Some Basic Concepts of Chemistry

Number of atoms $=\frac{\text { given mass }}{\text { molar mass }} \times$ atomicity $\times \mathrm{N}_{\mathrm{A}}$

A. $\frac{212}{106} \times 6 \times \mathrm{N}_{\mathrm{A}}=12 \mathrm{~N}_{\mathrm{A}}$

B. $\frac{248}{62} \times 3 \times \mathrm{N}_{\mathrm{A}}=12 \mathrm{~N}_{\mathrm{A}}$

C. $\frac{240}{40} \times 3 \times \mathrm{N}_{\mathrm{A}}=18 \mathrm{~N}_{\mathrm{A}}$

D. $\frac{12}{2} \times \mathrm{N}_{\mathrm{A}} \times 2=12 \mathrm{~N}_{\mathrm{A}}$

E. $\frac{220}{44} \times \mathrm{N}_{\mathrm{A}} \times 3=15 \mathrm{~N}_{\mathrm{A}}$

A, B and D have same number of atoms

Dalton's theory could explain the laws of chemical combination. However, it could not explain the laws of gaseous volumes.

To find the amount of glucose required to prepare $250 \, \mathrm{mL}$ of a $\frac{M}{20}$ aqueous solution, we need to follow these steps:

1. Convert the volume from milliliters to liters, because molarity is expressed in moles per liter (M).

2. Use the equation for molarity:

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

where:

• $M$ is the molarity (given as $\frac{1}{20} \, \mathrm{M}$)
• $n$ is the number of moles of solute
• $V$ is the volume of the solution in liters

3. Rearrange the equation to solve for the number of moles of glucose ($n$):

$n = M \times V$

4. Convert the moles of glucose to grams using the molar mass of glucose.

Let's proceed with the calculations:

1. Convert the volume to liters:

$250 \, \mathrm{mL} = \frac{250}{1000} \, \mathrm{L} = 0.25 \, \mathrm{L}$

2. Calculate the moles of glucose:

$n = \left( \frac{1}{20} \right) \times 0.25 \, \mathrm{L} = \frac{0.25}{20} \, \mathrm{mol} = 0.0125 \, \mathrm{mol}$

3. Convert the moles of glucose to grams using the molar mass of glucose:

$\text{mass} = n \times \text{molar mass}$

$\text{mass} = 0.0125 \, \mathrm{mol} \times 180 \, \mathrm{g} \, \mathrm{mol}^{-1} = 2.25 \, \mathrm{g}$

So, the amount of glucose required is:

Option A: 2.25 g

To determine which of the given options contains the same number of molecules as $1.0 \mathrm{~g}$ of $\mathrm{H}_2$, we need to use the concept of moles and Avogadro's number. First, let's calculate the number of moles in $1.0 \mathrm{~g}$ of $\mathrm{H}_2$.

The molar mass of $\mathrm{H}_2$ is $2 \mathrm{~g/mol}$. Therefore, the number of moles of $\mathrm{H}_2$ in $1.0 \mathrm{~g}$ can be calculated as:

$ \text{Number of moles of } \mathrm{H}_2 = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{1.0 \mathrm{~g}}{2 \mathrm{~g/mol}} = 0.5 \mathrm{~mol} $

Next, we need to find out which of the given options corresponds to the same number of moles. Let's calculate the number of moles for each option:

Option A: $14 \mathrm{~g} \text{ of } \mathrm{N}_2$.

The molar mass of $\mathrm{N}_2$ is $28 \mathrm{~g/mol}$. The number of moles of $\mathrm{N}_2$ in $14 \mathrm{~g}$ is:

$ \text{Number of moles of } \mathrm{N}_2 = \frac{14 \mathrm{~g}}{28 \mathrm{~g/mol}} = 0.5 \mathrm{~mol} $

Option B: $18 \mathrm{~g} \text{ of } \mathrm{H}_2 \mathrm{O}$.

The molar mass of $\mathrm{H}_2 \mathrm{O}$ is $18 \mathrm{~g/mol}$. The number of moles of $\mathrm{H}_2 \mathrm{O}$ in $18 \mathrm{~g}$ is:

$ \text{Number of moles of } \mathrm{H}_2 \mathrm{O} = \frac{18 \mathrm{~g}}{18 \mathrm{~g/mol}} = 1 \mathrm{~mol} $

Option C: $16 \mathrm{~g} \text{ of } \mathrm{CO}$.

The molar mass of $\mathrm{CO}$ is $28 \mathrm{~g/mol}$. The number of moles of $\mathrm{CO}$ in $16 \mathrm{~g}$ is:

$ \text{Number of moles of } \mathrm{CO} = \frac{16 \mathrm{~g}}{28 \mathrm{~g/mol}} = 0.571 \mathrm{~mol} $

Option D: $28 \mathrm{~g} \text{ of } \mathrm{N}_2$.

