NCERT Solutions for Class 9 Science Chapter 3 - Atoms And Molecules

Answer:

$$\begin{gathered} \left(\mathrm{i}\right) 2 \mathrm{moles} \mathrm{of} \mathrm{water} = 18\times 2 = 36 \mathrm{grams} \mathrm{of} \mathrm{water} \\ \left(\mathrm{ii}\right) 20 \mathrm{moles} \mathrm{of} \mathrm{water} = 18\times 20 = 360 \mathrm{grams} \mathrm{of} \mathrm{water} \\ \left(\mathrm{iii}\right) 6.022\times {10}^{23} \mathrm{molecules} \mathrm{of} \mathrm{water} = 1 \mathrm{mole} \mathrm{of} \mathrm{water} = 18 \mathrm{grams} \mathrm{of} \mathrm{water} \\ \left(\mathrm{iv}\right) 1.2044 \times {10}^{25} \mathrm{molecules} \mathrm{of} \mathrm{water} = \frac{1.2044\times {10}^{25}}{6.022\times {10}^{23}}=20 \mathrm{moles} \mathrm{of} \mathrm{water} = 360 \mathrm{grams} \mathrm{of} \mathrm{water} \end{gathered}$$
Hence, the correct answer is option d.

Answer:

Atoms of inert gases like neon and argon exist independently.
Hence, the correct answer is option a.

Answer:

Nitrogen exists as a diatomic molecule and the symbol is N₂.
Hence, the correct answer is option b.

Answer:

The symbol for sodium is Na.
Hence, the correct answer is option d.
 

Answer:

$\mathrm{Mass}=\mathrm{number} \mathrm{of} \mathrm{moles}\times \mathrm{molar} \mathrm{mass}$
$$\begin{gathered} \left(\mathrm{a}\right) 0.2 \mathrm{moles} \mathrm{of} \mathrm{sucrose} = 0.2 \times 342.3 \mathrm{grams}=68.46 \mathrm{grams} \\ \left(\mathrm{b}\right) 2 \mathrm{moles} \mathrm{of} {\mathrm{CO}}_2 = 2\times 44= 88 \mathrm{grams} \\ \left(\mathrm{c}\right) 2 \mathrm{moles} \mathrm{of} {\mathrm{CaCO}}_3= 2\times 100 = 200 \mathrm{grams} \\ \left(\mathrm{d}\right) 10 \mathrm{moles} \mathrm{of} {\mathrm{H}}_2\mathrm{O} = 10\times 18 =180 \mathrm{grams} \end{gathered}$$
Hence, the correct answer is option c.

Answer:

$$\begin{gathered} \mathrm{Number} \mathrm{of} \mathrm{atoms} = \frac{\mathrm{mass} \mathrm{of} \mathrm{the} \mathrm{substance} \times \mathrm{Number} \mathrm{of} \mathrm{atoms} \mathrm{in} \mathrm{a} \mathrm{molecule} \times {\mathrm{N}}_{\mathrm{a}}}{\mathrm{Molar} \mathrm{Mass}} \\ \left(\mathrm{a}\right) 18 \mathrm{grams} \mathrm{of} \mathrm{Water} = 1 \mathrm{mole} = 6.022 \times {10}^{23} \mathrm{molecules} = 3\times 6.022 \times {10}^{23} = 1.8066\times {10}^{24} \mathrm{atoms} \\ \left(\mathrm{b}\right) 18 \mathrm{grams} \mathrm{of} \mathrm{Oxygen} = \frac{18\times 2 }{32}\times 6.022 \times {10}^{23} =6.77\times {10}^{23} \mathrm{atoms} \\ \left(\mathrm{c}\right) 18 \mathrm{grams} \mathrm{of} {\mathrm{CO}}_2 =\frac{18\times 3}{44}\times 6.022 \times {10}^{23}= 7.39 \times {10}^{23} \mathrm{atoms} \\ \left(\mathrm{d}\right) 18 \mathrm{grams} \mathrm{of} {\mathrm{CH}}_4 = \frac{18\times 5}{16}\times 6.022 \times {10}^{23}=3.38 \times {10}^{24} \mathrm{atoms} \end{gathered}$$
Hence, the correct answer is option d.
 

