Some Basic Concepts of Chemistry

From combustion:

• Increase in mass of CaCl₂ tube = mass of H₂O formed

  $ m(\text{H}_2\text{O})=0.15\text{ g},\quad n(\text{H}_2\text{O})=\frac{0.15}{18}=0.008333\text{ mol} $

  Moles of H atoms:

  $ n(\text{H})=2n(\text{H}_2\text{O})=0.016667\text{ mol} $

  Mass of H:

  $ m(\text{H})=0.016667\times 1=0.016667\text{ g} $

• Increase in mass of potash tube = mass of CO₂ formed

  $ m(\text{CO}_2)=0.1837\text{ g},\quad n(\text{CO}_2)=\frac{0.1837}{44}=0.004175\text{ mol} $

  Moles of C atoms = moles of CO₂:

  $ n(\text{C})=0.004175\text{ mol} $

  Mass of C:

  $ m(\text{C})=0.004175\times 12=0.05010\text{ g} $

Sample mass = 0.25 g, so mass of oxygen in the compound:

$ m(\text{O}) = 0.25 - (0.016667+0.05010)=0.18323\text{ g} $

Percentage of oxygen:

$ \% \text{O}=\frac{0.18323}{0.25}\times 100=73.29\%\approx \boxed{73\%} $

Balanced equation:

$\mathrm{Ba(OH)_2 + 2HCl \rightarrow BaCl_2 + 2H_2O}$

Moles of $\mathrm{Ba(OH)_2}$ used:

$n_{\mathrm{Ba(OH)_2}}=MV=0.1\times 25.0\times 10^{-3}=2.5\times 10^{-3}\ \text{mol}$

From the equation, $1$ mol $\mathrm{Ba(OH)_2}$ reacts with $2$ mol $\mathrm{HCl}$, so

$n_{\mathrm{HCl}}=2\times 2.5\times 10^{-3}=5.0\times 10^{-3}\ \text{mol}$

Mass of $\mathrm{HCl}$:

$m_{\mathrm{HCl}}=n\times M=5.0\times 10^{-3}\times 36.5=0.1825\ \text{g}$

$0.1825\ \text{g}=182.5\ \text{mg}$

So $x=182.5\ \text{mg}=1825\times 10^{-1}\ \text{mg}$.

Nearest integer $= \boxed{1825}$.

$\begin{aligned} & \text { Moles of } \mathrm{I}^{-} \text {in } \mathrm{NaI}=\text { Moles of }\left(\mathrm{I}^{-}\right) \text {in } \mathrm{AgI}=\frac{4.74}{235} \\ & \text { Moles of } \mathrm{NaI}=\frac{4.74}{235} \\ & \text { Molarity }[\mathrm{NaI}]=\frac{4.74}{235 \times 0.02}=1.008 \end{aligned}$

The chemical reaction between butane ($\mathrm{C}_4\mathrm{H}_{10}$) and oxygen ($\mathrm{O}_2$) to produce carbon dioxide ($\mathrm{CO}_2$) and water ($\mathrm{H}_2\mathrm{O}$) can be represented by the following equation:

$ \mathrm{C}_4\mathrm{H}_{10}(\mathrm{g}) + \frac{13}{2} \mathrm{O}_2(\mathrm{g}) \rightarrow 4 \mathrm{CO}_2(\mathrm{g}) + 5 \mathrm{H}_2\mathrm{O}(\mathrm{l}) $

Given that we have 174.0 kg of butane and 320.0 kg of oxygen, we are tasked with calculating the volume of water produced in liters.

Mole Calculation for Water:

The stoichiometry of the reaction shows that 5 moles of water are produced per 13/2 moles of oxygen.

Using the moles of oxygen available from input, calculate the moles of water produced:

$ \text{Moles of } \mathrm{H}_2\mathrm{O} = 5 \times \frac{2}{13} \times (10 \times 10^3) $

Mass of Water Produced:

Calculate the mass of water using the molar mass of water (18 g/mol):

$ \text{Mass of } \mathrm{H}_2\mathrm{O} = \left(\frac{10^5}{13}\right) \times 18 $

The result will be:

$ = 1.3846 \times 10^5 \ \text{g} $

Volume of Water:

Convert the mass of water to volume, considering the density of water is $1 \ \text{g/mL}$ (or $1 \ \text{g/L}$):

$ \text{Volume of } \mathrm{H}_2\mathrm{O} = 138.46 \ \text{liters} $

Therefore, the volume of water formed is approximately 138 liters when rounded to the nearest integer.

Convert the mass of $\mathrm{CO}_2$ to moles:

The molar mass of $\mathrm{CO}_2$ is 44 g/mol. Thus, the number of moles of $\mathrm{CO}_2$ produced is calculated as follows:

$ \mathrm{n}_{\mathrm{CO}_2} = \frac{220 \times 10^{-3} \, \text{g}}{44 \, \text{g/mol}} = 5 \times 10^{-3} \text{ moles} $

Calculate the mass of carbon in $\mathrm{CO}_2$:

Since each mole of $\mathrm{CO}_2$ contains one mole of carbon, the moles of carbon are the same: $5 \times 10^{-3}$ moles. The molar mass of carbon is 12 g/mol, so the mass of carbon is:

$ \mathrm{m}_{\mathrm{C}} = 5 \times 10^{-3} \, \text{moles} \times 12 \, \text{g/mol} = 60 \times 10^{-3} \, \text{g} = 60 \, \text{mg} $

Calculate the percentage composition of carbon:

To find the percentage by mass of carbon in the compound, use the formula:

$ \% \, \text{Carbon} = \left( \frac{60 \, \text{mg}}{500 \, \text{mg}} \right) \times 100 = 12\% $

Thus, the percentage composition of carbon in the compound is approximately 12%.

$\rightarrow$ Molecular formula of Thyroxine $\Rightarrow$ $\mathrm{C}_{15} \mathrm{H}_{11} \mathrm{O}_4 \mathrm{NI}_4$

$\rightarrow$ Molecular mass of Thyroxine -

$\begin{aligned} & \mathrm{C} \rightarrow 15 \times 12=180 \\ & \mathrm{H} \rightarrow 11 \times 1=11 \\ & \mathrm{O} \rightarrow 16 \times 4=64 \\ & \mathrm{~N} \rightarrow 14 \times 1=14 \\ & \mathrm{I} \rightarrow 127 \times 4=508 \end{aligned}$

$\rightarrow$ Molecular mass of Thyroxine $\Rightarrow 777$

$\begin{aligned} & \rightarrow \% \text { of Iodine }=\frac{508}{777} \times 100 \\ & =65.38 \% \\ & \text { Nearest integer }=65 \end{aligned}$

$\begin{aligned} & \mathrm{CaCO}_3 \rightarrow \mathrm{CaO}+\mathrm{CO}_2 \\ & \text { mass of } \mathrm{CaCO}_3=\frac{150 \times 75}{100}=112.5 \mathrm{~kg} \\ & =112500 \mathrm{~g} \\ & \mathrm{n}_{\mathrm{CaCO}_3}=1125 \end{aligned}$