The molar mass of $\mathrm{N}_2$ is $28 \mathrm{~g/mol}$. The number of moles of $\mathrm{N}_2$ in $28 \mathrm{~g}$ is:

$ \text{Number of moles of } \mathrm{N}_2 = \frac{28 \mathrm{~g}}{28 \mathrm{~g/mol}} = 1 \mathrm{~mol} $

Comparing these calculated values, we can see that $0.5 \mathrm{~mol}$ of $\mathrm{N}_2$ (Option A) is equal to $0.5 \mathrm{~mol}$ of $\mathrm{H}_2$. Therefore, the correct answer is:

Option A: $14 \mathrm{~g}$ of $\mathrm{N}_2$.

To determine the percentage composition of carbon and hydrogen in the organic compound, we need to calculate the amounts of carbon in CO$_2$ and hydrogen in H$_2$O produced from the complete combustion of the compound.

First, let's calculate the amount of carbon in CO$_2$:

The molar mass of CO$_2$ is 44 g/mol, and the molar mass of carbon (C) is 12 g/mol. Given that 0.2 g of CO$_2$ is produced, we can use the following proportion to find the mass of carbon:

$\frac{12 \text{ g (C)}}{44 \text{ g (CO}_2\text{)}} = \frac{x}{0.2 \text{ g (CO}_2\text{)}}$

Solving for $x$ (mass of carbon):

$x = \frac{12 \times 0.2}{44} = \frac{2.4}{44} = 0.0545 \text{ g}$

Next, let's calculate the amount of hydrogen in H$_2$O:

The molar mass of H$_2$O is 18 g/mol, and the molar mass of hydrogen (H) in a single H$_2$O molecule is 2 g/mol. Given that 0.1 g of H$_2$O is produced, we can use the following proportion to find the mass of hydrogen:

$\frac{2 \text{ g (H)}}{18 \text{ g (H}_2\text{O})} = \frac{y}{0.1 \text{ g (H}_2\text{O})}$

Solving for $y$ (mass of hydrogen):

$y = \frac{2 \times 0.1}{18} = \frac{0.2}{18} = 0.0111 \text{ g}$

Now, we can calculate the percentage composition of carbon and hydrogen in the organic compound, which has a total mass of 0.3 g.

Percentage of carbon:

$\% \text{C} = \left( \frac{0.0545 \text{ g}}{0.3 \text{ g}} \right) \times 100 = 18.18\%$

Percentage of hydrogen:

$\% \text{H} = \left( \frac{0.0111 \text{ g}}{0.3 \text{ g}} \right) \times 100 = 3.70\%$

Therefore, the correct answer is:

Option B: 18.18% and 3.70%

To find the mass of sodium hydroxide (NaOH) left unreacted, we first need to determine the moles of NaOH and HCl originally present and compare them to see which one is in excess.

The molar mass of NaOH is approximately $40 \text{ g/mol}$, so the number of moles of NaOH in 1 gram can be calculated as follows:

$\text{Moles of NaOH} = \frac{\text{Mass of NaOH}}{\text{Molar mass of NaOH}} = \frac{1 \text{ g}}{40 \text{ g/mol}} = 0.025 \text{ moles}$

Next, we calculate the number of moles of HCl using its concentration and the volume of the solution. Recall that concentration (Molarity, M) is defined as moles of solute per liter of solution. Given that the concentration of HCl is $0.75 \text{ M}$ and the volume of the solution is $25 \text{ mL}$ or $0.025 \text{ L}$, we can find the moles of HCl:

$\text{Moles of HCl} = \text{Concentration} \times \text{Volume in liters} = 0.75 \text{ M} \times 0.025 \text{ L} = 0.01875 \text{ moles}$

Now, we compare the moles of NaOH and HCl. The stoichiometry of the reaction between NaOH and HCl is:

$\text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O}$

Each mole of NaOH reacts with one mole of HCl. Given that there are more moles of NaOH (0.025 moles) than HCl (0.01875 moles), HCl is the limiting reactant and will be completely consumed. The excess NaOH can be calculated:

$\text{Excess moles of NaOH} = \text{Moles of NaOH} - \text{Moles of HCl} = 0.025 \text{ moles} - 0.01875 \text{ moles} = 0.00625 \text{ moles}$

To find the mass of the unreacted NaOH:

$\text{Mass of unreacted NaOH} = \text{Moles of unreacted NaOH} \times \text{Molar mass of NaOH} = 0.00625 \text{ moles} \times 40 \text{ g/mol} = 0.25 \text{ g} = 250 \text{ mg}$

Therefore, the mass of sodium hydroxide left unreacted is 250 mg, which corresponds to Option B.

To determine which option contains the highest number of helium atoms, we need to analyze each option based on the amount of helium it represents and apply Avogadro's Law as required.