Answer:

$$\begin{gathered} \left(\mathrm{a}\right) 1 \mathrm{gram} \mathrm{of} {\mathrm{CO}}_2 = \frac{1}{44}\times 6.022 \times {10}^{23} = 1.368 \times {10}^{23} \mathrm{molecules} \\ \left(\mathrm{b}\right) 1 \mathrm{gram} \mathrm{of} {\mathrm{N}}_2= \frac{1}{28}\times 6.022 \times {10}^{23}= 2.15\times {10}^{22} \mathrm{molecules} \\ \left(\mathrm{c}\right) 1 \mathrm{gram} \mathrm{of} {\mathrm{H}}_2 = \frac{1}{2}\times 6.022 \times {10}^{23}= 3.011\times {10}^{23 }\mathrm{molecules} \\ \left(\mathrm{d}\right) 1 \mathrm{gram} \mathrm{of} {\mathrm{CH}}_4 = \frac{1}{16}\times 6.022 \times {10}^{23}= 3.76 \times {10}^{22 }\mathrm{molecules} \end{gathered}$$
Hence, the correct answer is option c.

Answer:

$\mathrm{Mass} \mathrm{of} \mathrm{one} \mathrm{atom} \mathrm{of} \mathrm{Oxygen} = \frac{\mathrm{Atomic} \mathrm{mass}}{6.022\times {10}^{23}}= \frac{16}{6.022\times {10}^{23}}$
Hence, the correct answer is option a.

Answer:

$$\begin{gathered} \mathrm{Moles} \mathrm{of} \mathrm{sucrose} = \frac{3.42}{342.3}=0.01 \mathrm{mol} \\ 1 \mathrm{mole} \mathrm{of} \mathrm{sucrose} \mathrm{has} 11\times 6.022\times {10}^{23} \mathrm{oxygen} \mathrm{atoms} \\ 0.01 \mathrm{mole} \mathrm{of} \mathrm{sucrose} \mathrm{will} \mathrm{have} 0.01\times 11\times 6.022\times {10}^{23} = 0.11\times 6.022\times {10}^{23} \mathrm{oxygen} \mathrm{atoms} \\ \mathrm{Number} \mathrm{of} \mathrm{moles} \mathrm{of} \mathrm{water} = \frac{18}{18}= 1 \mathrm{mol} \\ 1 \mathrm{mol} \mathrm{of} \mathrm{water} \mathrm{has} 1\times 6.022\times {10}^{23} \mathrm{Oxygen} \mathrm{atoms} \\ \mathrm{Total} \mathrm{oxygen} \mathrm{atoms} = 0.11\times 6.022\times {10}^{23} + 1\times 6.022\times {10}^{23} = 1.11\times 6.022\times {10}^{23} = 6.68 11\times {10}^{23} \mathrm{atoms} \end{gathered}$$
Hence, the correct answer is option a.

Answer:

A change in the physical state can be brought about when energy is either given to or taken out of the system. For example, during boiling of water heat energy is given out whereas in the case of melting of ice heat is taken by the system. The change in the state of matter occurs because of energy exchange between the system and the surroundings.

Answer:

(a) Valency of Ca is 2 and Cl is 1. The correct formula is CaCl₂.
(b) Valency of Bi is 3 and PO₄ is 3. The correct formula is BiPO₄.
(c) Valency of Na is 1 and SO4 is 2. The correct formula is Na₂SO₄.
(d) Valency of Na is 1 and S is 2. The correct formula is Na₂S.
Hence, the correct answer is option b.
 

Answer:

(a) CuBr₂
(b) Al(NO₃)₃
(c) Ca₃(PO₄)₂
(d) Fe₂S₃
(e) HgCl₂
(f) Mg(CH₃COO)₂
 

Answer:

Cations: Cu²⁺,  Na⁺, Fe³⁺
Anions: Cl^-, SO₄^2-, PO₄^3-

CuCl₂, CuSO₄, Cu₃(PO₄)₂
NaCl, Na₂SO₄, Na₃PO₄
FeCl₃, Fe₂(SO₄)₃, FePO₄

Answer:

(a) CH3COO^- and Na⁺
(b) Na⁺ and Cl^-
(c) No ions are present in H₂
(d) NH₄⁺ and NO₃^-

Answer:

(a) CaF₂
(b) H₂S
(c) NH₃
(d) CCl₄
(e) Na₂O
(f) CO
(g) CO₂

Answer:

