Some Basic Concepts of Chemistry

(A) If emperical formula of compound 3 is $P_3 Q_4$ then its molar ratio will be $\frac{40}{3} : \frac{60}{4}$

$ =\frac{40}{3} \times \frac{4}{60}=\frac{16}{18}=0.88 $

If empirical formula of compound 2 is $\mathrm{P}_3 \mathrm{Q}_5$, then its molar ratio

$ \begin{aligned} & =\frac{44.4}{2} : {\frac{55.6}{5}} \\\\ & =\frac{44.4}{2} \times \frac{5}{55.6}=2 \end{aligned} $

Since molar ratio of both the compound is not equal

So, option (A) is not correct.

(B) If empirical formula of compound 3 is $\mathrm{P}_3 \mathrm{Q}_2$, i.e.,

$ \begin{aligned} & \frac{40}{M_P}: \frac{60}{M_Q}=\frac{3}{2} \\\\ & \Rightarrow \frac{40}{M_P} \times \frac{M_Q}{60}=\frac{3}{2} \end{aligned} $

$ \begin{aligned} \Rightarrow \frac{4 \mathrm{M}_{\mathrm{Q}}}{6 \mathrm{M}_{\mathrm{P}}} & =\frac{3}{2} \\\\ \Rightarrow \frac{\mathrm{M}_{\mathrm{Q}}}{\mathrm{M}_{\mathrm{P}}} & =\frac{3}{2} \times \frac{6}{4}=\frac{9}{4} \end{aligned} $

and $\mathrm{M}_{\mathrm{P}}=20$ (given)

So, $ \frac{\mathrm{M}_{\mathrm{Q}}}{20}=\frac{9}{4}$

$ \begin{aligned} & \mathrm{M}_{\mathrm{Q}}=\frac{9 \times 26^5}{4} \\\\ & \mathrm{M}_{\mathrm{Q}}=45 \end{aligned} $

So, option (B) is correct.

(C) If empirical formula of compound 2 is PQ, So the molar ratio is

$ \frac{44.4}{1}: \frac{55.6}{1}=\frac{44.4}{55.6}=0.79 \sim 0.8 $

Empirical formula of compound 1 is $P_5 Q_4$

so the molar ratio is $\frac{50}{5} : { } \frac{50}{4}$.

$ =\frac{50}{5} \times \frac{4}{50}=\frac{4}{5}=0.8 $

Since, molar ratio of both the compound is equal hence, state $(\mathrm{C})$ is correct.

So, option (C) is correct.

(D) $\mathrm{M}_{\mathrm{P}}=70, \mathrm{M}_{\mathrm{Q}}=35$

Molar ratio of compound 1 is

$ \begin{aligned} \frac{50}{M_{\mathrm{P}}}: \frac{50}{\mathrm{M}_{\mathrm{Q}}} & =\frac{50}{70}: \frac{50}{35} \\\\ & =\frac{50}{70} \times \frac{35}{50}=\frac{1}{2} \end{aligned} $

Hence, empirical formula of compound $\mathrm{PQ}_2$.

So, option (D) is incorrect.

To determine the percentage change in the molarity of the solution, we need to follow these steps:

1. Calculate the initial volume of the solution at $+90^{\circ} \mathrm{C}$.
2. Calculate the final volume of the solution at $+10^{\circ} \mathrm{C}$.
3. Determine the initial and final molarities of the solution.
4. Compute the percentage change in the molarity.

Given information:

• Mass of the solute (substance) = 50 g
• Mass of the solvent (water) = 1 kg = 1000 g
• Initial density = $1.1 \mathrm{~g} \mathrm{~cc}^{-1}$
• Final density = $1.15 \mathrm{~g} \mathrm{~cc}^{-1}$

Step 1: Calculate the initial volume of the solution at $+90^{\circ} \mathrm{C}$.

The total mass of the solution is:

$\text{Total mass} = 50 \, g + 1000 \, g = 1050 \, g$

Using the initial density, the initial volume is:

$V_{\text{initial}} = \frac{\text{Total mass}}{\text{Initial density}} = \frac{1050 \, g}{1.1 \, \mathrm{g/cc}} \approx 954.545 \, \mathrm{cc}$

Step 2: Calculate the final volume of the solution at $+10^{\circ} \mathrm{C}$.

Using the final density, the final volume is:

$V_{\text{final}} = \frac{\text{Total mass}}{\text{Final density}} = \frac{1050 \, g}{1.15 \, \mathrm{g/cc}} \approx 913.043 \, \mathrm{cc}$

Step 3: Determine the initial and final molarities of the solution.