$\begin{aligned} & \text { So moles of } \mathrm{CaO}=1125 \\ & \qquad \text { mass of } \mathrm{CaO}=\frac{1125 \times 56}{1000}=63 \mathrm{~kg} \\ & \text { Correct answer } \Rightarrow 63 \end{aligned}$

Let mass of iron $=\mathrm{w} g$

$\begin{aligned} & \Rightarrow \frac{\mathrm{w}}{150 \times 10^3} \times 10^6=12 \\ & \Rightarrow \mathrm{w}=150 \times 12 \times 10^{-3}=1.8 \mathrm{gm} \end{aligned}$

Let mass of $\mathrm{FeSO}_4 \cdot 7 \mathrm{H}_2 \mathrm{O}=\mathrm{w}_1 \mathrm{gm}$

$\begin{aligned} & \Rightarrow \text { Moles of } \mathrm{Fe}=\frac{1.8}{56}=\left(\frac{\mathrm{w}_1}{56+96+7 \times 18}\right) \\ & \Rightarrow \mathrm{w}_1=8.935 \mathrm{gm} \end{aligned}$

$\begin{array}{ll} \mathrm{C}_6 \mathrm{H}_5 \mathrm{NO}_2 & \mathrm{MF}=\mathrm{C}_6 \mathrm{H}_4 \mathrm{~N}_2 \mathrm{O}_4 \\ \mathrm{MW}=123 & \mathrm{MW}=168 \\ & \therefore \frac{4.2}{168}=0.025 \mathrm{~mol} \end{array}$

$\because$ required gm of nitro benzene

$\begin{aligned} & =123 \times 0.025 \\ & =3.075 \end{aligned}$

$\therefore$ Nearest integer is 3

M.wt. of given compound = 86

$\begin{aligned} &\text { Applying POAC on ' } \mathrm{N} \text { ' }\\ &\begin{aligned} & \mathrm{n}_{\mathrm{X}} \times 2=\mathrm{n}_{\mathrm{N}_2} \times 2 \\ & \frac{0.42}{86}=\mathrm{n}_{\mathrm{N}_2} \\ & \Rightarrow(\text { Volume })_{\mathrm{N}_2} \text { at } \mathrm{STP}=\frac{0.42}{86} \times 22.4 \mathrm{~L} \\ & =0.1108 \mathrm{~L}=110.8 \mathrm{~m} \ell \end{aligned} \end{aligned}$

To determine the percentage of carbon in the organic compound, follow these steps:

Calculate Moles of CO₂:

Given: 1.46 g of CO₂ is produced.

Molar mass of CO₂ = 44 g/mol.

Moles of CO₂ = $\frac{1.46 \text{ g}}{44 \text{ g/mol}}$.

Determine Moles of Carbon Atoms:

Each molecule of CO₂ contains one atom of carbon.

Thus, moles of carbon = moles of CO₂.

Calculate Mass of Carbon:

Molar mass of carbon (C) = 12 g/mol.

Mass of carbon = Moles of carbon × 12 g/mol.

Mass of carbon = $\frac{1.46}{44} \times 12$.

Calculate Percentage of Carbon:

Total mass of the organic compound = 0.5 g.

Percentage of carbon = $\left(\frac{\text{Mass of carbon}}{0.5 \text{ g}}\right) \times 100\%$.

Hence, Percentage of carbon = $\frac{1.46}{44} \times \frac{12}{0.5} \times 100$.

Percentage of carbon = 79.63%.

Therefore, the percentage of carbon in the compound is approximately 80%.

Assuming 100 gm solution contain 70 gm solute.

Volume of 100 gm solution will be $\frac{100}{1.25} \mathrm{ml}$.

Molarity $=\frac{70 / 70}{100 / 1.25} \times 1000=12.5$ or $125 \times 10^{-1}$

Identify the given weight percentages from a 100 g sample:

Carbon (C): 14.5 g

Chlorine (Cl): 64.46 g

Hydrogen (H): 1.8 g

Since the sample must sum to 100 g, the remaining mass is from oxygen (O):

$\text{O: } 100 - (14.5 + 64.46 + 1.8) = 100 - 80.76 = 19.24\text{ g}.$

Next, convert these masses to moles using the given atomic masses:

Moles of C:

$\frac{14.5\text{ g}}{12\text{ g/mol}} \approx 1.2083\text{ mol}.$

Moles of Cl:

$\frac{64.46\text{ g}}{35.5\text{ g/mol}} \approx 1.8171\text{ mol}.$

Moles of H:

$\frac{1.8\text{ g}}{1\text{ g/mol}} = 1.8\text{ mol}.$

Moles of O:

$\frac{19.24\text{ g}}{16\text{ g/mol}} \approx 1.2025\text{ mol}.$

Now, find the simplest whole-number ratio by dividing each by the smallest number of moles (approximately 1.2025 mol):

Ratio for C:

$1.2083 / 1.2025 \approx 1.00$

Ratio for O:

$1.2025 / 1.2025 = 1.00$

Ratio for H:

$1.8 / 1.2025 \approx 1.50$

Ratio for Cl:

$1.8171 / 1.2025 \approx 1.51 \approx 1.50$

Since the ratios for H and Cl are about 1.5, multiplying all ratios by 2 will give whole numbers:

C: 1 × 2 = 2

O: 1 × 2 = 2

H: 1.5 × 2 = 3

Cl: 1.5 × 2 = 3

Thus, the empirical formula of the compound is:

$\mathrm{C_2H_3O_2Cl_3}.$

Calculate the empirical formula mass by summing the contributions:

C: $2 \times 12 = 24\text{ g/mol}$

H: $3 \times 1 = 3\text{ g/mol}$

O: $2 \times 16 = 32\text{ g/mol}$

Cl: $3 \times 35.5 = 106.5\text{ g/mol}$

Total mass:

$24 + 3 + 32 + 106.5 = 165.5\text{ g/mol}.$

The problem asks for the empirical formula mass in the form “__ × 10⁻¹”. We can express 165.5 g/mol as:

$165.5 = 1655 \times 10^{-1}\text{ g/mol}.$

Thus, the answer is: $1655 \times 10^{-1}\text{ g/mol}.$

moles of $\mathrm{Fe}_3 \mathrm{O}_4=\frac{2.32 \times 10^3 \times 10^3}{232}=10000 \mathrm{~mol}$

moles of $\mathrm{CO}=\frac{2.8 \times 10^2 \times 10^3}{28}=10000 \mathrm{~mol}$

$\begin{aligned} & \mathrm{Fe}_3 \mathrm{O}_4+4 \mathrm{CO} \longrightarrow 3 \mathrm{Fe}+4 \mathrm{CO}_2 \\ & 10^4 \mathrm{~mol} \quad 10^4 \mathrm{~mol} \end{aligned}$