(a) Cobalt Co
(b) Carbon C
(c) Aluminium Al
(d) Helium He
(e) Sodium Na

Answer:

Answer:

 $\begin{array}{rcl}\left(\mathrm{a}\right) {{\mathrm{CO}}_3}^{2-} & =& 1 \mathrm{atom} \mathrm{of} \mathrm{C}+\hspace{0.17em}3 \mathrm{atoms} \mathrm{of} \mathrm{O}\\ & =& 4 \mathrm{atoms}\\ \left(\mathrm{b}\right) {{\mathrm{PO}}_4}^{3-} & =& 1 \mathrm{atom} \mathrm{of} \mathrm{P} +4 \mathrm{atoms} \mathrm{of} \mathrm{O}\\ & =& 5 \mathrm{atoms}\\ \left(\mathrm{c}\right) {\mathrm{P}}_2{\mathrm{O}}_5 & =& 2 \mathrm{atoms} \mathrm{of} \mathrm{P} + 5 \mathrm{atoms} \mathrm{of} \mathrm{O}\\ & =& 7 \mathrm{atoms}\\ \left(\mathrm{d}\right) \mathrm{CO} & =& 1 \mathrm{atom} \mathrm{of} \mathrm{C}\hspace{0.17em}+ 1 \mathrm{atom} \mathrm{of} \mathrm{O}\\ & =& 2 \mathrm{atoms}\end{array}$
 

Answer:

$$\begin{gathered} \mathrm{Mass} \mathrm{of} \mathrm{one} \mathrm{mole}\left({\mathrm{N}}_{\mathrm{A}}\right)\mathrm{of} \mathrm{neutrons} = 1 \mathrm{gram} \\ \mathrm{Mass} \mathrm{of} 1 \mathrm{neutron} = \frac{1}{{\mathrm{N}}_{\mathrm{A}}} \mathrm{grams} \\ \mathrm{Mass} \mathrm{of} 1 \mathrm{mole} \mathrm{of} {\mathrm{H}}_2\mathrm{O} = 18 \mathrm{grams} \\ \mathrm{Mass} \mathrm{of} \mathrm{one} \mathrm{molecule} \mathrm{of} \mathrm{water} = \frac{18}{{\mathrm{N}}_{\mathrm{A}}} \mathrm{grams} \\ \mathrm{Number} \mathrm{of} \mathrm{neutrons} \mathrm{in} \mathrm{one} \mathrm{atom} \mathrm{of} \mathrm{H} = 0 \\ \mathrm{Number} \mathrm{of} \mathrm{neutrons} \mathrm{in} \mathrm{one} \mathrm{atom} \mathrm{of} \mathrm{O} = 8 \\ \mathrm{Mass} \mathrm{of} 8 \mathrm{neutrons} (\mathrm{in} {\mathrm{H}}_2\mathrm{O}) = \frac{8}{{\mathrm{N}}_{\mathrm{A}}} \\ \mathrm{Fraction} \mathrm{of} \mathrm{mass} \mathrm{of} \mathrm{water} \mathrm{due} \mathrm{to} \mathrm{neutrons} =\frac{8}{{\mathrm{N}}_{\mathrm{A}}}/\frac{18}{{\mathrm{N}}_{\mathrm{A}}} = \frac{8}{18}=\frac{4}{9} \end{gathered}$$

Answer:

Yes, the solubility of a substance changes with temperature. It generally increases with temperature. For example, more sugar can be dissolved in hot water as compared to cold water.

Answer:

Monoatomic: He, Ag
Diatomic: F₂, HCl
Polyatomic with atomicity > 2
Atomicity (3) = NO₂, N₂O, O₃
Atomicity (4) = P₄, H₂O₂
Atomicity (5) = CH₄
Atomicity (8) = C₂H₆
Atomicity (14) = P₄O₁₀
 

Answer:

Salt and sugar can be differentiated from each other, by heating or dissolving in water.
(i) Sugar powder is charred to black on heating, whereas salt do not change.
(ii) A salt solution conducts electricity due to the formation of ions in the solution, whereas a sugar solution does not do so.
 

