Some Basic Concepts of Chemistry

(B) is incorrect because Dalton said atoms are indivisible, but subatomic particles exist.

(D) is wrong as Dalton assumed all atoms of an element are identical, but isotopes disprove this.

(E) is actually correct, as chemical reactions involve reorganization of atoms.

Hence, correct incorrect statements are only (B) and (D).

Given:

C = 42.1%, H = 6.4%, O = 100 − 48.5 = 51.5%

Divide by atomic masses:

C → 42.1 ÷ 12 = 3.508

H → 6.4 ÷ 1 = 6.4

O → 51.5 ÷ 16 = 3.219

Divide all by 3.219:

C ≈ 1.09, H ≈ 1.99, O ≈ 1

Empirical formula ≈ C₁H₂O₁

Empirical mass = 12 + 2 + 16 = 30

n = 342 ÷ 30 = 11.4 ≈ 11

Molecular formula = C₁₁H₂₂O₁₁

Answer: Option D

Benzalacetone is synthesized from benzaldehyde (10.6 g) and acetone. Theoretical yield is 14.6 g from 0.1 moles of benzaldehyde. Given actual yield = 7.3 g, the percentage yield is $\frac{7.3}{14.6}\times 100=50\mathrm{\%}$

Correct answer: A (50%).

According to Dalton's Atomic Theory,
compounds are formed when atoms of different elements combine in a fixed (definite) ratio by mass and number.
So, saying they combine in "any" ratio is incorrect.

The given process represents phase transitions:

• Solid → Vapors is sublimation (conversion of solid directly to gas).
• Vapors → Solid is deposition (direct conversion of gas to solid).

Thus, the correct answer is:

(C) I → Sublimation, II → Deposition.

Correct answer is A. Dalton’s atomic theory states that compounds form when atoms combine in a fixed simple whole number ratio, not any ratio. So option A is incorrect. Option B matches Dalton’s postulate, though it is now known isotopes exist. Option C is Dalton’s idea; although atoms are divisible (electrons, protons exist), this was Dalton’s belief. Option D is also correct per Dalton—he proposed that chemical reactions are rearrangements of atoms. Thus, option A contradicts Dalton’s theory and is the only incorrect statement among the given choices.

When the temperature is changed,volume will be changed and they are directly proportional to each other. Molarity is defined as the number of moles per unit volume, so molarity is also affected by temperature.

Weight is the force of gravity on an object, while mass is the amount of matter.

More moles of neutralization → More heat released → Higher temperature rise. Largest temperature increase occurs when maximum moles are neutralized with a strong acid and strong base. Hence, option (d) is correct.

Since 75 % of the solution is nitric acid by mass, 30 g is 75 % of the total mass.
Let the total mass of the solution be M.

0.75 × M = 30g

$\mathrm{M}=\frac{30}{0.75}=40\mathrm{g}$

$\mathrm{V}=\frac{\text{ Mass }}{\text{ Density }}=\frac{40\mathrm{g}}{1.25\mathrm{g}/\mathrm{mL}}=32\mathrm{mL}$

$$\begin{gathered} {\mathrm{CH}}_3- \mathrm{C}\equiv \mathrm{CH} + \mathrm{Na}\to {\mathrm{CH}}_3-\mathrm{C} \equiv \stackrel{-}{\mathrm{C}} \stackrel{+}{\mathrm{Na}} +\frac{1}{2} {\mathrm{H}}_2 \\ 1 \mathrm{mole} \mathrm{of} \mathrm{propyne} \mathrm{gives} 0.5 \mathrm{mole} \mathrm{of} \mathrm{hydrogen} \mathrm{gas} \\ {\mathrm{CH}}_3-\mathrm{C}\equiv \mathrm{CH} +{\mathrm{NaNH}}_2\to {\mathrm{CH}}_3-\mathrm{C} \equiv \stackrel{-}{\mathrm{C}} \stackrel{+}{\mathrm{Na}} + {\mathrm{NH}}_3 \\ \mathrm{n} = \frac{4 \mathrm{g}}{40 \mathrm{g}} = 0.1 \mathrm{mole} \\ 1 \mathrm{mole} \mathrm{occupies} 22400 \mathrm{ml} \\ \mathrm{So}, 0.1 \mathrm{mole} \mathrm{occupies} 2240 \mathrm{ml} \end{gathered}$$

Benzalacetone is synthesized from benzaldehyde (10.6 g) and acetone. Theoretical yield is 14.6 g from 0.1 moles of benzaldehyde. Given actual yield = 7.3 g, the percentage yield is $\frac{7.3}{14.6}\times 100=50\mathrm{\%}$

Correct answer: A (50%).