CO is L.R.

mole of $\mathrm{Fe}=\frac{3}{4} \times 10^4$

mass of $\mathrm{Fe}=\frac{3}{4} \times \frac{10^4 \times 56}{1000} \mathrm{~kg}=420 \mathrm{~kg}$

$\begin{aligned} & \mathrm{C}_6 \mathrm{H}_5 \mathrm{COOH}+\mathrm{NaHCO}_3 \rightarrow \mathrm{C}_6 \mathrm{H}_5 \mathrm{COO} \mathrm{Na}^{+} \\ & +\mathrm{H}_2 \mathrm{O}+\mathrm{CO}_2\end{aligned}$

$\begin{aligned} & \mathrm{x} \mathrm{~gm} \quad \text{11.2 L}\\ & \text { mole of } \mathrm{C}_6 \mathrm{H}_5 \mathrm{COOH}=\text { mole of } \mathrm{CO}_2=\frac{11.2}{22.4}=0.5 \\ & \text { mass of } \mathrm{C}_6 \mathrm{H}_5 \mathrm{COOH}=\mathrm{x}=0.5 \times 122=61 \mathrm{gm} \end{aligned}$

Organic compound $\xrightarrow{\text { combustion }} \underset{\substack{0.9 \mathrm{gm}}}{\mathrm{H}_2 \mathrm{O}}$

$\begin{aligned} & \therefore \text { mole of } \mathrm{H}_2 \mathrm{O}=\frac{0.9}{18}=0.05 \text { mole } \\ & \begin{aligned} \therefore \text { mole of } \mathrm{H} \text { in } \mathrm{H}_2 \mathrm{O} & =0.05 \times 2=0.1 \text { mole } \\ & =\text { mole of } \mathrm{H} \text { in } 0.01 \text { mole } \\ & \text { Organic compound } \end{aligned} \end{aligned}$

$\begin{aligned} &\begin{aligned} \therefore \text { wt of } \mathrm{H} \text { atom in } 0.01 \text { mole compound } & =0.1 \times 1 \\ & =0.1 \mathrm{gm} \end{aligned}\\ &\therefore \text { wt of } \mathrm{H} \text { atom in one mole compound }\\ &\begin{aligned} & =\frac{0.1}{0.01}=10 \mathrm{gm} \\ & \because \text { wt. } \% \text { of } H=\frac{\text { wt. of } H \text { in one mole compound }}{\text { Molar mass of compound }} \times 1 \\ & 10=\frac{10}{M} \times 100 \\ & \therefore M=100 \end{aligned} \end{aligned}$

To determine the mass of aluminum oxide produced, we first consider the chemical reaction between aluminum and oxygen:

$ 4 \text{Al} + 3 \text{O}_2 \rightarrow 2 \text{Al}_2\text{O}_3 $

Given:

Molar mass of Al = 27.0 g/mol

Molar mass of O = 16.0 g/mol

81.0 g of Al and 128.0 g of O$_2$

Calculating moles of reactants:

For aluminum (Al), using its molar mass:

$ \frac{81 \text{ g}}{27 \text{ g/mol}} = 3 \text{ moles of Al} $

For oxygen (O$_2$), using its molar mass (O$_2$ is composed of two oxygen atoms):

$ \frac{128 \text{ g}}{32 \text{ g/mol}} = 4 \text{ moles of O}_2 $

Identifying the limiting reagent:

From the stoichiometry of the reaction:

4 moles of Al react with 3 moles of O$_2$

3 moles of Al require $ \frac{3}{4} \times 3 $ moles of O$_2$ = 2.25 moles of O$_2$

Since only 2.25 moles of O$_2$ are needed by 3 moles of Al, and we have 4 moles of O$_2$, aluminum is the limiting reagent.

Calculating moles of $\text{Al}_2\text{O}_3$ formed:

From the reaction, 4 moles of Al yield 2 moles of $\text{Al}_2\text{O}_3$. Thus:

$ \frac{3}{4} \times 2 = \frac{3}{2} \text{ moles of } \text{Al}_2\text{O}_3 $

Calculating mass of $\text{Al}_2\text{O}_3$:

The molar mass of $\text{Al}_2\text{O}_3$ is:

$ (2 \times 27) + (3 \times 16) = 102 \text{ g/mol} $

So, the mass of $\frac{3}{2}$ moles of $\text{Al}_2\text{O}_3$ is:

$ \frac{3}{2} \times 102 = 153 \text{ grams} $

Thus, the mass of aluminum oxide produced is 153 grams.

$\begin{aligned} & \text { Millimoles of } \mathrm{BaSO}_4=\frac{466}{233}=2 \mathrm{~m} \mathrm{~mol} \\ & \% \mathrm{~S}=\frac{\frac{466}{233} \times 32}{160} \times 100=40 \% \end{aligned}$

To determine the final concentration of the NaOH solution, we use the formula for mixing solutions :

$ M_F = \frac{M_1 \times V_1 + M_2 \times V_2}{V_1 + V_2} $

Where :

$ M_1 = 2 \, \text{M} $ and $ V_1 = 20 \, \text{mL} $: Concentration and volume of the first solution.

$ M_2 = 0.5 \, \text{M} $ and $ V_2 = 400 \, \text{mL} $: Concentration and volume of the second solution.

Substitute the values into the equation:

$ M_F = \frac{2 \times 20 + 0.5 \times 400}{420} $

Calculating each term:

$ 2 \times 20 = 40 $

$ 0.5 \times 400 = 200 $

Add these results:

$ M_F = \frac{40 + 200}{420} = \frac{240}{420} \approx 0.571 \, \text{M} $

Convert this to scientific notation as the problem specifies the answer should be in $ \times 10^{-2} $ form:

$ M_F = 57.1 \times 10^{-2} \, \text{M} $

Rounded to the nearest integer, the final concentration is:

$ 57 $

To find the molality of the solution, we need to follow these steps:

1. Determine the number of moles of anhydrous $\mathrm{CuSO}_4$ in the solution.

The molarity (M) is given as $2 \times 10^{-1}$ M in a 500 mL solution. Hence, the number of moles of $\mathrm{CuSO}_4$ is:

$ \text{Moles of } \mathrm{CuSO}_4 = \text{Molarity} \times \text{Volume in liters} = 2 \times 10^{-1} \times 0.5 = 0.1 \text{ moles} $

2. Calculate the mass of the solution using the given density.

The density of the solution is given as $1.25 \mathrm{~g/mL}$. The volume of the solution is 500 mL. Therefore, the mass of the solution is:

$ \text{Mass of solution} = \text{Density} \times \text{Volume} = 1.25 \times 500 = 625 \text{ grams} $

3. Find the mass of the solvent (water) in the solution by subtracting the mass of the solute (anhydrous $\mathrm{CuSO}_4$) from the total mass of the solution.