Answer:

The atomic mass of Mg = 24 g/mol
24 grams of Magnesium = 1 mol
$\begin{array}{rcl}\mathrm{Number} \mathrm{of} \mathrm{moles} \mathrm{in} 12\mathrm{g} \mathrm{of} \mathrm{Mg} & =& \frac{\mathrm{Given} \mathrm{mass}}{\mathrm{Molar} \mathrm{mass}}\\ & =& \frac{12}{24}\\ & =& 0.5 \mathrm{mol}\end{array}$

Answer:

$$\begin{gathered} \left(\mathrm{a}\right) \mathrm{Mass} \mathrm{of} 1 \mathrm{mole} \mathrm{of} {\mathrm{CO}}_2 = 44 \mathrm{grams} \\ \mathrm{Mass} \mathrm{of} 5 \mathrm{moles} \mathrm{of} {\mathrm{CO}}_2 = 44\times 5 = 220 \mathrm{grams} \\ \mathrm{Mass} \mathrm{of} 1 \mathrm{mole} \mathrm{of} {\mathrm{H}}_2\mathrm{O} = 18 \mathrm{grams} \\ \mathrm{Mass} \mathrm{of} 5 \mathrm{moles} \mathrm{of} {\mathrm{H}}_2\mathrm{O} = 18\times 5 = 90 \mathrm{grams} \\ \left(\mathrm{b}\right) \mathrm{Molar} \mathrm{mass} \mathrm{of} \mathrm{Ca} = 40 \mathrm{grams} = 1 \mathrm{mole} \\ \mathrm{Number} \mathrm{of} \mathrm{moles} \mathrm{in} 240 \mathrm{grams} \mathrm{of} \mathrm{Ca} = \frac{240}{40}=6 \\ \mathrm{Molar} \mathrm{mass} \mathrm{of} \mathrm{Mg} = 24 \mathrm{grams} = 1 \mathrm{mole} \\ \mathrm{Number} \mathrm{of} \mathrm{moles} \mathrm{in} 240 \mathrm{grams} \mathrm{of} \mathrm{Mg} =\frac{240}{24}=10 \\ \mathrm{Mole} \mathrm{ratio} \mathrm{of} \mathrm{Ca} \mathrm{and} \mathrm{Mg} = 6:10 = 3:5 \end{gathered}$$

Answer:

$$\begin{gathered} \left(\mathrm{a}\right) \mathrm{Ca} : \mathrm{C} : \mathrm{O} = 40 : 12 : 16\times 3 = 40 : 12 : 48 = 10 : 3 :12 \\ \left(\mathrm{b}\right) \mathrm{Mg}: \mathrm{Cl} = 24 : 35.5\times 2 = 24: 71 \\ \left(\mathrm{c}\right) \mathrm{H}:\mathrm{S}:\mathrm{O} = 1\times 2: 32: 4\times 16 = 2:32: 64 = 1:16: 32 \\ \left(\mathrm{d}\right) \mathrm{C}:\mathrm{H}:\mathrm{O} = 12:1\times 6:16 = 12:6:16 = 6:3:8 \\ \left(\mathrm{e}\right) \mathrm{N}:\mathrm{H} = 14:3 \\ \left(\mathrm{f}\right) \mathrm{Ca}:\mathrm{O}:\mathrm{H} = 40 : 32: 2 = 20:16:1 \end{gathered}$$

Answer:

CaCl₂ (aq) → Ca²⁺ (aq) + 2Cl^– (aq)
$$\begin{gathered} 111 \mathrm{grams} (1 \mathrm{mole}) \to 1 \mathrm{mole} + 2 \mathrm{moles} \\ 222 \mathrm{grams} (2 \mathrm{moles}) \to 2 \mathrm{moles} + 4 \mathrm{moles} \\ \mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{ions} \mathrm{given} \mathrm{by} 2 \mathrm{moles} \mathrm{of} {\mathrm{CaCl}}_2 =6 \\ \mathrm{Number} \mathrm{of} \mathrm{ions} = \mathrm{Number} \mathrm{of} \mathrm{moles} \mathrm{of} \mathrm{ions} \times {\mathrm{N}}_{\mathrm{A}} = 6 \times 6.022\times {10}^{23} = 3.6132 \times {10}^{24} \mathrm{ions} \end{gathered}$$

Answer:

Number of electrons in Na atom = 11 
Number of electrons in Na⁺ = 10
The difference of the number of electrons for 1 mole of Na atom and Na⁺ = 1 mole
For 100 moles the difference will be 100 moles of electrons
Mass of 100 moles of electrons= 5.48002 g
$$\begin{gathered} \mathrm{Mass} \mathrm{of} 1 \mathrm{mole} \mathrm{of} \mathrm{electron} =\frac{5.48002}{100}\mathrm{g} \\ \mathrm{Mass} \mathrm{of} \mathrm{one} \mathrm{electron} = \frac{5.48002}{100\times 6.022\times {10}^{23}} = 9.1 \times {10}^{-28} \mathrm{g} = 9.1 \times {10}^{-31 }\mathrm{kg} \end{gathered}$$

Answer:

The molar mass of HgS = 200.6 + 32 = 232.6 grams
Mass of Hg in 232.6 of HgS = 200.6 grams
$\mathrm{Mass} \mathrm{of} \mathrm{Hg} \mathrm{in} 225 \mathrm{grams} \mathrm{of} \mathrm{HgS} =\frac{200.6}{232.6}\times 225 = 194.04 \mathrm{grams}$
 

Answer:

Mass of 1 screws = 4.11 grams
$$\begin{gathered} \mathrm{Mass} \mathrm{of} 1 \mathrm{mole} \mathrm{of} \mathrm{screws} = 4.11 \times 6.022\times {10}^{23} = 2.475 \times {10}^{24} \mathrm{grams} = 2.475 \times {10}^{21} \mathrm{kg} \\ \frac{\mathrm{Mass} \mathrm{of} \mathrm{earth}}{\mathrm{Mass} \mathrm{of} 1 \mathrm{mole} \mathrm{screws}}= \frac{5.98 \times {10}^{24} \mathrm{kg}}{2.475 \times {10}^{21} \mathrm{kg}} = 2416 \\ \mathrm{It} \mathrm{shows} \mathrm{that} \mathrm{earth} \mathrm{is} 2416 \mathrm{times} \mathrm{heavier} \mathrm{than} \mathrm{one} \mathrm{mole} \mathrm{of} \mathrm{screws}. \end{gathered}$$

Answer:

$$\begin{gathered} \mathrm{Number} \mathrm{of} \mathrm{oxygen} \mathrm{atoms} \mathrm{in} \mathrm{the} \mathrm{sample} = 2.58 \times {10}^{24} \\ 6.022\times {10}^{23} \mathrm{atoms} \mathrm{of} \mathrm{oxygen} = 1 \mathrm{mole} \\ 2.58 \times {10}^{24} \mathrm{atoms} \mathrm{of} \mathrm{oxygen} =\frac{2.58\times {10}^{24}}{6.022 \times {10}^{23}}\mathrm{moles} = 4.28 \mathrm{moles} \end{gathered}$$

Answer:

$$\begin{gathered} \left(\mathrm{a}\right) \mathrm{Mass} \mathrm{of} \mathrm{container} \mathrm{containing} 5 \mathrm{moles} \mathrm{of} \mathrm{C} \mathrm{atoms} = 5\times 12 = 60 \mathrm{g} \\ \mathrm{Mass} \mathrm{of} \mathrm{container} \mathrm{containing} 5 \mathrm{moles} \mathrm{of} \mathrm{Na} \mathrm{atoms} = 5 \times 23 = 115 \mathrm{g} \\ \mathrm{Hence}, \mathrm{container} \mathrm{of} \mathrm{krish} \mathrm{is} \mathrm{heavier}. \\ \left(\mathrm{b}\right) \mathrm{Both} \mathrm{containers} \mathrm{have} \mathrm{same} \mathrm{number} \mathrm{of} \mathrm{atoms} \mathrm{since} \mathrm{they} \mathrm{contain} \mathrm{same} \mathrm{number} \\ \mathrm{of} \mathrm{moles}. \end{gathered}$$

Answer:

Answer:

$$\begin{gathered} 1 \mathrm{Mole} \mathrm{of} \mathrm{stars} = 6.022 \times {10}^{23} \mathrm{stars} \\ \mathrm{Number} \mathrm{of} \mathrm{moles} \mathrm{of} \mathrm{stars} = \frac{{10}^{22}}{6.023 \times {10}^{23}} \\ = 0.0166 \mathrm{moles} \end{gathered}$$