Molarity (M) is defined as the number of moles of solute per liter of solution. Let's assume the molar mass (M_s) of the solute (substance) is required (denote it as M_s).

Number of moles of the solute:

$\text{Number of moles} = \frac{50 \, g}{M_s \, \mathrm{g/mol}} = \frac{50}{M_s} \, \mathrm{mol}$

Initial molarity:

$M_{\text{initial}} = \frac{\text{Number of moles}}{\text{Initial volume in liters}} = \frac{\frac{50}{M_s}}{\frac{954.545}{1000}} = \frac{50 \times 1000}{M_s \times 954.545} \approx \frac{50 \times 1.047}{M_s}$

Final molarity:

$M_{\text{final}} = \frac{\text{Number of moles}}{\text{Final volume in liters}} = \frac{\frac{50}{M_s}}{\frac{913.043}{1000}} = \frac{50 \times 1000}{M_s \times 913.043} \approx \frac{50 \times 1.095}{M_s}$

Step 4: Compute the percentage change in molarity.

Percentage change in molarity:

$\text{Percentage change} = \frac{M_{\text{final}} - M_{\text{initial}}}{M_{\text{initial}}} \times 100$

Substituting the values:

$\text{Percentage change} = \frac{\frac{50 \times 1.095}{M_s} - \frac{50 \times 1.047}{M_s}}{\frac{50 \times 1.047}{M_s}} \times 100 = \frac{50 \times (1.095 - 1.047)}{50 \times 1.047} \times 100$

$\text{Percentage change} = \frac{(1.095 - 1.047)}{1.047} \times 100 = \frac{0.048}{1.047} \times 100 \approx 4.58\%$

Since the closest provided option is 4.5, the correct answer is:

Option B: 4.5

Calgon is a water softener. The molecular formula of calgon is $\mathrm{Na}_6 \mathrm{P}_6 \mathrm{O}_{18}$. Thus, the empirical formula of calgon is $\mathrm{NaPO}_3$.

Law of definite proportion states that the individual elements that constitute a chemical compound are always present in a fixed mass ratio.

(a) The ratio of oxygen in $\mathrm{H}_2 \mathrm{O}$ and $\mathrm{H}_2 \mathrm{O}_2$ with respect to fixed mass of hydrogen atom is a whole number. This is the statement of "law of multiple proportion."

(b) As the ratio of elements is fixed, so percentage of oxygen in $\mathrm{H}_2 \mathrm{O}$ is constant or fixed irrespective of the source. This statement is with reference to law of definite proportion. Thus, this is the correct option.

(c) Avogadro's law states that the "equal volume of all gases at the same temperature and pressure should contain equal number of molecules."

(d) "Mass can neither be created nor be destroyed" is the statement of "law of conservation of mass."

The balanced equation is

$2 \mathrm{Pb}_3 \mathrm{O}_4 \longrightarrow 6 \mathrm{PbO}+\mathrm{O}_2$

So, value of $x$ and $y$ in the given reactions are 2 and 6 respectively.

Millimoles of H₃PO₃ = M $\times$ V = 1 $\times$ 50 = 50

For 1 millimole of H₃PO₃, we require 2 millimoles of NaOH.

For 50 millimole of H₃PO₃, we require (2 $\times$ 50) = 100 millimoles of NaOH.

Millimoles of NaOH = M $\times$ V = 100

1 $\times$ V = 100

V = 100 mL

Millimoles of H₃PO₂ = M $\times$ V = 2 $\times$ 100 = 200

For 1 millimole of H₃PO₂, we require 1 millimoles of NaOH.

For 200 millimoles of H₃PO₂, we require 200 millimoles of NaOH.

So, volume of NaOH = 200 mL

C_xH_yO_z + O₂ $ \to $ xCO₂ + ${y \over 2}{H_2}O$

2.64 g of CO₂ contains 0.72 g C.

1.08 g of H₂O contains 0.12 g H.

$ \therefore $ Mass of oxygen present = 1.80 – (0.72 +0.12) = 0.96 g

% of O = ${{0.96} \over {1.80}} \times 100$ = 53.33 %

$\begin{aligned} M & =15.9 \mathrm{~g} \text { (Given }) \\ V & =10 \mathrm{~mL} \end{aligned}$

$\begin{aligned} & \text { Density }=\frac{\text { Mass }}{\text { Volume }}=\frac{15.9 \mathrm{~g}}{10 \mathrm{~mL}} \\ & \text { Density }=1.59 \mathrm{~g} \mathrm{~mL}^{-1} . \end{aligned}$