We know the number of moles of $\mathrm{CuSO}_4$, and we can find its molar mass.

$ \text{Molar mass of } \mathrm{CuSO}_4 = 63.5 + 32 + 4 \times 16 = 159.5 \text{ g/mol} $

Therefore, the mass of $\mathrm{CuSO}_4$ is:

$ \text{Mass of } \mathrm{CuSO}_4 = \text{Number of moles} \times \text{Molar mass} = 0.1 \times 159.5 = 15.95 \text{ grams} $

4. Calculate the mass of the solvent (water):

$ \text{Mass of water} = \text{Mass of solution} - \text{Mass of } \mathrm{CuSO}_4 = 625 - 15.95 = 609.05 \text{ grams} $

Convert the mass of water to kilograms:

$ \text{Mass of water} = 609.05 \text{ grams} = 0.60905 \text{ kilograms} $

5. Calculate the molality using the formula:

$ \text{Molality} (\text{m}) = \frac{\text{Moles of solute}}{\text{Mass of solvent in kg}} = \frac{0.1}{0.60905} = 0.164 \text{ mol/kg} $

Converting to the desired unit:

$ \text{Molality} (\text{m}) = 0.164 \times 10^3 = 164 \times 10^{-3} $

Therefore, the molality of the solution is approximately 164 $\times 10^{-3} \mathrm{~m}$.

To determine the mass percent of the solute (ethyl alcohol) in the solution, we start by calculating the masses of ethyl alcohol and water using their respective molar masses.

The molar mass of ethyl alcohol (C₂H₅OH) is given as 46 g/mol and the molar mass of water (H₂O) is given as 18 g/mol.

First, we calculate the mass of ethyl alcohol:

$\text{Mass of ethyl alcohol} = \text{Number of moles} \times \text{Molar mass}$

$\text{Mass of ethyl alcohol} = 1 \, \text{mol} \times 46 \, \text{g/mol} = 46 \, \text{g}$

Next, we calculate the mass of water:

$\text{Mass of water} = \text{Number of moles} \times \text{Molar mass}$

$\text{Mass of water} = 9 \, \text{mol} \times 18 \, \text{g/mol} = 162 \, \text{g}$

Now, we have the masses of both components of the solution. The total mass of the solution is the sum of the masses of ethyl alcohol and water:

$\text{Total mass of solution} = 46 \, \text{g} + 162 \, \text{g} = 208 \, \text{g}$

Then we calculate the mass percent of the solute (ethyl alcohol) using the formula:

$\text{Mass percent of solute} = \left( \frac{\text{Mass of solute}}{\text{Total mass of solution}} \right) \times 100\%$

$\text{Mass percent of solute} = \left( \frac{46 \, \text{g}}{208 \, \text{g}} \right) \times 100\%$

$\text{Mass percent of solute} = \left( \frac{46}{208} \right) \times 100\% \approx 22.12\%$

Since the integer answer is requested, we round 22.12% to the nearest whole number. Therefore, the mass percent of the solute (ethyl alcohol) in the solution is 22%.

To determine the mole fraction of urea in a solution where the molality is $4.44 \ \mathrm{m}$, we first need to understand the definitions and relationships involved.

Molality ($\mathrm{m}$) is given by:

$ \mathrm{m} = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} $

Here, the molality is $4.44 \ \mathrm{m}$, which means there are $4.44$ moles of urea (solute) per kilogram of water (solvent).

The next step involves calculating the mole fraction. Mole fraction ($X$) of urea is given by:

$ X_{\text{urea}} = \frac{\text{moles of urea}}{\text{moles of urea} + \text{moles of water}} $

Since we have $4.44$ moles of urea, let's determine the moles of water. The molar mass of water ($\mathrm{H_2O}$) is approximately $18 \ \mathrm{g/mol}$. Therefore, the moles of water in $1 \ \mathrm{kg}$ (or $1000 \ \mathrm{g}$) of water are:

$ \text{Moles of water} = \frac{1000 \ \mathrm{g}}{18 \ \mathrm{g/mol}} \approx 55.56 \ \text{moles} $

We then substitute these values into the mole fraction formula:

$ X_{\text{urea}} = \frac{4.44}{4.44 + 55.56} $

Simplifying the expression inside the denominator first:

$ 4.44 + 55.56 = 60 $

So, the mole fraction of urea becomes:

$ X_{\text{urea}} = \frac{4.44}{60} $

Dividing the values gives us:

$ X_{\text{urea}} \approx 0.074 $

The problem asks for the mole fraction in the form $x \times 10^{-3}$, so we convert our result to this form:

$ X_{\text{urea}} = 74 \times 10^{-3} $

Therefore, the value of $x$ is:

$ \boxed{74} $

To determine the maximum amount of acetanilide that can be prepared from 6.55 g of aniline, we need to use stoichiometry. Let's go through the process step by step.

1. Molecular weights calculation:

The molecular weight of aniline (C₆H₅NH₂) is calculated as follows:

$\text{Molecular weight of aniline} = 6 \times 12 + 5 \times 1 + 14 + 2 \times 1 = 93 \text{ g/mol}$

The molecular weight of acetanilide (C₈H₉NO) is calculated as follows:

$\text{Molecular weight of acetanilide} = 8 \times 12 + 9 \times 1 + 14 + 16 = 135 \text{ g/mol}$

2. Mole calculation:

Moles of aniline:

$\text{Moles of aniline} = \frac{\text{Mass of aniline}}{\text{Molecular weight of aniline}} = \frac{6.55 \text{ g}}{93 \text{ g/mol}} = 0.0704 \text{ mol}$

3. Stoichiometry of the reaction:

The reaction between aniline and acetic anhydride to form acetanilide follows a 1:1 mole ratio.

4. Mass calculation:

Theoretical mass of acetanilide formed:

$\text{Mass of acetanilide} = \text{Moles of aniline} \times \text{Molecular weight of acetanilide} = 0.0704 \text{ mol} \times 135 \text{ g/mol} = 9.504 \text{ g}$

5. Convert to the desired unit:

Given the unit required is $\times 10^{-1} \mathrm{~g}$, we express 9.504 g as:

$9.504 \text{ g} = 95.04 \times 10^{-1} \text{ g}$

Therefore, the maximum amount of acetanilide that can be prepared from 6.55 g of aniline is $95.04 \times 10^{-1} \mathrm{~g}$.

Moles of $\mathrm{N}_2=0.1$

$\begin{aligned} \text { Mass of } \mathrm{C}_2 \mathrm{H}_5 \mathrm{NH}_2 & =(0.1) \times 45 \\ & =4.5 \mathrm{~gm} \\ & =45 \times 10^{-1} \\ & =45 \end{aligned} $

To solve this problem, we will use the fact that the volume ratio of gases in a reaction at the same conditions of temperature and pressure represents their mole ratio according to Avogadro's law. Thus, since the gas volumes given are at the same conditions, we can directly relate them to their stoichiometric coefficients in the balanced equation for combustion of a hydrocarbon.

