228231
What will be the mass of one atom of ${ }^{12} \mathrm{C}$?
1 $1 \mathrm{amu}$
2 $1.9923 \times 10^{-23} \mathrm{~g}$
3 $1.6603 \times 10^{-22} \mathrm{~g}$
4 $6 \mathrm{amu}$
Explanation:
Mass of ${ }^{12} \mathrm{C}=12 \mathrm{gm}$ $6.022 \times 10^{23}$ atoms are present in $12 \mathrm{~g}$ of carbon -12 element. Mass of $6.022 \times 10^{23}$ atom $=12 \mathrm{gm}$. $\begin{aligned} \text { Mass of } 1 \text { Atom } & =\frac{12}{6.023 \times 10^{23}} \\ & =1.993 \times 10^{-23} \mathrm{gm} \end{aligned}$
WB-JEE-2020
Some Basic Concepts of Chemistry
228232
In a flask, the weight ratio of $\mathrm{CH}_4(\mathrm{~g})$ and $\mathrm{SO}_2(\mathrm{~g})$ at $298 \mathrm{~K}$ and 1 bar is $1: 2$. The ratio of the number of molecules of $\mathrm{SO}_2(\mathrm{~g})$ and $\mathrm{CH}_4(\mathrm{~g})$ is
1 $1: 4$
2 $4: 1$
3 $1: 2$
4 $2: 1$
Explanation:
Let mass of $\mathrm{CH}_4(\mathrm{~g})=1 \mathrm{~g}$ Number of moles of $\mathrm{CH}_4\left(\mathrm{n}_{\mathrm{CH}_4}\right)=\frac{1}{16}$ Number of molecules of $\mathrm{CH}_4(\mathrm{~g})=\frac{1}{16} \times \mathrm{N}_{\mathrm{A}}$ Let the mass of $\mathrm{SO}_2(\mathrm{~g})=2 \mathrm{~g}$ Number of moles of $\mathrm{SO}_2(\mathrm{~g})\left(\mathrm{n}_{\mathrm{so}_2}\right)=\frac{2}{64}$ Number of molecules of $\mathrm{SO}_2(\mathrm{~g})=\frac{2}{64} \times \mathrm{N}_{\mathrm{A}}$ Ratio of number of molecules of $\mathrm{SO}_2(\mathrm{~g})$ and number of molecules of $\mathrm{CH}_4(\mathrm{~g})=$ $\frac{2}{64} \times \mathrm{N}_{\mathrm{A}}: \frac{1}{16} \times \mathrm{N}_{\mathrm{A}} \Rightarrow \frac{1}{32}: \frac{1}{16} \Rightarrow 1: 2$
COMEDK-2020
Some Basic Concepts of Chemistry
228233
Equivalent mass of $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ in acidic solution is equal to
1 molecular mass $/ 2$
2 molecular mass $/ 4$
3 molecular mass
4 molecular mass/ 6.
Explanation:
The given reaction occurred in acidic medium is : $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7+14 \mathrm{H}^{+}+6 \mathrm{e}^{-} \longrightarrow 2 \mathrm{~K}^{+}+2 \mathrm{Cr}^{3+}+7 \mathrm{H}_2 \mathrm{O}$ In the above reaction, you can see that 1 molecule of $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ is releasing 6 electrons and molecular weight of $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7=294 \mathrm{~g} / \mathrm{mol}$ $\therefore$ Calculation of equivalent weight $=\frac{\text { Molecular wt. of } \mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7}{\text { Acidity }}$ Equivalent weight $=\frac{\text { Molecular wt. of } \mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7}{6}$
COMEDK-2019
Some Basic Concepts of Chemistry
228234
Equivalent weight of $\mathrm{KMnO}_4$ is equal to
1 one-sixth its molecular weight
2 its molecular weight
3 one-fifth its molecular weight
4 half is molecular weight.
Explanation:
$2 \mathrm{KMnO}_4+3 \mathrm{H}_2 \mathrm{SO}_4 \rightarrow$ $\mathrm{K}_2 \mathrm{SO}_4+2 \mathrm{MnSO}_4+3 \mathrm{H}_2 \mathrm{O}+5[\mathrm{O}]$ Since, mass of oxygen are available from $\frac{2 \times 8}{5 \times 16}=\frac{1}{5}$ mol. wt. of $\mathrm{KMnO}_4$ Therefore, equivalent weight of $\mathrm{KMnO}_4$ $=\frac{1}{5} \times$ molecular weight. Thus equivalent weight of $\mathrm{KMnO}_4$ is one fifth its molecular weight.
