Some Basic Concepts of Chemistry

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

Weight of $\mathrm{C}$ in $0.792 \mathrm{gm} $ $\mathrm{CO}_2$

$ \begin{aligned} & =\frac{12}{44} \times 0.792=0.216 \\\\ & \% \text { of } C \text { in compound }=\frac{0.216}{0.492} \times 100 \\\\ & =43.90 \% \end{aligned} $

$\mathrm{H_2O_2\longrightarrow H_2O+\frac{1}{2}O_2}$

$\frac{50}{22.7}$

$\therefore$ Moles of $\mathrm{H_2O_2}$ in solution $=\frac{50}{22.7}\times2$

$\therefore$ Strength $=\frac{\frac{50\times2}{22.7}\times34}{1}=149.78\approx150$

Mass of $\mathrm{CHCl_3=671.141\times1.49=1000}$ gm

$2.6 \times {10^{ - 3}} = {{moles\,of\,C{H_2}C{l_2}} \over {0.671141}}$

$\Rightarrow$ moles of $C{H_2}C{l_2} = 1.74496 \times {10^{ - 3}}$

mass of $C{H_2}C{l_2} = 148.32 \times {10^{ - 3}}$ gm

Composition of $C{H_2}C{l_2} = {{148.32 \times {{10}^{ - 3}}} \over {1000}} \times {10^6}$

= 148.32 ppm

$\approx$ 148

$\mathrm{\mathop {Na + {H_2}O}\limits_{0.69\,g} \to \mathop {NaOH + {1 \over 2}{H_2}}\limits_{0.03\,moles}} $

= 0.03 moles

$\therefore 0.03=2\times\mathrm{V}$

$\mathrm{V=\frac{0.03}{2}L}$

= 15 mL

Number of moles of $C{O_2} = {{4.4} \over {44}} = 0.1$

$\therefore$ Number of moles of C in 1 mole of compound = 10

$\therefore$ $120 = {{60} \over {100}} \times (x)$ [where x is molar mass of OC]

Molar mass = 200 g mol$^{-1}$

Molar mass of a hydrocarbon $(A)=84 \mathrm{~g} / \mathrm{mol}$

Mass of carbon in $1 \mathrm{~mol}$ of $(A)=\frac{85.8}{100} \times 84$

$ =72 ~\mathrm{gm} $

Mass of hydrogen in $1 \mathrm{~mol}$ of $(A)=12 ~\mathrm{gm}$

$\therefore$ Number of $\mathrm{H}$-atoms in a molecule of $(\mathrm{A})=12$.

$ \begin{aligned} \% \text { of sulphur } & =\frac{32}{233} \times \frac{\text { weight of } \mathrm{BaSO}_4 \text { formed }}{\text { weight of organic compound }} \times 100 \\\\ & =\frac{32}{233} \times \frac{1.4439}{0.471} \times 100 \\\\ & =42.10 \end{aligned} $

Mass percent, mole fraction, molarity, ppm & molality is used to express concentration. So, the number of units = 5

Let the number of $\mathrm{Fe}^{2+}$ ions be $x$ and number of $\mathrm{Fe}^{3+}$ ions be $0.93-x$.

According to the charge neutrality,

$ \begin{aligned} & 2 x+3(0.93-x)=2 \Rightarrow 2 x-3 x+2.79=2 \Rightarrow x=0.79 \\\\ & \mathrm{Fe}^{2+} \text { ion }=x=0.79 \\\\ & \mathrm{Fe}^{3+} \text { ion }=0.93-x=0.93-0.79=0.14 \\\\ & \% \text { of } \mathrm{Fe}^{2+} \text { ion }=\frac{0.79}{0.93} \times 100=84.95 \% \approx 85 \% \\\\ & \% \text { of } \mathrm{Fe}^{3+} \text { ion }=\frac{0.14}{0.93} \times 100=15.05 \% \end{aligned} $

As it is a metal deficiency defect,

so, 1 mol of $\mathrm{Fe}_{0.93} \mathrm{O}$ contains $1 \mathrm{~mol}$ of oxide ions and $0.93 \mathrm{~mol}$ of iron ions as $\mathrm{Fe}^{2+}$ and $\mathrm{Fe}^{3+}$.

So, moles of $\mathrm{Fe}^{2+}=0.79$

and moles of $\mathrm{Fe}^{3+}=0.14$

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

$F{e^{2 + }} \to F{e^{3 + }} + {e^ - }$

For one $\mathrm{Fe}^{2+}, n$-factor $=1$

$\therefore $ For $0.79 \,\mathrm{Fe}^{2+}, n$-factor $=0.79$

Out of $0.93 \mathrm{~mol}$, there are $0.79 \mathrm{~mol} \,\mathrm{Fe}^{2+}$ ions are present.

Molarity of solution $=\frac{5}{(40)} \frac{(1000)}{(450)}$

$\Rightarrow M \times V=500 \times .1$

$\Rightarrow \frac{5}{40} \times \frac{1000}{450} \times V=500 \times 0.1$

$\mathrm{V}=180 \mathrm{~mL}$