The general equation for complete combustion of a hydrocarbon with a formula $\mathrm{C}_x\mathrm{H}_y$ can be represented as:

$ \mathrm{C}_x\mathrm{H}_y + (x + \frac{y}{4})\mathrm{O}_2 \rightarrow x\mathrm{CO}_2 + \frac{y}{2}\mathrm{H}_2\mathrm{O} $

Given:

• $10 \mathrm{~mL}$ of hydrocarbon ($\mathrm{C}_x\mathrm{H}_y$)
• $40 \mathrm{~mL}$ of $\mathrm{CO}_2$
• $50 \mathrm{~mL}$ of $\mathrm{H}_2\mathrm{O}$

Since the volume ratios represent the mole ratios for gases, we can write the following ratios for the coefficients:

$ \frac{x}{1} = \frac{40 \mathrm{~mL} \mathrm{CO}_2}{10 \mathrm{~mL} \mathrm{C}_x\mathrm{H}_y} = 4 $

$ \frac{y}{2} = \frac{50 \mathrm{~mL} \mathrm{H}_2\mathrm{O}}{10 \mathrm{~mL} \mathrm{C}_x\mathrm{H}_y} = 5 $

From the first ratio, we see that:

$ x = 4 $

This tells us that there are four carbon atoms in the hydrocarbon molecule.

From the second ratio, by multiplying both sides by 2, we get:

$ y = 5 \times 2 = 10 $

This means there are ten hydrogen atoms in the hydrocarbon molecule.

So, the total number of carbon and hydrogen atoms in the hydrocarbon is:

$ \text{Total atoms} = x + y = 4 + 10 = 14 $

Therefore, the total number of carbon and hydrogen atoms in the hydrocarbon is 14.

To solve this problem, we will use the stoichiometry of the balanced chemical reaction. The balanced equation shows that 3 moles of $\mathrm{PbCl}_2$ react with 2 moles of $\left(\mathrm{NH}_4\right)_3 \mathrm{PO}_4$ to produce 1 mole of $\mathrm{Pb}_3\left(\mathrm{PO}_4\right)_2$.

The reaction is:

$ 3 \mathrm{PbCl}_2 + 2\left(\mathrm{NH}_4\right)_3 \mathrm{PO}_4 \rightarrow \mathrm{Pb}_3\left(\mathrm{PO}_4\right)_2 + 6 \mathrm{NH}_4 \mathrm{Cl} $

The molar ratio of $\mathrm{PbCl}_2$ to $\mathrm{Pb}_3\left(\mathrm{PO}_4\right)_2$ is 3:1, and the molar ratio of $\left(\mathrm{NH}_4\right)_3 \mathrm{PO}_4$ to $\mathrm{Pb}_3\left(\mathrm{PO}_4\right)_2$ is 2:1. We need to determine which reactant is the limiting reagent because it will dictate the amount of $\mathrm{Pb}_3\left(\mathrm{PO}_4\right)_2$ produced.

The stoichiometric calculations are as follows:

For $\mathrm{PbCl}_2$:

$ \text{Moles of }\mathrm{Pb}_3\left(\mathrm{PO}_4\right)_2\text{ formed from } \mathrm{PbCl}_2 = \frac{72 \text{ mmol of } \mathrm{PbCl}_2}{3 \text{ mmol of } \mathrm{PbCl}_2\text{/mmol of } \mathrm{Pb}_3\left(\mathrm{PO}_4\right)_2} = 24 \text{ mmol} $

For $\left(\mathrm{NH}_4\right)_3 \mathrm{PO}_4$:

$ \text{Moles of }\mathrm{Pb}_3\left(\mathrm{PO}_4\right)_2\text{ formed from } \left(\mathrm{NH}_4\right)_3 \mathrm{PO}_4 = \frac{50 \text{ mmol of } \left(\mathrm{NH}_4\right)_3 \mathrm{PO}_4}{2 \text{ mmol of } \left(\mathrm{NH}_4\right)_3 \mathrm{PO}_4\text{/mmol of } \mathrm{Pb}_3\left(\mathrm{PO}_4\right)_2} = 25 \text{ mmol} $

Now we can identify the limiting reagent by comparing the two amounts of $\mathrm{Pb}_3\left(\mathrm{PO}_4\right)_2$ that could be produced. The smaller quantity will be the actual amount produced since the limiting reagent restricts the reaction.

Since the $\mathrm{PbCl}_2$ can produce only 24 mmol of $\mathrm{Pb}_3\left(\mathrm{PO}_4\right)_2$ versus the 25 mmol that could be produced by $\left(\mathrm{NH}_4\right)_3 \mathrm{PO}_4$, $\mathrm{PbCl}_2$ is the limiting reagent.

Therefore, the amount of $\mathrm{Pb}_3\left(\mathrm{PO}_4\right)_2$ formed is 24 mmol (as a nearest integer).

Specific gravity (density) $=1.54 \mathrm{~g} / \mathrm{cc}$.

Volume $=1 \mathrm{~L}=1000 \mathrm{~ml}$

Mass of solution $=1.54 \times 1000$

$=1540 \mathrm{~g}$

$\%$ purity of $\mathrm{H}_2 \mathrm{SO}_4$ is $70 \%$

So weight of $\mathrm{H}_3 \mathrm{PO}_4=0.7 \times 1540=1078 \mathrm{~g}$

Mole of $\mathrm{H}_3 \mathrm{PO}_4=\frac{1078}{98}=11$

Molarity $=\frac{11}{1 \mathrm{~L}}=11$

$\mathrm{CH}_{4(\mathrm{~g})}+2 \mathrm{O}_{2(\mathrm{~g})} \rightarrow \mathrm{CO}_{2(\mathrm{~g})}+2 \mathrm{H}_2 \mathrm{O}_{(\ell)}$

$\mathrm{n}_{\mathrm{CO}_2}=\frac{22}{44}=0.5 \text { moles }$

So moles of $\mathrm{CH}_4$ required $=0.5$ moles i.e. $50 \times 10^{-2} \mathrm{~mole}$

$\mathrm{x}=50$

$\mathbf{C a F}_2$ does not evolve any gas with concentrated $\mathrm{H}_2 \mathrm{SO}_4$.

