228333
What is the weight of oxygen required for the complete combustion of $2.8 \mathrm{~kg}$ of ethylene?
1 $2.8 \mathrm{~kg}$
2 $6.4 \mathrm{~kg}$
3 $9.6 \mathrm{~kg}$
4 $96 \mathrm{~kg}$
Explanation:
$\mathrm{C}_2 \mathrm{H}_4 (ethylene) +3 \mathrm{O}_2 \longrightarrow 2 \mathrm{CO}_2+2 \mathrm{H}_2 \mathrm{O}$ To oxidise $1 \mathrm{~mol}$ of ethylene we required 3 moles of oxygen. Then, For oxidising $28 \mathrm{~g}$ of $\mathrm{C}_2 \mathrm{H}_4$, We need $3 \times 32=96 \mathrm{~g}$ of oxygen. For $2.8 \mathrm{~kg}$ of $\mathrm{C}_2 \mathrm{H}_4$, We need $=\frac{96}{28} \times 2.8=9.6 \mathrm{~kg}$ of oxygen.
NEET-1989
Some Basic Concepts of Chemistry
228335
A commercially cold conc. $\mathrm{HCl}$ is $35 \% \mathrm{HCl}$ by mass. If the density of this commercial acid is $1.46 \mathrm{~g} / \mathrm{mL}$, the molarity of this solution is: (Atomic mass: $\mathrm{Cl}=35.5 \mathrm{amu}, \mathrm{H}=1 \mathrm{amu}$ )
1 $10.2 \mathrm{M}$
2 $12.5 \mathrm{M}$
3 $14.0 \mathrm{M}$
4 $18.2 \mathrm{M}$
Explanation:
Given, Concentration of $\mathrm{HCl}=35 \%$ by mass Density $=1.46 \mathrm{~g} / \mathrm{mL}$ We know that- $\begin{aligned} \text { Molarity } & =\frac{(\mathrm{W} / \mathrm{W} \%) \times \mathrm{d} \times 10}{(\text { Molar mass })_{\text {solute }}} \\ & =\frac{35 \times 1.46 \times 10}{36.5} \\ & =14.0 \mathrm{M} \end{aligned}$
Shift-I
Some Basic Concepts of Chemistry
228336
Which one of the following contains the highest number of oxygen atoms?
1 One mole of aluminum sulphate
2 Three moles of ferrous sulphate $=3 \mathrm{FeSO}_4$ It contain 12 mole of oxygen $\therefore 12$ mole of oxygen $=12 \mathrm{~N}_{\mathrm{A}}$ atoms.
3 Three moles of hydrogen peroxide $=3 \mathrm{H}_2 \mathrm{O}_2$ It contain 6 mole of oxygen $\therefore 6$ mole of oxygen $=6 \mathrm{~N}_{\mathrm{A}}$ atoms
4 Two moles of potassium permanganate $=2 \mathrm{KMnO}_4$ It contain 8 mole of oxygen $\therefore 8$ mole of oxygen $=8 \mathrm{~N}_{\mathrm{A}}$ atoms
5 One mole of potassium dichromate $=\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ It contain 7 mole of oxygen $\therefore 7$ mole of oxygen $=7 \mathrm{~N}_{\mathrm{A}}$ atoms Here, $\mathrm{N}_{\mathrm{A}}=$ Avogadro number So, option (a) one mole of aluminum sulphate contains the highest number of oxygen atoms.
Explanation:
One mole of aluminum sulphate $=\mathrm{Al}_2\left(\mathrm{SO}_4\right)_3$ It contain 12 mole of oxygen 1 mole of oxygen $=\mathrm{N}_{\mathrm{A}}$ atoms $\therefore 12$ mole of oxygen $=12 \mathrm{~N}_{\mathrm{A}}$ atoms (b.) Three moles of ferrous sulphate $=3 \mathrm{FeSO}_4$ It contain 12 mole of oxygen $\therefore 12$ mole of oxygen $=12 \mathrm{~N}_{\mathrm{A}}$ atoms. (c.) Three moles of hydrogen peroxide $=3 \mathrm{H}_2 \mathrm{O}_2$ It contain 6 mole of oxygen $\therefore 6$ mole of oxygen $=6 \mathrm{~N}_{\mathrm{A}}$ atoms (d.) Two moles of potassium permanganate $=2 \mathrm{KMnO}_4$ It contain 8 mole of oxygen $\therefore 8$ mole of oxygen $=8 \mathrm{~N}_{\mathrm{A}}$ atoms (e.) One mole of potassium dichromate $=\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ It contain 7 mole of oxygen $\therefore 7$ mole of oxygen $=7 \mathrm{~N}_{\mathrm{A}}$ atoms Here, $\mathrm{N}_{\mathrm{A}}=$ Avogadro number So, option (a) one mole of aluminum sulphate contains the highest number of oxygen atoms.