Answer:

$$\begin{gathered} \left(\mathrm{a}\right) {10}^3 = \mathrm{kilo} \\ \left(\mathrm{b}\right) {10}^{-1} = \mathrm{deci} \\ \left(\mathrm{c}\right) {10}^{-2} = \mathrm{centi} \\ \left(\mathrm{d}\right) {10}^{-6} = \mathrm{micro} \\ \left(\mathrm{e}\right) {10}^{-9} = \mathrm{nano} \\ \left(\mathrm{f}\right) {10}^{-12} = \mathrm{pico} \end{gathered}$$

Answer:

$$\begin{gathered} \left(\mathrm{a}\right) 5.84 \times {10}^{-3}\mathrm{mg} \\ =\frac{ 5.84 \times {10}^{-3}}{{10}^3\times {10}^3}\mathrm{kg} = 5.84 \times {10}^{-9} \mathrm{kg} \\ \left(\mathrm{b}\right) 58.34 \mathrm{g} \\ = \frac{58.34}{{10}^3} = 5.834 \times {10}^{-2} \mathrm{kg} \\ \left(\mathrm{c}\right) 0.584 \mathrm{g} \\ =\frac{0.584}{{10}^3} \mathrm{kg} = 5.84 \times {10}^{-4} \mathrm{kg} \\ \left(\mathrm{d}\right) 5.873 \times {10}^{-21} \mathrm{g} \\ = \frac{5.873 \times {10}^{-21}}{{10}^3} = 5.873 \times {10}^{-24} \mathrm{kg} \end{gathered}$$

Answer:

$$\begin{gathered} {\mathrm{Mg}}^{2+}\mathrm{ion} = 10 \mathrm{electrons} \\ \mathrm{Mg} \mathrm{atom} = 12 \mathrm{electrons} \\ \mathrm{Difference} \mathrm{between} {\mathrm{Mg}}^{2+} \mathrm{and} \mathrm{Mg} = 2 \mathrm{moles} \mathrm{of} \mathrm{electrons} \\ \mathrm{The} \mathrm{difference} \mathrm{between} {10}^3 \mathrm{moles} \mathrm{of} \mathrm{Mg} \mathrm{and} {10}^3 \mathrm{moles} \mathrm{of} {\mathrm{Mg}}^{2+} = {10}^3 \times 2 \mathrm{moles} \mathrm{of} \mathrm{electrons} \\ \mathrm{Mass} 0\mathrm{f} {10}^3 \times 2 \mathrm{moles} \mathrm{of} \mathrm{eletrons} = 2 \times {10}^3 \times 6.022 \times {10}^{23} \times 9.1 \times {10}^{-31} \mathrm{kg} = 1.096 \times {10}^{-3} \mathrm{kg} \end{gathered}$$

Answer:

$$\begin{gathered} \left(\mathrm{i}\right) 100 \mathrm{grams} \mathrm{of} {\mathrm{N}}_2 = \frac{100}{28} \mathrm{moles} \\ \mathrm{Number} \mathrm{of} \mathrm{molecules} = \frac{100}{28}\times 6.022 \times {10}^{23} \mathrm{molecules} \\ \mathrm{Number} \mathrm{of} \mathrm{atoms} = \frac{2\times 100}{28}\times 6.022 \times {10}^{23} \mathrm{atoms} = 43.01 \times {10}^{23 } \mathrm{atoms} \\ \left(\mathrm{ii}\right) 100 \mathrm{grams} \mathrm{of} {\mathrm{NH}}_3 = \frac{100}{17} \mathrm{moles} \\ \mathrm{Number} \mathrm{of} \mathrm{molecules} = \frac{100}{17}\times 6.022 \times {10}^{23} \mathrm{molecules} \\ \mathrm{Number} \mathrm{of} \mathrm{atoms} = \frac{100}{17}\times 6.022\times {10}^{23}\times 4 \mathrm{atoms} = 141.69 \times {10}^{23} \mathrm{atoms} \\ \mathrm{Thus},{\mathrm{NH}}_3 \mathrm{would} \mathrm{have} \mathrm{more} \mathrm{atoms}. \end{gathered}$$