$\mathrm{NaBr} \rightarrow$ evolve $\mathrm{Br}_2$

$\mathrm{NaNO}_3 \rightarrow$ evolve $\mathrm{NO}_2$

$\mathrm{KI} \rightarrow$ evolve $\mathrm{I}_2$

$\begin{aligned} & \text { Moles }=\text { Molarity } \times \text { Volume in litres } \\ & =0.35 \times 0.25 \\ & \text { Mass }=\text { moles } \times \text { molar mass } \\ & =0.35 \times 0.25 \times 82.02=7.18 \mathrm{~g} \end{aligned}$

Ans. 7

$\begin{aligned} &\begin{aligned} & \text { Volume of silver coating }=0.05 \times 0.05 \times 10000 \\\\ & =25 \mathrm{~cm}^3 \\\\ & \text { Mass of silver deposited }=25 \times 7.9 \mathrm{~g} \\\\ & \text { Moles of silver atoms }=\frac{25 \times 7.9}{108} \\\\ & \text { Number of silver atoms }=\frac{25 \times 7.9}{108} \times 6.023 \times 10^{23} \\\\ & =11.01 \times 10^{23} \end{aligned}\\\\ &\text { Ans. } 11 \end{aligned}$

Equivalent of Oxalic acid $=$ Equivalents of $\mathrm{NaOH}$

$\begin{aligned} & 50 \times 0.5 \times 2=25 \times \mathrm{M} \times 1 \\ & \mathrm{M}_{\mathrm{NaOH}}=2 \mathrm{M} \\ & \mathrm{W}_{\mathrm{NaOH}} \text { in } 50 \mathrm{ml}=2 \times 50 \times 40 \times 10^{-3} \mathrm{~g}=4 \mathrm{g} \end{aligned}$

$\mathrm{m}=\frac{\mathrm{M} \times 1000}{\mathrm{~d}_{\text {sol }} \times 1000-\mathrm{M} \times \text { Molar mass }_{\text {solute }}}$

$815 \times 10^{-3} \mathrm{~m}$

$\begin{aligned} & M=\frac{n_{\text {solute }}}{V} \times 1000 \\ & =\frac{\left(\frac{31.4}{98}\right)}{\left(\frac{100}{1.25}\right)} \times 1000 \\ & =4.005 \approx 4 \end{aligned}$

$\begin{aligned} & \mathrm{n}_{\text {Acetan ilide }}=\mathrm{n}_{\text {Aniline }} \\ & \Rightarrow \frac{\mathrm{m}}{135}=\frac{9.3}{93} \\ & \Rightarrow \mathrm{m}=13.5 \mathrm{~g} \end{aligned}$

First, let's calculate the number of moles of NaOH that can be prepared from $84\ \text{g}$ of NaOH. The molar mass of NaOH is given as $40\ \text{g/mol}$.

The number of moles (n) is calculated using the formula:

$ n = \frac{mass}{molar\ mass} $

So for our case:

$ n = \frac{84\ \text{g}}{40\ \text{g/mol}} = 2.1\ \text{mol} $

Now that we know the number of moles, we can find out the volume of a $3\ \text{M}$ NaOH solution that can be prepared from it. The concentration (C) of a solution is related to the number of moles (n) and volume (V) by the following formula:

$ C = \frac{n}{V} $

Where: C = concentration in molarity (M) n = number of moles V = volume in liters (L) - Note that $1\ \text{dm}^3 = 1\ \text{L}$.

Since we want to find the volume (V), we can rearrange the formula to solve for V:

$ V = \frac{n}{C} $

Using the moles of NaOH and the concentration for the preparation:

$ V = \frac{2.1\ \text{mol}}{3\ \text{M}} $

Calculate the volume:

$ V = \frac{2.1}{3} = 0.7\ \text{L} $

To convert this volume to $\text{dm}^3$ (which is equivalent to liters), we use the conversion factor $1\ \text{L} = 1\ \text{dm}^3$. Therefore:

$ V = 0.7\ \text{dm}^3 $

To express this volume as $\times 10^{-1}\ \text{dm}^3$, we can write:

$ V = 7 \times 10^{-1}\ \text{dm}^3 $

Therefore, the volume of $3\ \text{M}$ NaOH solution which can be prepared from $84\ \text{g}$ of NaOH is $7 \times 10^{-1}\ \text{dm}^3$.

To solve this problem, we can use stoichiometry. First, we need to write down the balanced chemical equation for the complete combustion of methane ($\mathrm{CH}_4$).

The balanced equation for combustion of methane is:

$\mathrm{CH}_4+2 \mathrm{O}_2 \rightarrow \mathrm{CO}_2+2 \mathrm{H}_2 \mathrm{O}$

This equation tells us that one mole of methane reacts with two moles of oxygen to produce one mole of carbon dioxide and two moles of water.

Next, we should find the molar mass of methane ($\mathrm{CH}_4$) using the given molar masses:

Molar mass of $\mathrm{CH}_4$ = molar mass of C + 4 * molar mass of H

Molar mass of $\mathrm{CH}_4$ = $12.0\ \mathrm{g/mol}$ (for C) + $4 * 1.0\ \mathrm{g/mol}$ (for H)

Molar mass of $\mathrm{CH}_4$ = $12.0\ \mathrm{g/mol} + 4.0\ \mathrm{g/mol}$

Molar mass of $\mathrm{CH}_4$ = $16.0\ \mathrm{g/mol}$

Now, to determine the mass of methane required to produce $22\ \mathrm{g}$ of $\mathrm{CO}_2$, we should find out how many moles of $\mathrm{CO}_2$ there are in $22\ \mathrm{g}$ and use the mole ratio from the balanced equation to find the moles of methane required.

Moles of $\mathrm{CO}_2$ = mass of $\mathrm{CO}_2$ / molar mass of $\mathrm{CO}_2$

Molar mass of $\mathrm{CO}_2$ = molar mass of C + 2 * molar mass of O

Molar mass of $\mathrm{CO}_2$ = $12.0\ \mathrm{g/mol}$ + $2 * 16.0\ \mathrm{g/mol}$

Molar mass of $\mathrm{CO}_2$ = $12.0\ \mathrm{g/mol} + 32.0\ \mathrm{g/mol}$

Molar mass of $\mathrm{CO}_2$ = $44.0\ \mathrm{g/mol}$

Moles of $\mathrm{CO}_2$ = $\frac{22\ \mathrm{g}}{44.0\ \mathrm{g/mol}}$

Moles of $\mathrm{CO}_2$ = $0.5\ \mathrm{moles}$

We will use the stoichiometric ratio from the balanced chemical equation to find the moles of $\mathrm{CH}_4$ required to produce $0.5\ \mathrm{moles}$ of $\mathrm{CO}_2$:

$1\ \mathrm{mole\ of\ CH}_4 : 1\ \mathrm{mole\ of\ CO}_2$

This means that we also require $0.5\ \mathrm{moles}$ of $\mathrm{CH}_4$.

Finally, we need to convert moles of methane to grams to find the mass:

Mass of $\mathrm{CH}_4$ = moles of $\mathrm{CH}_4$ * molar mass of $\mathrm{CH}_4$

Mass of $\mathrm{CH}_4$ = $0.5\ \mathrm{moles} * 16.0\ \mathrm{g/mol}$

Mass of $\mathrm{CH}_4$ = $8.0\ \mathrm{g}$

Therefore, $8.0\ \mathrm{g}$ of methane is required to produce $22\ \mathrm{g}$ of carbon dioxide after complete combustion.