Kerala CEE -03.07.2022
Some Basic Concepts of Chemistry
228338
If a rocket runs on a fuel $\left(\mathrm{C}_{15} \mathrm{H}_{30}\right)$ and liquid oxygen, the weight of oxygen required and $\mathrm{CO}_2$ released for every litre of fuel respectively are: (Given: density of the fuel is $0.756 \mathrm{~g} / \mathrm{mL}$ )
1 $1188 \mathrm{~g}$ and $1296 \mathrm{~g}$
2 $2376 \mathrm{~g}$ and $2592 \mathrm{~g}$
3 $2592 \mathrm{~g}$ and $2376 \mathrm{~g}$
4 $3429 \mathrm{~g}$ and $3142 \mathrm{~g}$
Explanation:
Given, Density of fuel $=0.756 \mathrm{~g} / \mathrm{mL}$ Molecular formula of fuel $=\mathrm{C}_{15} \mathrm{H}_{30}$ Then, molar mass of fuel $=15 \times 12+30 \times 1$ $\begin{array}{r}=210 \mathrm{~g}\end{array}$ $\begin{aligned} \text { Density }=\frac{\text { mass }}{\text { volume }}\end{aligned}$ $\begin{aligned} & \text { Mass }=\text { density } \times \text { volume } \\ &=0.756 \times 1000=756 \mathrm{gm} .\end{aligned}$ The reaction is - $\mathrm{C}_{15} \mathrm{H}_{30}(\mathrm{l})+\frac{45}{2} \mathrm{O}_2(\mathrm{~g}) \longrightarrow 15 \mathrm{CO}_2(\mathrm{~g})+15 \mathrm{H}_2 \mathrm{O}$ Weight of oxygen $=\frac{45}{2}\left[\frac{756}{210}\right] \times 32=2592 \mathrm{~g}$ Weight of $\mathrm{CO}_2=15\left[\frac{756}{210}\right] \times 44=2376 \mathrm{~g}$.
228333
What is the weight of oxygen required for the complete combustion of $2.8 \mathrm{~kg}$ of ethylene?
1 $2.8 \mathrm{~kg}$
2 $6.4 \mathrm{~kg}$
3 $9.6 \mathrm{~kg}$
4 $96 \mathrm{~kg}$
Explanation:
$\mathrm{C}_2 \mathrm{H}_4 (ethylene) +3 \mathrm{O}_2 \longrightarrow 2 \mathrm{CO}_2+2 \mathrm{H}_2 \mathrm{O}$ To oxidise $1 \mathrm{~mol}$ of ethylene we required 3 moles of oxygen. Then, For oxidising $28 \mathrm{~g}$ of $\mathrm{C}_2 \mathrm{H}_4$, We need $3 \times 32=96 \mathrm{~g}$ of oxygen. For $2.8 \mathrm{~kg}$ of $\mathrm{C}_2 \mathrm{H}_4$, We need $=\frac{96}{28} \times 2.8=9.6 \mathrm{~kg}$ of oxygen.
NEET-1989
Some Basic Concepts of Chemistry
228335
A commercially cold conc. $\mathrm{HCl}$ is $35 \% \mathrm{HCl}$ by mass. If the density of this commercial acid is $1.46 \mathrm{~g} / \mathrm{mL}$, the molarity of this solution is: (Atomic mass: $\mathrm{Cl}=35.5 \mathrm{amu}, \mathrm{H}=1 \mathrm{amu}$ )
1 $10.2 \mathrm{M}$
2 $12.5 \mathrm{M}$
3 $14.0 \mathrm{M}$
4 $18.2 \mathrm{M}$
Explanation:
Given, Concentration of $\mathrm{HCl}=35 \%$ by mass Density $=1.46 \mathrm{~g} / \mathrm{mL}$ We know that- $\begin{aligned} \text { Molarity } & =\frac{(\mathrm{W} / \mathrm{W} \%) \times \mathrm{d} \times 10}{(\text { Molar mass })_{\text {solute }}} \\ & =\frac{35 \times 1.46 \times 10}{36.5} \\ & =14.0 \mathrm{M} \end{aligned}$
Shift-I
Some Basic Concepts of Chemistry
228336
Which one of the following contains the highest number of oxygen atoms?