Balanced reaction:

CH₄ + 2O₂ → CO₂ + 2H₂O

CO₂ produced = 11 g

Molar mass of CO₂ = 44 g/mol

Moles of CO₂ = 11 / 44 = 0.25 mol

From equation, 1 mol CH₄ gives 1 mol CO₂

So, moles of CH₄ needed = 0.25 mol

Final Answer: 0.25

$$\begin{gathered} Z{r}_3{\left(P{O}_4\right)}_4 \left(s\right) \rightleftharpoons 3{\mathrm{Zr}}^{4+}\left(\mathrm{aq}\right) +4{{\mathrm{PO}}_4}^{3-}\left(\mathrm{aq}\right) \\ 3\mathrm{s} 4\mathrm{s} \\ {\mathrm{K}}_{\mathrm{sp}} = {\left(3\mathrm{s}\right)}^3{\left(4\mathrm{s}\right)}^{4 }=6912 {\mathrm{s}}^7 \\ \mathrm{s} ={\left(\frac{{\mathrm{K}}_{\mathrm{sp}}}{6912}\right)}^{1/7} \end{gathered}$$

$200mL(0.2L)$ of KI solution $(0.1M)$
$\to$ Moles of Kl (I^–)

$=0.1\times 0.2=0.02\text{ moles }.$

From reaction: 5 moles of $I^{-}$react with 1 mole of $I{O}_3^{-}$.
Required $I{O}_3^{-}:0.02\times 1/5=0.004$ moles.

Equivalent weight formula:

$\text{ Equivalent weight }=\frac{\text{ Molecular weight }}{n-\text{ factor }}$

For $I{O}_3^{-}$in this reaction, the change in oxidation state ( n -factor) is 5 .
Hence, Equivalent weight =( Molecular weight)/5.

$$\begin{gathered} \mathrm{Initial} \mathrm{moles} =\frac{\mathrm{x} \times {10}^{-3}}{44} \\ \mathrm{removed} \mathrm{moles} =\frac{{10}^{21}}{6.02\times {10}^{23}} \\ \mathrm{Moles} \mathrm{left} = \mathrm{initial} \mathrm{moles} -\mathrm{removed} \mathrm{moles} \\ 2.8 \times {10}^{-3}=\frac{\mathrm{x}\times {10}^{-3}}{44}-\frac{{10}^{21}}{6.02\times {10}^{23}} \\ \mathrm{x} =196.2 \mathrm{mg} \end{gathered}$$

The relation between molarity and molality which includes density is:

$\text{ Molality }=\frac{\mathrm{M}\times 1000}{1000\mathrm{d}-\mathrm{M}\times {\mathrm{M}}_{\mathrm{w}}}$

Substitute the values in the above expression:

$=\frac{3000}{1250-175.5}=2.79\mathrm{m}$

\(
\begin{array}{|c|c|c|c|c|}
\hline \text { Element } & \text { mass% } & \text { Atomic mass(g) } & \frac{\text { mass%}}{\text{ atomic mass} } & \text { Integral ratio } \\
\hline \mathrm{C} & 54.2 & 12 & 4.52 & 2 \\
\hline \mathrm{H} & 9.2 & 1 & 9.2 & 4 \\
\hline \mathrm{O} & 36.6 & 16 & 2.3 & 1 \\
\hline
\end{array}
\)

The empirical formula is ${\mathrm{C}}_2{\mathrm{H}}_4\mathrm{O}$

Molecular formula = ${\left({\mathrm{C}}_2{\mathrm{H}}_4\mathrm{O}\right)}_{\mathrm{x}}$

$\mathrm{x}= \frac{\mathrm{Moleuclar} \mathrm{mass}}{\mathrm{Empirical} \mathrm{mass} }=\frac{132}{44}=3$

${\left({\mathrm{C}}_2{\mathrm{H}}_4\mathrm{O}\right)}_3= C_6{\mathrm{H}}_{12}{\mathrm{O}}_3$

The (n+l) value determines the energy of the orbital. The lower the value of (n+l), the lower will be the energy of orbitals. If the value of (n+l) is the same for the two orbitals, then lower the value of n , the lower will be the energy of the orbital.

(A) n+l =4

(B) n+l = 3

(C) n+l = 5

(D) n+l = 5

For (C) and (D) the value of (n+l) =5, hence they both have the same energy