Answer:

$$\begin{gathered} 1 \mathrm{mole} \mathrm{of} \mathrm{NaCl} = 23 + 35.5 = 58.5 \mathrm{grams} \mathrm{of} \mathrm{NaCl} \\ \mathrm{Number} \mathrm{of} \mathrm{moles} \mathrm{in} 5085 \mathrm{grams} \mathrm{of} \mathrm{NaCl} =\frac{5085}{58.5} = 0.1 \mathrm{moles} \\ \mathrm{Each} \mathrm{NaCl} \mathrm{formula} \mathrm{unit} = {\mathrm{Na}}^{+} + {\mathrm{Cl}}^{-} = 2 \mathrm{ions} \\ \mathrm{Number} \mathrm{of} \mathrm{ions} = 0.2\times 6.022\times {10}^{23} = 1.2044\times {10}^{23} \mathrm{ions} \end{gathered}$$
 

Answer:

$$\begin{gathered} 1 \mathrm{g} \mathrm{of} \mathrm{gold} \mathrm{sample} \mathrm{contains} = \frac{90}{100} = 0.9 \mathrm{g} \mathrm{gold} \\ \mathrm{Number} \mathrm{of} \mathrm{moles} \mathrm{of} \mathrm{gold} = \frac{\mathrm{Mass} \mathrm{of} \mathrm{gold}}{\mathrm{Atomic} \mathrm{mass} \mathrm{of} \mathrm{gold}} = \frac{0.9}{197} = 0.0046 \\ 1 \mathrm{mole} \mathrm{of} \mathrm{gold} \mathrm{contains} = 6.022 \times {10}^{23} \mathrm{atoms} \\ 0.0046 \mathrm{mole} \mathrm{of} \mathrm{gold} = 6.022 \times {10}^{23} \times 0.0046 = 2.77 \times {10}^{21} \mathrm{atoms} \end{gathered}$$

Answer:

An ionic compound is a compound formed due to the transfer of electrons between the reacting species.  An ionic compound contains a cation which is a positive ion and an anion which is a negative ion. For example, sodium chloride is an ionic compound made up of  Na⁺ and Cl^- ions. On the other hand, a molecular compound is a compound that is formed by sharing of electrons. It mainly consists of molecules. For example, ammonia (NH₃), carbon dioxide (CO₂).

Answer:

$$\begin{gathered} \mathrm{Mass} \mathrm{of} \mathrm{Aluminium} = 27 \mathrm{g} {\mathrm{mol}}^{-1} \\ \mathrm{Al}\to {\mathrm{Al}}^{3+}+3{\mathrm{e}}^{-} (3 \mathrm{moles} \mathrm{of} \mathrm{electrons}) \\ \mathrm{Mass} \mathrm{of} 3 \mathrm{moles} \mathrm{of} \mathrm{electrons} = 3 \times (9.1 \times {10}^{-28}) \times 6.022 \times {10}^{23} \mathrm{g} = 0.00164 \mathrm{g} \\ \mathrm{Molar} \mathrm{mass} \mathrm{of} {\mathrm{Al}}^{3+} = 27 - 0.00164 \mathrm{g} {\mathrm{mol}}^{-1} = 26.9984 \mathrm{g} {\mathrm{mol}}^{-1} \\ \therefore \mathrm{Difference} = 27 - 0.00164 \mathrm{g} {\mathrm{mol}}^{-1} = 26.9984 \mathrm{g} {\mathrm{mol}}^{-1} \end{gathered}$$

Answer:

$$\begin{gathered} \mathrm{Mass} \mathrm{of} \mathrm{gold} = \frac{\mathrm{m}}{100} \mathrm{grams} \\ \mathrm{Number} \mathrm{of} \mathrm{atoms} \mathrm{of} \mathrm{silver} = \frac{\mathrm{Mass}}{\mathrm{Atomic} \mathrm{mass}}\times {\mathrm{N}}_{\mathrm{A}} = \frac{\mathrm{m}}{108}\times {\mathrm{N}}_{\mathrm{A}} \\ \mathrm{Number} \mathrm{of} \mathrm{atoms} \mathrm{of} \mathrm{gold} = \frac{\mathrm{m}}{100\times 197}\times {\mathrm{N}}_{\mathrm{A}} \\ \mathrm{Ratio} \mathrm{of} \mathrm{number} \mathrm{of} \mathrm{atoms} \mathrm{of} \mathrm{gold} \mathrm{to} \mathrm{silver} = \mathrm{Au}:\mathrm{Ag} = \frac{\mathrm{m}}{100\times 197}\times {\mathrm{N}}_{\mathrm{A}} : \frac{\mathrm{m}}{108}\times {\mathrm{N}}_{\mathrm{A}} \\ =108:100\times 197 \\ = 108: 19700 \\ = 1:182.41 \end{gathered}$$