1 One mole of aluminum sulphate
2 Three moles of ferrous sulphate $=3 \mathrm{FeSO}_4$ It contain 12 mole of oxygen $\therefore 12$ mole of oxygen $=12 \mathrm{~N}_{\mathrm{A}}$ atoms.
3 Three moles of hydrogen peroxide $=3 \mathrm{H}_2 \mathrm{O}_2$ It contain 6 mole of oxygen $\therefore 6$ mole of oxygen $=6 \mathrm{~N}_{\mathrm{A}}$ atoms
4 Two moles of potassium permanganate $=2 \mathrm{KMnO}_4$ It contain 8 mole of oxygen $\therefore 8$ mole of oxygen $=8 \mathrm{~N}_{\mathrm{A}}$ atoms
5 One mole of potassium dichromate $=\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ It contain 7 mole of oxygen $\therefore 7$ mole of oxygen $=7 \mathrm{~N}_{\mathrm{A}}$ atoms Here, $\mathrm{N}_{\mathrm{A}}=$ Avogadro number So, option (a) one mole of aluminum sulphate contains the highest number of oxygen atoms.
Explanation:
One mole of aluminum sulphate $=\mathrm{Al}_2\left(\mathrm{SO}_4\right)_3$ It contain 12 mole of oxygen 1 mole of oxygen $=\mathrm{N}_{\mathrm{A}}$ atoms $\therefore 12$ mole of oxygen $=12 \mathrm{~N}_{\mathrm{A}}$ atoms (b.) Three moles of ferrous sulphate $=3 \mathrm{FeSO}_4$ It contain 12 mole of oxygen $\therefore 12$ mole of oxygen $=12 \mathrm{~N}_{\mathrm{A}}$ atoms. (c.) Three moles of hydrogen peroxide $=3 \mathrm{H}_2 \mathrm{O}_2$ It contain 6 mole of oxygen $\therefore 6$ mole of oxygen $=6 \mathrm{~N}_{\mathrm{A}}$ atoms (d.) Two moles of potassium permanganate $=2 \mathrm{KMnO}_4$ It contain 8 mole of oxygen $\therefore 8$ mole of oxygen $=8 \mathrm{~N}_{\mathrm{A}}$ atoms (e.) One mole of potassium dichromate $=\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ It contain 7 mole of oxygen $\therefore 7$ mole of oxygen $=7 \mathrm{~N}_{\mathrm{A}}$ atoms Here, $\mathrm{N}_{\mathrm{A}}=$ Avogadro number So, option (a) one mole of aluminum sulphate contains the highest number of oxygen atoms.
Kerala CEE -03.07.2022
Some Basic Concepts of Chemistry
228338
If a rocket runs on a fuel $\left(\mathrm{C}_{15} \mathrm{H}_{30}\right)$ and liquid oxygen, the weight of oxygen required and $\mathrm{CO}_2$ released for every litre of fuel respectively are: (Given: density of the fuel is $0.756 \mathrm{~g} / \mathrm{mL}$ )
1 $1188 \mathrm{~g}$ and $1296 \mathrm{~g}$
2 $2376 \mathrm{~g}$ and $2592 \mathrm{~g}$
3 $2592 \mathrm{~g}$ and $2376 \mathrm{~g}$
4 $3429 \mathrm{~g}$ and $3142 \mathrm{~g}$
Explanation:
Given, Density of fuel $=0.756 \mathrm{~g} / \mathrm{mL}$ Molecular formula of fuel $=\mathrm{C}_{15} \mathrm{H}_{30}$ Then, molar mass of fuel $=15 \times 12+30 \times 1$ $\begin{array}{r}=210 \mathrm{~g}\end{array}$ $\begin{aligned} \text { Density }=\frac{\text { mass }}{\text { volume }}\end{aligned}$ $\begin{aligned} & \text { Mass }=\text { density } \times \text { volume } \\ &=0.756 \times 1000=756 \mathrm{gm} .\end{aligned}$ The reaction is - $\mathrm{C}_{15} \mathrm{H}_{30}(\mathrm{l})+\frac{45}{2} \mathrm{O}_2(\mathrm{~g}) \longrightarrow 15 \mathrm{CO}_2(\mathrm{~g})+15 \mathrm{H}_2 \mathrm{O}$ Weight of oxygen $=\frac{45}{2}\left[\frac{756}{210}\right] \times 32=2592 \mathrm{~g}$ Weight of $\mathrm{CO}_2=15\left[\frac{756}{210}\right] \times 44=2376 \mathrm{~g}$.