Option A: $4 \, \text{mol}$ of helium

Using Avogadro's number, which is approximately $6.022 \times 10^{23}$ atoms per mole, the number of helium atoms in 4 moles can be calculated as:

$4 \, \text{mol} \times 6.022 \times 10^{23} \, \text{atoms/mol} = 24.088 \times 10^{23} \, \text{atoms}$

Option B: $4 \, \text{u}$ of helium

The atomic mass of helium is approximately 4 u (atomic mass units). Therefore, $4 \, \text{u}$ represents about 1 mole of helium atoms (since the molar mass of helium is approximately 4 g/mol, which equals 4 u). Thus, this option represents:

$1 \, \text{mol} \times 6.022 \times 10^{23} \, \text{atoms/mol} = 6.022 \times 10^{23} \, \text{atoms}$

Option C: $4 \, \text{g}$ of helium

Similarly, as we've established that the molar mass of helium is 4 g/mol, $4 \, \text{g}$ of helium equates exactly to:

$1 \, \text{mol} \times 6.022 \times 10^{23} \, \text{atoms/mol} = 6.022 \times 10^{23} \, \text{atoms}$

Option D: $2.271098 \, \text{L}$ of helium at STP (Standard Temperature and Pressure)

At STP, one mole of any ideal gas occupies 22.4 L. Therefore, the amount of helium in moles for $2.271098 \, \text{L}$ can be derived from:

$\frac{2.271098 \, \text{L}}{22.4 \, \text{L/mol}} \approx 0.101 \, \text{mole}$

Using this to find the number of atoms:

$0.101 \, \text{mol} \times 6.022 \times 10^{23} \, \text{atoms/mol} \approx 6.082 \times 10^{22} \, \text{atoms}$

Conclusion:

Comparing the numbers:

• Option A: $24.088 \times 10^{23} \, \text{atoms}$


• Option B: $6.022 \times 10^{23} \, \text{atoms}$


• Option C: $6.022 \times 10^{23} \, \text{atoms}$


• Option D: $6.082 \times 10^{22} \, \text{atoms}$

Option A clearly contains the highest number of helium atoms, which is $24.088 \times 10^{23} \, \text{atoms}.$

So, empirical formula of

$ \therefore \text { The correct empirical formula of compound } \mathrm{X} \text { is } \mathrm{ABC}_3 $

$\begin{aligned} & \mathrm{m}=\frac{1000 \times \mathrm{M}}{1000 \times \mathrm{d}-\mathrm{MM}_{\mathrm{w}}} \\ & \mathrm{m}=\frac{1000 \times 1}{1000 \times 1.25-1 \times 85} \\ & \mathrm{~m}=\frac{1000}{1165}=0.858\end{aligned}$

Weight of impure limestone $=20 \mathrm{~g}$

Weight of pure limestone $\left(\mathrm{CaCO}_{3}\right)=20 \%$ of $20 \mathrm{~g}$

$=\frac{20}{100} \times 20$

$=4 \mathrm{~g}$

$\mathrm{n}_{\mathrm{CaCO}_{3}}=\frac{4}{100}=0.04$

Image

$\mathrm{n}_{\mathrm{CO}_{2}}=0.04$

$\mathrm{W}_{\mathrm{CO}_{2}}=0.04 \times 44$

$=1.76 \mathrm{~g}$

Mass = Volume $\times$ Density

= 2.5 $\times$ 2.15

= 5.375 g

Since 2.5 has two significant figures, so the mass of solution in correct significant figures will be 5.4 g.

Fe_0.96O

Let Fe(II) present in Fe_0.96O = x

Fe(III) present = (0.96 $-$ x)

Total charge on Fe = 2x + (0.96 $-$ x)3

Total charge on O = $-$2

2x + (0.96 $-$ x)3 = 2

2x + 2.88 $-$ 3x = 2

$-$x = $-$0.88

x = 0.88

Fe²⁺ = 0.88, Fe³⁺ = 0.08

Fraction of Fe³⁺ = ${{0.08} \over {0.96}} = {1 \over {12}}$

Molality is the moles of solute dissolved per kg of solvent therefore 500 g, 1 molal solution contains 0.5 of solute, as

$m = {{Moles\,of\,solute} \over {Mass\,of\,solvent\,(in\,kg)}}$

$1 = {{0.5} \over {Mass\,of\,solvent\,(in\,kg)}}$

$\therefore$ Mass of solvent (in kg) = 0.5

= 500 g

Let m gram mass of CaCO₃ is required

Pure CaCO₃ in m gram $ = {{95} \over {100}} \times m$

Moles of CaCO₃ $ = {{95} \over {100}} \times {m \over {100}}$

Moles of HCl required = 2 $\times$ moles of CaCO₃

$ = 2 \times {{95} \over {100}} \times {m \over {100}}$

$2 \times {{95} \over {100}} \times {m \over {100}} = {{50} \over {1000}} \times 0.5$

$m = 1.315\,g \approx 1.32\,g$