times 10^{16}$

Now, total surface sites available $= 6.023 \times 10^{14} \times 1000 = 6.023 \times 10^{17}$

Surface sites used in adsorption $= \frac{20}{100} \times 6.023 \times 10^{17} = 6.023 \times 10^{16}$

$\Rightarrow$ Sites occupied per molecules $= \frac{\text{Number of sites}}{\text{Number of molecules}} = \frac{6.023 \times 10^{16}}{6.023 \times 10^{16}} = 2$

23. Galena (an ore) is partially oxidised by passing air through it at high temperature. After some time, the passage of air is stopped, but the heating is continued in a closed furnace such that the contents undergo selfreduction. The weight (in kg) of Pb produced per kg of $\text{O}_2$ consumed is ______. (Atomic weights in $\text{g mol}^{-1}$: $\text{O} = 16, \text{S} = 32, \text{Pb} = 207$) [JEE Adv. 2018 P-II]

Answer: 6.47

Explanation:

$$2\text{PbS} + 3\text{O}_2 \longrightarrow 2\text{PbO} + 2\text{SO}_2$$

$$2\text{PbO} + \text{PbS} \longrightarrow 3\text{Pb} + \text{SO}_2$$

3 moles of $\text{O}_2$ produce 3 moles of lead

96 kg of oxygen produce 621 kg of lead.

1 kg of oxygen produce $\frac{621}{96} = 6.468 \approx 6.47\text{ kg}$

24. To measure the quantity of $\text{MnCl}_2$ dissolved in an aqueous solution, it was completely converted to $\text{KMnO}_4$ using the reaction, $\text{MnCl}_2 + \text{K}_2\text{S}_2\text{O}_8 + \text{H}_2\text{O} \longrightarrow \text{KMnO}_4 + \text{H}_2\text{SO}_4 + \text{HCl}$ (equation not balanced). Few drops of concentrated $\text{HCl}$ were added to this solution and gently warmed. Further, oxalic acid (225 mg) was added in portions till the colour of the permanganate ion disappeared. The quantity of $\text{MnCl}_2$ (in mg) present in the initial solution is _____. [JEE Adv. 2018 P-II] (Atomic weights in $\text{g mol}^{-1}$: $\text{Mn} = 55, \text{Cl} = 35.5$)

Answer: 126

Explanation: From POAC,

m moles of $\text{MnCl}_2 = \text{m moles of KMnO}_4 = x$ (let) and, meq of $\text{KMnO}_4 = \text{meq of oxalic acid}$

$$x \times 5 = \frac{225}{90} \times 2$$

$$x = 1$$

$$\therefore \text{m moles of MnCl}_2 = 1 \implies \text{mg of MnCl}_2 = 1 \times (55 + 71) = 126\text{ mg}$$

25. On dissolving 0.5 g of non-volatile, non-ionic to 39 g of benzene, its vapour pressure decreases from 650 mm of Hg. The depression of freezing point of benzene (in K) upon addition of the solute is ______. [JEE Adv. 2019 P-II]

Answer: 1.02

Explanation:

$$\frac{P^o - P_s}{P_s} = \frac{n_{\text{solute}}}{n_{\text{solvent}}}$$

$$\frac{650 - 640}{640} = \frac{0.5 / M}{39 / 78}$$

$\implies M = 64\text{ gm solute}$

$$\Delta T_f = \frac{0.5 \times 78}{64 \times 39} \times 5.12 = 1.02\text{ K}$$

26. 1 mole of Rhombic sulphur is treated with conc. $\text{HNO}_3$. Find the mass of $\text{H}_2\text{O}$ formed. [JEE Adv. 2019 P-II]

Answer: 288

Explanation:

$$\text{S}_8 + 32\text{H}_2\text{O} \longrightarrow 8\text{H}_2\text{SO}_4 + 48\text{H}^+ + 48e^-$$

$$48[\text{HNO}_3 + \text{H}^+ + e^- \longrightarrow \text{NO}_2 + \text{H}_2\text{O}]$$

$$\text{S}_8 + 48\text{HNO}_3 \longrightarrow 8\text{H}_2\text{SO}_4 + 48\text{NO}_2 + 16\text{H}_2\text{O}$$

$\implies \text{Mass of H}_2\text{O}\text{ formed} = (16) \times (18)\text{ g} = 288\text{ g}$

Fill in the Blanks

27. The value of n in the molecular formula $\text{Be}_3\text{Al}_2\text{Si}_6\text{O}_n$ is …………… [JEE 2010]

Answer: 18

Explanation: $\text{Be}_3\text{Al}_2\text{Si}_6\text{O}_n \implies (3 \times 2) + (2 \times 3) + (6 \times 4) + (n \times -2) = 0$

Or $6 + 6 + 24 - 2n = 0 \implies 2n = 36 \implies n = 18$

28. A student performs a titration with different burettes and finds titre values of 25.2 mL, 25.25 mL, and 25.0 mL. The number of significant figures in the average titre value is …………… [JEE 2010]

Answer: 3

Explanation: A student performs a titration with different burettes and finds titre values of 25.2 mL, 25.25 mL and 25.0 mL. The number of significant figures in the average titre value is 3. This is according to the rules of significant figures.

29. Silver (atomic weight $108\text{ g mol}^{-1}$) has a density of $10.5\text{ g cm}^{-3}$. The number of silver atoms on a surface of area $10^{-12}\text{ m}^2$ can be expressed in scientific notation as $Y \times 10^{-x}$. The value of x is …………… [JEE 2010]

Answer: 7

Explanation: Density $= 10.5\text{ g cm}^{-3}$. This means 10.5 g silver is present in $1\text{ cm}^3$. $\frac{10.5}{108}$ mol silver is present in $1\text{ cm}^3$. On solving, we get $1.5 \times 10^{-7} = Y \times 10^{-x} \therefore x = 7$.

30. Among the following, the number of elements showing only one non-zero oxidation state is…………… O, Cl, F, N, P, Sn, Tl, Na, Ti [JEE 2010]

Answer: 2

Explanation: $\text{Na}$ shows only $+1$. $\text{F}$ shows only $-1$. So, the number of elements showing only one non-zero oxidation state is 2.

31. The total number of electrons present in 18 mL of water is …………… [JEE 2010]

Answer: $6.023 \times 10^{24}$

Explanation: Number of electron in one molecule of $\text{H}_2\text{O}$ is $2 + 8 = 10$. $\text{Density} = 1$, 18 mL means 18 g. $\text{Moles} = \frac{18}{18} = 1\text{ molecule} = 6.023 \times 10^{23}$. $\text{Electrons} = 6.023 \times 10^{23} \times 10 = 6.023 \times 10^{24}$.