228333
What is the weight of oxygen required for the complete combustion of $2.8 \mathrm{~kg}$ of ethylene?
1 $2.8 \mathrm{~kg}$
2 $6.4 \mathrm{~kg}$
3 $9.6 \mathrm{~kg}$
4 $96 \mathrm{~kg}$
Explanation:
$\mathrm{C}_2 \mathrm{H}_4 (ethylene) +3 \mathrm{O}_2 \longrightarrow 2 \mathrm{CO}_2+2 \mathrm{H}_2 \mathrm{O}$ To oxidise $1 \mathrm{~mol}$ of ethylene we required 3 moles of oxygen. Then, For oxidising $28 \mathrm{~g}$ of $\mathrm{C}_2 \mathrm{H}_4$, We need $3 \times 32=96 \mathrm{~g}$ of oxygen. For $2.8 \mathrm{~kg}$ of $\mathrm{C}_2 \mathrm{H}_4$, We need $=\frac{96}{28} \times 2.8=9.6 \mathrm{~kg}$ of oxygen.
NEET-1989
Some Basic Concepts of Chemistry
228335
A commercially cold conc. $\mathrm{HCl}$ is $35 \% \mathrm{HCl}$ by mass. If the density of this commercial acid is $1.46 \mathrm{~g} / \mathrm{mL}$, the molarity of this solution is: (Atomic mass: $\mathrm{Cl}=35.5 \mathrm{amu}, \mathrm{H}=1 \mathrm{amu}$ )
1 $10.2 \mathrm{M}$
2 $12.5 \mathrm{M}$
3 $14.0 \mathrm{M}$
4 $18.2 \mathrm{M}$
Explanation:
Given, Concentration of $\mathrm{HCl}=35 \%$ by mass Density $=1.46 \mathrm{~g} / \mathrm{mL}$ We know that- $\begin{aligned} \text { Molarity } & =\frac{(\mathrm{W} / \mathrm{W} \%) \times \mathrm{d} \times 10}{(\text { Molar mass })_{\text {solute }}} \\ & =\frac{35 \times 1.46 \times 10}{36.5} \\ & =14.0 \mathrm{M} \end{aligned}$
Shift-I
Some Basic Concepts of Chemistry
228336
Which one of the following contains the highest number of oxygen atoms?
1 One mole of aluminum sulphate
2 Three moles of ferrous sulphate $=3 \mathrm{FeSO}_4$ It contain 12 mole of oxygen $\therefore 12$ mole of oxygen $=12 \mathrm{~N}_{\mathrm{A}}$ atoms.
3 Three moles of hydrogen peroxide $=3 \mathrm{H}_2 \mathrm{O}_2$ It contain 6 mole of oxygen $\therefore 6$ mole of oxygen $=6 \mathrm{~N}_{\mathrm{A}}$ atoms
4 Two moles of potassium permanganate $=2 \mathrm{KMnO}_4$ It contain 8 mole of oxygen $\therefore 8$ mole of oxygen $=8 \mathrm{~N}_{\mathrm{A}}$ atoms
5 One mole of potassium dichromate $=\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ It contain 7 mole of oxygen $\therefore 7$ mole of oxygen $=7 \mathrm{~N}_{\mathrm{A}}$ atoms Here, $\mathrm{N}_{\mathrm{A}}=$ Avogadro number So, option (a) one mole of aluminum sulphate contains the highest number of oxygen atoms.