The given solution contains a complex ion $\left(\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_5 \mathrm{Cl}\right) \mathrm{Cl}_2$ which contains one chloride ion per complex. Therefore, when this complex is treated with silver nitrate, each mole of the complex will consume 2 moles of silver nitrate to form two moles of silver chloride.

The first step is to write the balanced chemical equation for the reaction between the complex and silver nitrate, as follows:

$\left(\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_5 \mathrm{Cl}\right) \mathrm{Cl}_2 + 2 \mathrm{AgNO}_3 \longrightarrow \left(\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_5 \mathrm{Cl}\right) + 2 \mathrm{AgCl} + 2 \mathrm{NO}_3^-$

From the equation, we can see that 1 mole of the complex consumes 2 moles of silver nitrate to form 2 moles of silver chloride. Therefore, the number of millimoles of chloride ions present in the given solution can be calculated as:

$\text{Millimoles of } \mathrm{Cl}^- \text{ ions} = \text{concentration} \times \text{volume} = 0.01\ \mathrm{M} \times 2 \times 10\ \mathrm{mL} = 0.2\ \mathrm{mmol}$

To calculate the volume of 0.1 M silver nitrate required, we can use the formula:

$\text{Millimoles of silver nitrate required} = \text{Millimoles of chloride ions} \times 2 = 0.4\ \mathrm{mmol}$

We can then use the formula:

$\text{Volume of silver nitrate solution} = \frac{\text{Millimoles of silver nitrate required}}{\text{Molarity of silver nitrate solution}}$

Substituting the values, we get:

$\text{Volume of silver nitrate solution} = \frac{0.4\ \mathrm{mmol}}{0.1\ \mathrm{M}} = 4\ \mathrm{mL}$

Therefore, the volume of 0.1 M silver nitrate required for complete precipitation of chloride ions present in 20 mL of 0.01 M $\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_5 \mathrm{Cl}\right] \mathrm{Cl}_{2}$ solution as silver chloride is $\boxed{4\ \mathrm{mL}}$.

When the carbonate reacts with HCl, it produces CO₂, as shown in the following balanced equation:

$\mathrm{M}_2\mathrm{CO}_3 + 2\mathrm{HCl} \rightarrow 2\mathrm{MCl} + \mathrm{H}_2\mathrm{O} + \mathrm{CO}_2$

From the problem, we know that 1 g of the carbonate produces 0.01 mol of CO₂.

To determine the molar mass of the carbonate, we can use the stoichiometry of the reaction:

1 mol of $\mathrm{M}_2\mathrm{CO}_3$ produces 1 mol of CO2.

So, 0.01 mol of CO₂ corresponds to 0.01 mol of $\mathrm{M}_2\mathrm{CO}_3$.

Now, we can find the molar mass of $\mathrm{M}_2\mathrm{CO}_3$:

$\frac{\text{mass of }\mathrm{M}_2\mathrm{CO}_3}{\text{moles of }\mathrm{M}_2\mathrm{CO}_3} = \text{molar mass of }\mathrm{M}_2\mathrm{CO}_3$

$\frac{1 \text{ g}}{0.01 \text{ mol}} = 100 \text{ g mol}^{-1}$

So, the molar mass of $\mathrm{M}_2\mathrm{CO}_3$ is approximately 100 g/mol.

To find the percentage of hydrogen in the compound, we first need to determine the ratio of moles of carbon and hydrogen in the products.

The moles of CO₂ produced can be calculated by dividing the mass of CO₂ produced by its molar mass (44.01 g/mol):

$\text{moles of CO}_{2} = \frac{0.220 \,\mathrm{g}}{44.01 \,\mathrm{g/mol}} = 0.005 \, \mathrm{mol}$

Since each mole of CO₂ contains one mole of carbon, there are 0.005 moles of carbon in the compound.

Now, let's calculate the moles of H₂O produced:

$\text{moles of H}_{2}\text{O} = \frac{0.126 \,\mathrm{g}}{18.02 \,\mathrm{g/mol}} = 0.007 \, \mathrm{mol}$

Since each mole of H₂O contains two moles of hydrogen, there are 0.014 moles of hydrogen in the compound.

Now, let's find the masses of carbon and hydrogen in the compound:

Mass of carbon = moles of carbon × molar mass of carbon

$\text{Mass of carbon} = 0.005 \, \mathrm{mol} \times 12.01 \,\mathrm{g/mol} = 0.060 \,\mathrm{g}$

Mass of hydrogen = moles of hydrogen × molar mass of hydrogen

$\text{Mass of hydrogen} = 0.014 \, \mathrm{mol} \times 1.008 \,\mathrm{g/mol} = 0.014 \,\mathrm{g}$

Now, we are given that the percentage of carbon is 24%. Let's find the total mass of the compound:

$\text{Total mass of compound} = \frac{\text{Mass of carbon}}{\% \text{ of carbon}} = \frac{0.060 \,\mathrm{g}}{0.24} = 0.250 \,\mathrm{g}$

Finally, let's find the percentage of hydrogen:

$\% \text{ of hydrogen} = \frac{\text{Mass of hydrogen}}{\text{Total mass of compound}} \times 100 = \frac{0.014 \,\mathrm{g}}{0.250 \,\mathrm{g}} \times 100 = 5.6$

So, the percentage of hydrogen is 5.6% or 56 × 10^-1.

Reaction with HCl:

$\mathrm{Ca}(\mathrm{OH})_2+2 \mathrm{HCl} \rightarrow \mathrm{CaCl}_2+2 \mathrm{H}_2 \mathrm{O}$

Volume of $\mathrm{Ca}(\mathrm{OH})_2=10 \mathrm{ml}$

Volume of $\mathrm{HCl}=20 \mathrm{ml}$

Concentration of $\mathrm{HCl}=0.5 \mathrm{M}$.

No. of milli moles of $\mathrm{HCl}=10$

No. of milli moles of $\mathrm{Ca}(\mathrm{OH})_2=5$.

i.e. $\mathrm{M}_{\mathrm{Ca}(\mathrm{OH})_2}=\frac{\text { no. of milli moles }}{\mathrm{V}(\mathrm{ml})}=\frac{5}{10}$ $=0.5 \mathrm{M}$.

Reaction with $\mathrm{H}_2 \mathrm{SO}_4$:

$\mathrm{Ca}(\mathrm{OH})_2+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{CaSO}_4+2 \mathrm{H}_2 \mathrm{O} \text {. }$

No. of milli moles of $\mathrm{Ca}(\mathrm{OH})_2=20 \times 0.5$ $=10$

i.e. no. of milli moles of $\mathrm{H}_2 \mathrm{SO}_4=10$

$ \begin{aligned} \Rightarrow & \mathrm{M}_{\mathrm{H}_2 \mathrm{SO}_4}=\frac{\text { no, of milli moles }}{\mathrm{V}(\mathrm{ml})} \\\\ & =\frac{10}{10} \\\\ & =1 \mathrm{M} \end{aligned} $

So, the concentration of $\mathrm{H}_2 \mathrm{SO}_4$ is 1 M.