Answer:

$$\begin{gathered} \mathrm{Molar} \mathrm{mass} \mathrm{of} {\mathrm{CH}}_4 = 16 \mathrm{grams} = 6.022\times {10}^{23} \mathrm{molecules} \\ \mathrm{Thus}, 6.022 \times {10}^{23} \mathrm{molecules} \mathrm{of} {\mathrm{CH}}_4 \mathrm{will} \mathrm{have} \mathrm{mass} = 16 \mathrm{grams} \\ \mathrm{Therefore}, 1.5\times {10}^{20} \mathrm{molecules} \mathrm{of} {\mathrm{CH}}_4 \mathrm{will} \mathrm{have} \mathrm{mass} = \frac{16}{6.022\times {10}^{23}}\times 1.5\times {10}^{20} \\ =4\times {10}^{-3 } \mathrm{grams} \\ \mathrm{Molar} \mathrm{mass} \mathrm{of} {\mathrm{C}}_2{\mathrm{H}}_6 = 30 \mathrm{g}/\mathrm{mol} = 6.022 \times {10}^{23 }\mathrm{molecules} \\ \mathrm{Thus}, 30 \mathrm{grams} \mathrm{of} {\mathrm{C}}_2{\mathrm{H}}_6 \mathrm{have} \mathrm{molecules} = 6.022\times {10}^{23 } \\ \mathrm{Therefore}, 4\times {10}^{-3} \mathrm{grams} \mathrm{of} {\mathrm{C}}_2{\mathrm{H}}_6 \mathrm{will} \mathrm{have} \mathrm{molecules} =\frac{6.022\times {10}^{23}}{30}\times 4\times {10}^{-3} \\ = 0.803 \times {10}^{20} \\ = 8.03 \times {10}^{19} \end{gathered}$$

Answer:

(a) Law of conservation of mass
(b) Polyatomic ion
$$\begin{gathered} \left(\mathrm{c}\right)\left(3\times 40 + 2\times 31 + 8\times 16 \right) \\ = 120 + 62 + 128 \\ = 310 \mathrm{u} \end{gathered}$$
(d) Na₂CO₃, (NH₄)₂SO₄

Answer:

1. Silver
2. Gold
3. Iron
4. Copper
5. Phosphorous
6. Mercury
7. Hydrogen
8. Lead

Answer:

ure(a) 1. Chlorine

Answer:

(a) Caustic Potash, KOH = (39+16+1) = 56 g/mol
(b) Baking powder, NaHCO₃ = (23 +1+12+48)= 84 g/mol 
(c) Limestone, CaCO₃ = (40+12+48) =100 g/mol
(d) Caustic soda, NaOH = (23+16+1) = 40 g/mol
(e) Ethanol, C₂H₅OH = (24 + 6 + 16) = 46 g/mol
(f) Common salt, NaCl = (23 + 35.5) = 58.5 g/mol

Answer:

$$\begin{gathered} 6{\mathrm{CO}}_2 + 6{\mathrm{H}}_2\mathrm{O} \underset{\mathrm{Sunlight}}{\overset{\mathrm{Chlorophyll}}{\to }} {\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6 + 6{\mathrm{O}}_2 \\ 1 \mathrm{mole}(180 \mathrm{grams}) \mathrm{of} \mathrm{glucose} \mathrm{needs} 6 \mathrm{moles} (6\times 18 \mathrm{grams}) \mathrm{of} {\mathrm{H}}_2\mathrm{O} \\ 1 \mathrm{gram} \mathrm{of} \mathrm{glucose} \mathrm{will} \mathrm{need} \frac{108}{180} \mathrm{of} {\mathrm{H}}_2\mathrm{O} \\ 18 \mathrm{grams} \mathrm{of} \mathrm{glucose} \mathrm{will} \mathrm{need} \frac{108}{180}\times 18 = 10.8 \mathrm{grams} \\ \mathrm{Volume} \mathrm{of} \mathrm{water} \mathrm{used} = \frac{\mathrm{Mass}}{\mathrm{Density}} =\frac{10.8}{1}= 10.8 {\mathrm{cm}}^3 \end{gathered}$$