Explanation:
One mole of aluminum sulphate $=\mathrm{Al}_2\left(\mathrm{SO}_4\right)_3$ It contain 12 mole of oxygen 1 mole of oxygen $=\mathrm{N}_{\mathrm{A}}$ atoms $\therefore 12$ mole of oxygen $=12 \mathrm{~N}_{\mathrm{A}}$ atoms (b.) Three moles of ferrous sulphate $=3 \mathrm{FeSO}_4$ It contain 12 mole of oxygen $\therefore 12$ mole of oxygen $=12 \mathrm{~N}_{\mathrm{A}}$ atoms. (c.) Three moles of hydrogen peroxide $=3 \mathrm{H}_2 \mathrm{O}_2$ It contain 6 mole of oxygen $\therefore 6$ mole of oxygen $=6 \mathrm{~N}_{\mathrm{A}}$ atoms (d.) Two moles of potassium permanganate $=2 \mathrm{KMnO}_4$ It contain 8 mole of oxygen $\therefore 8$ mole of oxygen $=8 \mathrm{~N}_{\mathrm{A}}$ atoms (e.) One mole of potassium dichromate $=\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ It contain 7 mole of oxygen $\therefore 7$ mole of oxygen $=7 \mathrm{~N}_{\mathrm{A}}$ atoms Here, $\mathrm{N}_{\mathrm{A}}=$ Avogadro number So, option (a) one mole of aluminum sulphate contains the highest number of oxygen atoms.
Kerala CEE -03.07.2022
Some Basic Concepts of Chemistry
228338
If a rocket runs on a fuel $\left(\mathrm{C}_{15} \mathrm{H}_{30}\right)$ and liquid oxygen, the weight of oxygen required and $\mathrm{CO}_2$ released for every litre of fuel respectively are: (Given: density of the fuel is $0.756 \mathrm{~g} / \mathrm{mL}$ )
1 $1188 \mathrm{~g}$ and $1296 \mathrm{~g}$
2 $2376 \mathrm{~g}$ and $2592 \mathrm{~g}$
3 $2592 \mathrm{~g}$ and $2376 \mathrm{~g}$
4 $3429 \mathrm{~g}$ and $3142 \mathrm{~g}$
Explanation:
Given, Density of fuel $=0.756 \mathrm{~g} / \mathrm{mL}$ Molecular formula of fuel $=\mathrm{C}_{15} \mathrm{H}_{30}$ Then, molar mass of fuel $=15 \times 12+30 \times 1$ $\begin{array}{r}=210 \mathrm{~g}\end{array}$ $\begin{aligned} \text { Density }=\frac{\text { mass }}{\text { volume }}\end{aligned}$ $\begin{aligned} & \text { Mass }=\text { density } \times \text { volume } \\ &=0.756 \times 1000=756 \mathrm{gm} .\end{aligned}$ The reaction is - $\mathrm{C}_{15} \mathrm{H}_{30}(\mathrm{l})+\frac{45}{2} \mathrm{O}_2(\mathrm{~g}) \longrightarrow 15 \mathrm{CO}_2(\mathrm{~g})+15 \mathrm{H}_2 \mathrm{O}$ Weight of oxygen $=\frac{45}{2}\left[\frac{756}{210}\right] \times 32=2592 \mathrm{~g}$ Weight of $\mathrm{CO}_2=15\left[\frac{756}{210}\right] \times 44=2376 \mathrm{~g}$.
228333
What is the weight of oxygen required for the complete combustion of $2.8 \mathrm{~kg}$ of ethylene?
1 $2.8 \mathrm{~kg}$
2 $6.4 \mathrm{~kg}$
3 $9.6 \mathrm{~kg}$
4 $96 \mathrm{~kg}$
Explanation:
$\mathrm{C}_2 \mathrm{H}_4 (ethylene) +3 \mathrm{O}_2 \longrightarrow 2 \mathrm{CO}_2+2 \mathrm{H}_2 \mathrm{O}$ To oxidise $1 \mathrm{~mol}$ of ethylene we required 3 moles of oxygen. Then, For oxidising $28 \mathrm{~g}$ of $\mathrm{C}_2 \mathrm{H}_4$, We need $3 \times 32=96 \mathrm{~g}$ of oxygen. For $2.8 \mathrm{~kg}$ of $\mathrm{C}_2 \mathrm{H}_4$, We need $=\frac{96}{28} \times 2.8=9.6 \mathrm{~kg}$ of oxygen.