The reaction between magnesium and hydrochloric acid is as follows :

`Mg(s) + 2HCl(aq) → MgCl<sub>2</sub>(aq) + H<sub>2</sub>(g)`

From the equation, we see that 1 mole of magnesium (Mg) produces 1 mole of hydrogen gas (H₂).

The molar mass of magnesium (Mg) is given as 24 g/mol. Therefore, 2.4 g of magnesium would correspond to (2.4 g)/(24 g/mol) = 0.1 mol of Mg.

Since 1 mole of Mg produces 1 mole of H₂, 0.1 mol of Mg would produce 0.1 mol of H₂.

At STP (Standard Temperature and Pressure), the molar volume of a gas is 22.4 L/mol.

Therefore, the volume of 0.1 mol of H₂ would be (0.1 mol)*(22.4 L/mol) = 2.24 L.

So, the volume of hydrogen liberated at STP by treating 2.4 g of magnesium with excess of hydrochloric acid is 2.24 L.

In the terms of the question where the volume is expressed as ______ × 10^-2 L, we convert 2.24 L into the desired form, i.e., 2.24 L = 2.24 $ \times $ 10² $ \times $ 10^-2 L, so the answer is 224 × 10^-2 L.

Let's first calculate the amount of sugar in each of the solutions :

1. In the 25% sugar solution, the amount of sugar is 25% of 200 g, which equals 50 g.

2. In the 40% sugar solution, the amount of sugar is 40% of 500 g, which equals 200 g.

Now, let's mix these two solutions together:

The total amount of sugar in the combined solution is 50 g (from the 25% solution) + 200 g (from the 40% solution) = 250 g.

The total weight of the combined solution is 200 g (of the 25% solution) + 500 g (of the 40% solution) = 700 g.

Therefore, the mass percentage of sugar in the combined solution is $ =\frac{250}{700} \times 100 $ = 35.71%

Rounding to the nearest integer gives 36%. So, the mass percentage of the resulting sugar solution is 36%.

Mass of Carbon $=12$

Molar Mass of $\mathrm{CO}_2=12+(16 \times 2)=44$

Mass of Compound $=0.5 \mathrm{~g}$

$ \begin{aligned} &\% \text { of } \mathrm{C} =\frac{\text { Molar mass of } \mathrm{C} \times \text { Mass Of } \mathrm{CO}_2}{\text { Mass Of Compound } \times \text { Molar Mass } \mathrm{Of} ~ \mathrm{CO}_2} \\\\ &\frac{60}{100} =\frac{12 \times x}{0.5 \times 44} \\\\ &1.1 =x \\\\ &x =11 \times 10^{-1} \end{aligned} $

Given the balanced chemical equation:

$3 \mathrm{BaCl}_2 + 2 \mathrm{Na}_3\mathrm{PO}_4 \rightarrow \mathrm{Ba}_3\mathrm{(PO}_4)_2 + 6 \mathrm{NaCl}$

We can see that 3 moles of $\mathrm{BaCl}_2$ react with 2 moles of $\mathrm{Na}_3\mathrm{PO}_4$ to produce 1 mole of $\mathrm{Ba}_3\mathrm{(PO}_4)_2$.

Given that you have 5 moles of $\mathrm{BaCl}_2$ and 2 moles of $\mathrm{Na}_3\mathrm{PO}_4$, let's calculate the maximum number of moles of $\mathrm{Ba}_3\mathrm{(PO}_4)_2$ that can be formed:

From $\mathrm{BaCl}_2$, using the molar ratio from the balanced equation:

$5 \text{ moles } \mathrm{BaCl}_2 \times \frac{1 \text{ mole } \mathrm{Ba}_3\mathrm{(PO}_4)_2}{3 \text{ moles } \mathrm{BaCl}_2} = \frac{5}{3} \text{ moles } \mathrm{Ba}_3\mathrm{(PO}_4)_2$

From $\mathrm{Na}_3\mathrm{PO}_4$, using the molar ratio from the balanced equation:

$2 \text{ moles } \mathrm{Na}_3\mathrm{PO}_4 \times \frac{1 \text{ mole } \mathrm{Ba}_3\mathrm{(PO}_4)_2}{2 \text{ moles } \mathrm{Na}_3\mathrm{PO}_4} = 1 \text{ mole } \mathrm{Ba}_3\mathrm{(PO}_4)_2$

The limiting reactant is $\mathrm{Na}_3\mathrm{PO}_4$ because it produces 1 mole of $\mathrm{Ba}_3\mathrm{(PO}_4)_2$, which is less than the $\frac{5}{3}$ moles produced by $\mathrm{BaCl}_2$.

Therefore, the maximum number of moles of $\mathrm{Ba}_3\mathrm{(PO}_4)_2$ formed is 1 mole. Rounding to the nearest integer, the answer is 1.

Mass of $10 \mathrm{~mL}$ of $\mathrm{Br}_2=10 \times 3.2=32 \mathrm{gm}$

Mass of $90 \mathrm{~mL}^{\mathrm{L}} \mathrm{CCl}_4=90 \times 1.6=144 \mathrm{gm}$

Molality of $\mathrm{Br}_2$ solution in $\mathrm{CCl}_4$

$ \begin{aligned} & =\frac{32 \times 1000}{160 \times 144} \\\\ & =1.39 \mathrm{M} \\\\ & =139 \times 10^{-2} \end{aligned} $

$\mathrm{m}=\frac{1000 \mathrm{M}}{1000 \rho-\mathrm{M} . \mathrm{m_w}}=\frac{1000 \times 3}{1000-3 \times(58.5)}$ $=\frac{3000}{(1000-175.5)}=3.638$

$=363.8 \times 10^{-2}$

Nearest integer $=364$

Limiting reagent is carbon. One mole carbon produces one mole CO. Hence, volume at STP is 227 $ \times $ 10^-1 litre.

Let, metal $\mathrm{M}$ present x and (0.83 - x) in oxidation states $+2$ and $+3$ respectively.

$ \begin{aligned} & 2 x+3(0.83-x)=2 \\\\ & x=0.49 \\\\ & \% M^{2+}=\frac{0.49}{0.83} \times 100 \\\\ & =59 \% \end{aligned} $

$ \mathrm{Zn}+2 \mathrm{HCl} \rightarrow \mathrm{ZnCl}_2+\mathrm{H}_2 \uparrow $

Moles of $\mathrm{Zn}$ used $=\frac{11.5}{65.4}=$ Moles of $\mathrm{H}_2$ evolved

Volume of $\mathrm{H}_2=\frac{11.5}{65.4} \times 22.7 \mathrm{~L}=3.99 \mathrm{~L}$