228231
What will be the mass of one atom of ${ }^{12} \mathrm{C}$?
1 $1 \mathrm{amu}$
2 $1.9923 \times 10^{-23} \mathrm{~g}$
3 $1.6603 \times 10^{-22} \mathrm{~g}$
4 $6 \mathrm{amu}$
Explanation:
Mass of ${ }^{12} \mathrm{C}=12 \mathrm{gm}$ $6.022 \times 10^{23}$ atoms are present in $12 \mathrm{~g}$ of carbon -12 element. Mass of $6.022 \times 10^{23}$ atom $=12 \mathrm{gm}$. $\begin{aligned} \text { Mass of } 1 \text { Atom } & =\frac{12}{6.023 \times 10^{23}} \\ & =1.993 \times 10^{-23} \mathrm{gm} \end{aligned}$
WB-JEE-2020
Some Basic Concepts of Chemistry
228232
In a flask, the weight ratio of $\mathrm{CH}_4(\mathrm{~g})$ and $\mathrm{SO}_2(\mathrm{~g})$ at $298 \mathrm{~K}$ and 1 bar is $1: 2$. The ratio of the number of molecules of $\mathrm{SO}_2(\mathrm{~g})$ and $\mathrm{CH}_4(\mathrm{~g})$ is
1 $1: 4$
2 $4: 1$
3 $1: 2$
4 $2: 1$
Explanation:
Let mass of $\mathrm{CH}_4(\mathrm{~g})=1 \mathrm{~g}$ Number of moles of $\mathrm{CH}_4\left(\mathrm{n}_{\mathrm{CH}_4}\right)=\frac{1}{16}$ Number of molecules of $\mathrm{CH}_4(\mathrm{~g})=\frac{1}{16} \times \mathrm{N}_{\mathrm{A}}$ Let the mass of $\mathrm{SO}_2(\mathrm{~g})=2 \mathrm{~g}$ Number of moles of $\mathrm{SO}_2(\mathrm{~g})\left(\mathrm{n}_{\mathrm{so}_2}\right)=\frac{2}{64}$ Number of molecules of $\mathrm{SO}_2(\mathrm{~g})=\frac{2}{64} \times \mathrm{N}_{\mathrm{A}}$ Ratio of number of molecules of $\mathrm{SO}_2(\mathrm{~g})$ and number of molecules of $\mathrm{CH}_4(\mathrm{~g})=$ $\frac{2}{64} \times \mathrm{N}_{\mathrm{A}}: \frac{1}{16} \times \mathrm{N}_{\mathrm{A}} \Rightarrow \frac{1}{32}: \frac{1}{16} \Rightarrow 1: 2$
COMEDK-2020
Some Basic Concepts of Chemistry
228233
Equivalent mass of $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ in acidic solution is equal to
1 molecular mass $/ 2$
2 molecular mass $/ 4$
3 molecular mass
4 molecular mass/ 6.
Explanation:
The given reaction occurred in acidic medium is : $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7+14 \mathrm{H}^{+}+6 \mathrm{e}^{-} \longrightarrow 2 \mathrm{~K}^{+}+2 \mathrm{Cr}^{3+}+7 \mathrm{H}_2 \mathrm{O}$ In the above reaction, you can see that 1 molecule of $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ is releasing 6 electrons and molecular weight of $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7=294 \mathrm{~g} / \mathrm{mol}$ $\therefore$ Calculation of equivalent weight $=\frac{\text { Molecular wt. of } \mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7}{\text { Acidity }}$ Equivalent weight $=\frac{\text { Molecular wt. of } \mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7}{6}$
COMEDK-2019
Some Basic Concepts of Chemistry
228234
Equivalent weight of $\mathrm{KMnO}_4$ is equal to
1 one-sixth its molecular weight
2 its molecular weight
3 one-fifth its molecular weight
4 half is molecular weight.
Explanation:
$2 \mathrm{KMnO}_4+3 \mathrm{H}_2 \mathrm{SO}_4 \rightarrow$ $\mathrm{K}_2 \mathrm{SO}_4+2 \mathrm{MnSO}_4+3 \mathrm{H}_2 \mathrm{O}+5[\mathrm{O}]$ Since, mass of oxygen are available from $\frac{2 \times 8}{5 \times 16}=\frac{1}{5}$ mol. wt. of $\mathrm{KMnO}_4$ Therefore, equivalent weight of $\mathrm{KMnO}_4$ $=\frac{1}{5} \times$ molecular weight. Thus equivalent weight of $\mathrm{KMnO}_4$ is one fifth its molecular weight.