NEET-1989
Some Basic Concepts of Chemistry
228335
A commercially cold conc. $\mathrm{HCl}$ is $35 \% \mathrm{HCl}$ by mass. If the density of this commercial acid is $1.46 \mathrm{~g} / \mathrm{mL}$, the molarity of this solution is: (Atomic mass: $\mathrm{Cl}=35.5 \mathrm{amu}, \mathrm{H}=1 \mathrm{amu}$ )
1 $10.2 \mathrm{M}$
2 $12.5 \mathrm{M}$
3 $14.0 \mathrm{M}$
4 $18.2 \mathrm{M}$
Explanation:
Given, Concentration of $\mathrm{HCl}=35 \%$ by mass Density $=1.46 \mathrm{~g} / \mathrm{mL}$ We know that- $\begin{aligned} \text { Molarity } & =\frac{(\mathrm{W} / \mathrm{W} \%) \times \mathrm{d} \times 10}{(\text { Molar mass })_{\text {solute }}} \\ & =\frac{35 \times 1.46 \times 10}{36.5} \\ & =14.0 \mathrm{M} \end{aligned}$
Shift-I
Some Basic Concepts of Chemistry
228336
Which one of the following contains the highest number of oxygen atoms?
1 One mole of aluminum sulphate
2 Three moles of ferrous sulphate $=3 \mathrm{FeSO}_4$ It contain 12 mole of oxygen $\therefore 12$ mole of oxygen $=12 \mathrm{~N}_{\mathrm{A}}$ atoms.
3 Three moles of hydrogen peroxide $=3 \mathrm{H}_2 \mathrm{O}_2$ It contain 6 mole of oxygen $\therefore 6$ mole of oxygen $=6 \mathrm{~N}_{\mathrm{A}}$ atoms
4 Two moles of potassium permanganate $=2 \mathrm{KMnO}_4$ It contain 8 mole of oxygen $\therefore 8$ mole of oxygen $=8 \mathrm{~N}_{\mathrm{A}}$ atoms
5 One mole of potassium dichromate $=\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ It contain 7 mole of oxygen $\therefore 7$ mole of oxygen $=7 \mathrm{~N}_{\mathrm{A}}$ atoms Here, $\mathrm{N}_{\mathrm{A}}=$ Avogadro number So, option (a) one mole of aluminum sulphate contains the highest number of oxygen atoms.
Explanation:
One mole of aluminum sulphate $=\mathrm{Al}_2\left(\mathrm{SO}_4\right)_3$ It contain 12 mole of oxygen 1 mole of oxygen $=\mathrm{N}_{\mathrm{A}}$ atoms $\therefore 12$ mole of oxygen $=12 \mathrm{~N}_{\mathrm{A}}$ atoms (b.) Three moles of ferrous sulphate $=3 \mathrm{FeSO}_4$ It contain 12 mole of oxygen $\therefore 12$ mole of oxygen $=12 \mathrm{~N}_{\mathrm{A}}$ atoms. (c.) Three moles of hydrogen peroxide $=3 \mathrm{H}_2 \mathrm{O}_2$ It contain 6 mole of oxygen $\therefore 6$ mole of oxygen $=6 \mathrm{~N}_{\mathrm{A}}$ atoms (d.) Two moles of potassium permanganate $=2 \mathrm{KMnO}_4$ It contain 8 mole of oxygen $\therefore 8$ mole of oxygen $=8 \mathrm{~N}_{\mathrm{A}}$ atoms (e.) One mole of potassium dichromate $=\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ It contain 7 mole of oxygen $\therefore 7$ mole of oxygen $=7 \mathrm{~N}_{\mathrm{A}}$ atoms Here, $\mathrm{N}_{\mathrm{A}}=$ Avogadro number So, option (a) one mole of aluminum sulphate contains the highest number of oxygen atoms.
Kerala CEE -03.07.2022
Some Basic Concepts of Chemistry
228338
If a rocket runs on a fuel $\left(\mathrm{C}_{15} \mathrm{H}_{30}\right)$ and liquid oxygen, the weight of oxygen required and $\mathrm{CO}_2$ released for every litre of fuel respectively are: (Given: density of the fuel is $0.756 \mathrm{~g} / \mathrm{mL}$ )
1 $1188 \mathrm{~g}$ and $1296 \mathrm{~g}$
2 $2376 \mathrm{~g}$ and $2592 \mathrm{~g}$
3 $2592 \mathrm{~g}$ and $2376 \mathrm{~g}$
4 $3429 \mathrm{~g}$ and $3142 \mathrm{~g}$
Explanation:
Given, Density of fuel $=0.756 \mathrm{~g} / \mathrm{mL}$ Molecular formula of fuel $=\mathrm{C}_{15} \mathrm{H}_{30}$ Then, molar mass of fuel $=15 \times 12+30 \times 1$ $\begin{array}{r}=210 \mathrm{~g}\end{array}$ $\begin{aligned} \text { Density }=\frac{\text { mass }}{\text { volume }}\end{aligned}$ $\begin{aligned} & \text { Mass }=\text { density } \times \text { volume } \\ &=0.756 \times 1000=756 \mathrm{gm} .\end{aligned}$ The reaction is - $\mathrm{C}_{15} \mathrm{H}_{30}(\mathrm{l})+\frac{45}{2} \mathrm{O}_2(\mathrm{~g}) \longrightarrow 15 \mathrm{CO}_2(\mathrm{~g})+15 \mathrm{H}_2 \mathrm{O}$ Weight of oxygen $=\frac{45}{2}\left[\frac{756}{210}\right] \times 32=2592 \mathrm{~g}$ Weight of $\mathrm{CO}_2=15\left[\frac{756}{210}\right] \times 44=2376 \mathrm{~g}$.
228333
What is the weight of oxygen required for the complete combustion of $2.8 \mathrm{~kg}$ of ethylene?
1 $2.8 \mathrm{~kg}$
2 $6.4 \mathrm{~kg}$
3 $9.6 \mathrm{~kg}$
4 $96 \mathrm{~kg}$
Explanation:
$\mathrm{C}_2 \mathrm{H}_4 (ethylene) +3 \mathrm{O}_2 \longrightarrow 2 \mathrm{CO}_2+2 \mathrm{H}_2 \mathrm{O}$ To oxidise $1 \mathrm{~mol}$ of ethylene we required 3 moles of oxygen. Then, For oxidising $28 \mathrm{~g}$ of $\mathrm{C}_2 \mathrm{H}_4$, We need $3 \times 32=96 \mathrm{~g}$ of oxygen. For $2.8 \mathrm{~kg}$ of $\mathrm{C}_2 \mathrm{H}_4$, We need $=\frac{96}{28} \times 2.8=9.6 \mathrm{~kg}$ of oxygen.
NEET-1989
Some Basic Concepts of Chemistry
228335
A commercially cold conc. $\mathrm{HCl}$ is $35 \% \mathrm{HCl}$ by mass. If the density of this commercial acid is $1.46 \mathrm{~g} / \mathrm{mL}$, the molarity of this solution is: (Atomic mass: $\mathrm{Cl}=35.5 \mathrm{amu}, \mathrm{H}=1 \mathrm{amu}$ )
1 $10.2 \mathrm{M}$
2 $12.5 \mathrm{M}$
3 $14.0 \mathrm{M}$
4 $18.2 \mathrm{M}$
Explanation:
Given, Concentration of $\mathrm{HCl}=35 \%$ by mass Density $=1.46 \mathrm{~g} / \mathrm{mL}$ We know that- $\begin{aligned} \text { Molarity } & =\frac{(\mathrm{W} / \mathrm{W} \%) \times \mathrm{d} \times 10}{(\text { Molar mass })_{\text {solute }}} \\ & =\frac{35 \times 1.46 \times 10}{36.5} \\ & =14.0 \mathrm{M} \end{aligned}$
Shift-I
Some Basic Concepts of Chemistry
228336
Which one of the following contains the highest number of oxygen atoms?
1 One mole of aluminum sulphate
2 Three moles of ferrous sulphate $=3 \mathrm{FeSO}_4$ It contain 12 mole of oxygen $\therefore 12$ mole of oxygen $=12 \mathrm{~N}_{\mathrm{A}}$ atoms.