228231
What will be the mass of one atom of ${ }^{12} \mathrm{C}$?
1 $1 \mathrm{amu}$
2 $1.9923 \times 10^{-23} \mathrm{~g}$
3 $1.6603 \times 10^{-22} \mathrm{~g}$
4 $6 \mathrm{amu}$
Explanation:
Mass of ${ }^{12} \mathrm{C}=12 \mathrm{gm}$ $6.022 \times 10^{23}$ atoms are present in $12 \mathrm{~g}$ of carbon -12 element. Mass of $6.022 \times 10^{23}$ atom $=12 \mathrm{gm}$. $\begin{aligned} \text { Mass of } 1 \text { Atom } & =\frac{12}{6.023 \times 10^{23}} \\ & =1.993 \times 10^{-23} \mathrm{gm} \end{aligned}$
WB-JEE-2020
Some Basic Concepts of Chemistry
228232
In a flask, the weight ratio of $\mathrm{CH}_4(\mathrm{~g})$ and $\mathrm{SO}_2(\mathrm{~g})$ at $298 \mathrm{~K}$ and 1 bar is $1: 2$. The ratio of the number of molecules of $\mathrm{SO}_2(\mathrm{~g})$ and $\mathrm{CH}_4(\mathrm{~g})$ is
1 $1: 4$
2 $4: 1$
3 $1: 2$
4 $2: 1$
Explanation:
Let mass of $\mathrm{CH}_4(\mathrm{~g})=1 \mathrm{~g}$ Number of moles of $\mathrm{CH}_4\left(\mathrm{n}_{\mathrm{CH}_4}\right)=\frac{1}{16}$ Number of molecules of $\mathrm{CH}_4(\mathrm{~g})=\frac{1}{16} \times \mathrm{N}_{\mathrm{A}}$ Let the mass of $\mathrm{SO}_2(\mathrm{~g})=2 \mathrm{~g}$ Number of moles of $\mathrm{SO}_2(\mathrm{~g})\left(\mathrm{n}_{\mathrm{so}_2}\right)=\frac{2}{64}$ Number of molecules of $\mathrm{SO}_2(\mathrm{~g})=\frac{2}{64} \times \mathrm{N}_{\mathrm{A}}$ Ratio of number of molecules of $\mathrm{SO}_2(\mathrm{~g})$ and number of molecules of $\mathrm{CH}_4(\mathrm{~g})=$ $\frac{2}{64} \times \mathrm{N}_{\mathrm{A}}: \frac{1}{16} \times \mathrm{N}_{\mathrm{A}} \Rightarrow \frac{1}{32}: \frac{1}{16} \Rightarrow 1: 2$
COMEDK-2020
Some Basic Concepts of Chemistry
228233
Equivalent mass of $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ in acidic solution is equal to
1 molecular mass $/ 2$
2 molecular mass $/ 4$
3 molecular mass
4 molecular mass/ 6.
Explanation:
The given reaction occurred in acidic medium is : $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7+14 \mathrm{H}^{+}+6 \mathrm{e}^{-} \longrightarrow 2 \mathrm{~K}^{+}+2 \mathrm{Cr}^{3+}+7 \mathrm{H}_2 \mathrm{O}$ In the above reaction, you can see that 1 molecule of $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ is releasing 6 electrons and molecular weight of $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7=294 \mathrm{~g} / \mathrm{mol}$ $\therefore$ Calculation of equivalent weight $=\frac{\text { Molecular wt. of } \mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7}{\text { Acidity }}$ Equivalent weight $=\frac{\text { Molecular wt. of } \mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7}{6}$
COMEDK-2019
Some Basic Concepts of Chemistry
228234
Equivalent weight of $\mathrm{KMnO}_4$ is equal to
1 one-sixth its molecular weight
2 its molecular weight
3 one-fifth its molecular weight
4 half is molecular weight.
Explanation:
$2 \mathrm{KMnO}_4+3 \mathrm{H}_2 \mathrm{SO}_4 \rightarrow$ $\mathrm{K}_2 \mathrm{SO}_4+2 \mathrm{MnSO}_4+3 \mathrm{H}_2 \mathrm{O}+5[\mathrm{O}]$ Since, mass of oxygen are available from $\frac{2 \times 8}{5 \times 16}=\frac{1}{5}$ mol. wt. of $\mathrm{KMnO}_4$ Therefore, equivalent weight of $\mathrm{KMnO}_4$ $=\frac{1}{5} \times$ molecular weight. Thus equivalent weight of $\mathrm{KMnO}_4$ is one fifth its molecular weight.
228231
What will be the mass of one atom of ${ }^{12} \mathrm{C}$?
1 $1 \mathrm{amu}$
2 $1.9923 \times 10^{-23} \mathrm{~g}$
3 $1.6603 \times 10^{-22} \mathrm{~g}$
4 $6 \mathrm{amu}$
Explanation:
Mass of ${ }^{12} \mathrm{C}=12 \mathrm{gm}$ $6.022 \times 10^{23}$ atoms are present in $12 \mathrm{~g}$ of carbon -12 element. Mass of $6.022 \times 10^{23}$ atom $=12 \mathrm{gm}$. $\begin{aligned} \text { Mass of } 1 \text { Atom } & =\frac{12}{6.023 \times 10^{23}} \\ & =1.993 \times 10^{-23} \mathrm{gm} \end{aligned}$
WB-JEE-2020
Some Basic Concepts of Chemistry
228232
In a flask, the weight ratio of $\mathrm{CH}_4(\mathrm{~g})$ and $\mathrm{SO}_2(\mathrm{~g})$ at $298 \mathrm{~K}$ and 1 bar is $1: 2$. The ratio of the number of molecules of $\mathrm{SO}_2(\mathrm{~g})$ and $\mathrm{CH}_4(\mathrm{~g})$ is
1 $1: 4$
2 $4: 1$
3 $1: 2$
4 $2: 1$
Explanation:
Let mass of $\mathrm{CH}_4(\mathrm{~g})=1 \mathrm{~g}$ Number of moles of $\mathrm{CH}_4\left(\mathrm{n}_{\mathrm{CH}_4}\right)=\frac{1}{16}$ Number of molecules of $\mathrm{CH}_4(\mathrm{~g})=\frac{1}{16} \times \mathrm{N}_{\mathrm{A}}$ Let the mass of $\mathrm{SO}_2(\mathrm{~g})=2 \mathrm{~g}$ Number of moles of $\mathrm{SO}_2(\mathrm{~g})\left(\mathrm{n}_{\mathrm{so}_2}\right)=\frac{2}{64}$ Number of molecules of $\mathrm{SO}_2(\mathrm{~g})=\frac{2}{64} \times \mathrm{N}_{\mathrm{A}}$ Ratio of number of molecules of $\mathrm{SO}_2(\mathrm{~g})$ and number of molecules of $\mathrm{CH}_4(\mathrm{~g})=$ $\frac{2}{64} \times \mathrm{N}_{\mathrm{A}}: \frac{1}{16} \times \mathrm{N}_{\mathrm{A}} \Rightarrow \frac{1}{32}: \frac{1}{16} \Rightarrow 1: 2$
COMEDK-2020
Some Basic Concepts of Chemistry
228233
Equivalent mass of $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ in acidic solution is equal to
1 molecular mass $/ 2$
2 molecular mass $/ 4$
3 molecular mass
4 molecular mass/ 6.
Explanation:
The given reaction occurred in acidic medium is : $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7+14 \mathrm{H}^{+}+6 \mathrm{e}^{-} \longrightarrow 2 \mathrm{~K}^{+}+2 \mathrm{Cr}^{3+}+7 \mathrm{H}_2 \mathrm{O}$ In the above reaction, you can see that 1 molecule of $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ is releasing 6 electrons and molecular weight of $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7=294 \mathrm{~g} / \mathrm{mol}$ $\therefore$ Calculation of equivalent weight $=\frac{\text { Molecular wt. of } \mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7}{\text { Acidity }}$ Equivalent weight $=\frac{\text { Molecular wt. of } \mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7}{6}$
COMEDK-2019
Some Basic Concepts of Chemistry
228234
Equivalent weight of $\mathrm{KMnO}_4$ is equal to
1 one-sixth its molecular weight
2 its molecular weight
3 one-fifth its molecular weight
4 half is molecular weight.
Explanation:
$2 \mathrm{KMnO}_4+3 \mathrm{H}_2 \mathrm{SO}_4 \rightarrow$ $\mathrm{K}_2 \mathrm{SO}_4+2 \mathrm{MnSO}_4+3 \mathrm{H}_2 \mathrm{O}+5[\mathrm{O}]$ Since, mass of oxygen are available from $\frac{2 \times 8}{5 \times 16}=\frac{1}{5}$ mol. wt. of $\mathrm{KMnO}_4$ Therefore, equivalent weight of $\mathrm{KMnO}_4$ $=\frac{1}{5} \times$ molecular weight. Thus equivalent weight of $\mathrm{KMnO}_4$ is one fifth its molecular weight.