3 Three moles of hydrogen peroxide $=3 \mathrm{H}_2 \mathrm{O}_2$ It contain 6 mole of oxygen $\therefore 6$ mole of oxygen $=6 \mathrm{~N}_{\mathrm{A}}$ atoms
4 Two moles of potassium permanganate $=2 \mathrm{KMnO}_4$ It contain 8 mole of oxygen $\therefore 8$ mole of oxygen $=8 \mathrm{~N}_{\mathrm{A}}$ atoms
5 One mole of potassium dichromate $=\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ It contain 7 mole of oxygen $\therefore 7$ mole of oxygen $=7 \mathrm{~N}_{\mathrm{A}}$ atoms Here, $\mathrm{N}_{\mathrm{A}}=$ Avogadro number So, option (a) one mole of aluminum sulphate contains the highest number of oxygen atoms.
Explanation:
One mole of aluminum sulphate $=\mathrm{Al}_2\left(\mathrm{SO}_4\right)_3$ It contain 12 mole of oxygen 1 mole of oxygen $=\mathrm{N}_{\mathrm{A}}$ atoms $\therefore 12$ mole of oxygen $=12 \mathrm{~N}_{\mathrm{A}}$ atoms (b.) Three moles of ferrous sulphate $=3 \mathrm{FeSO}_4$ It contain 12 mole of oxygen $\therefore 12$ mole of oxygen $=12 \mathrm{~N}_{\mathrm{A}}$ atoms. (c.) Three moles of hydrogen peroxide $=3 \mathrm{H}_2 \mathrm{O}_2$ It contain 6 mole of oxygen $\therefore 6$ mole of oxygen $=6 \mathrm{~N}_{\mathrm{A}}$ atoms (d.) Two moles of potassium permanganate $=2 \mathrm{KMnO}_4$ It contain 8 mole of oxygen $\therefore 8$ mole of oxygen $=8 \mathrm{~N}_{\mathrm{A}}$ atoms (e.) One mole of potassium dichromate $=\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ It contain 7 mole of oxygen $\therefore 7$ mole of oxygen $=7 \mathrm{~N}_{\mathrm{A}}$ atoms Here, $\mathrm{N}_{\mathrm{A}}=$ Avogadro number So, option (a) one mole of aluminum sulphate contains the highest number of oxygen atoms.
Kerala CEE -03.07.2022
Some Basic Concepts of Chemistry
228338
If a rocket runs on a fuel $\left(\mathrm{C}_{15} \mathrm{H}_{30}\right)$ and liquid oxygen, the weight of oxygen required and $\mathrm{CO}_2$ released for every litre of fuel respectively are: (Given: density of the fuel is $0.756 \mathrm{~g} / \mathrm{mL}$ )
1 $1188 \mathrm{~g}$ and $1296 \mathrm{~g}$
2 $2376 \mathrm{~g}$ and $2592 \mathrm{~g}$
3 $2592 \mathrm{~g}$ and $2376 \mathrm{~g}$
4 $3429 \mathrm{~g}$ and $3142 \mathrm{~g}$
Explanation:
Given, Density of fuel $=0.756 \mathrm{~g} / \mathrm{mL}$ Molecular formula of fuel $=\mathrm{C}_{15} \mathrm{H}_{30}$ Then, molar mass of fuel $=15 \times 12+30 \times 1$ $\begin{array}{r}=210 \mathrm{~g}\end{array}$ $\begin{aligned} \text { Density }=\frac{\text { mass }}{\text { volume }}\end{aligned}$ $\begin{aligned} & \text { Mass }=\text { density } \times \text { volume } \\ &=0.756 \times 1000=756 \mathrm{gm} .\end{aligned}$ The reaction is - $\mathrm{C}_{15} \mathrm{H}_{30}(\mathrm{l})+\frac{45}{2} \mathrm{O}_2(\mathrm{~g}) \longrightarrow 15 \mathrm{CO}_2(\mathrm{~g})+15 \mathrm{H}_2 \mathrm{O}$ Weight of oxygen $=\frac{45}{2}\left[\frac{756}{210}\right] \times 32=2592 \mathrm{~g}$ Weight of $\mathrm{CO}_2=15\left[\frac{756}{210}\right] \times 44=2376 \mathrm{~g}$.