285213
\(\quad 1.0 \mathrm{~g}\) of Mg is burnt with 0.28 g of \(\mathrm{O}_2\) in a closed vessel. Which reactant is left in excess and how much?
1 \(\mathrm{Mg}, 5.8 \mathrm{~g}\)
2 \(\mathrm{Mg}, 0.58 \mathrm{~g}\)
3 \(\mathrm{O}_2, 0.24 \mathrm{~g}\)
4 \(\mathrm{O}_2, 2.4 \mathrm{~g}\)
Explanation:
(b) Balanced reaction can be given as
\(2 \mathrm{Mg}+\mathrm{O}_2 \rightarrow 2 \mathrm{MgO}\)
\(24 \times 2=48 \quad 32\)
32 g of \(\mathrm{O}_2\) is required to burn 48 g Mg
So, \(0.28 \mathrm{~g} \mathrm{O}_2\) will be required for
\(\frac{48}{32} \times 0.28 \mathrm{~g} \mathrm{Mg}\)
\(=0.42 \mathrm{~g}\) of Mg
Thus, Mg will remain in excess
\(=1-0.42=0.58 \mathrm{~g}\)
Karnataka CET 2018
CHEMISTRY(KCET)
285214
If\(3.01 \times 10^{20}\) molecules are removed from 98 mg of \(\mathrm{H}_2 \mathrm{SO}_4\), then number of moles of \(\mathrm{H}_2 \mathrm{SO}_4\) left are
1 \(0.5 \times 10^{-3} \mathrm{~mol}\)
2 \(0.1 \times 10^{-3} \mathrm{~mol}\)
3 \(9.95 \times 10^{-2} \mathrm{~mol}\)
4 \(1.66 \times 10^{-3} \mathrm{~mol}\)
Explanation:
(a) No. of moles of\(\mathrm{H}_2 \mathrm{SO}_4\) in 98 mg
\(=\frac{98 \times 10^{-3}}{98}=1 \times 10^{-3}\)
No. of moles of \(3.01 \times 10^{20}\) molecules of \(\mathrm{H}_2 \mathrm{SO}_4\)
\(=\frac{3.01 \times 10^{20}}{6.02 \times 10^{23}}=\frac{1}{2} \times 10^{-3}\)
Moles of \(\mathrm{H}_2 \mathrm{SO}_4\) left
\(=\left(1 \times 10^{-3}-0.5 \times 10^{-3}\right)\)
\(=0.5 \times 10^{-3}\)
Karnataka CET 2017
CHEMISTRY(KCET)
285215
An organic compound contains\(\mathrm{C}=40 \%\), \(\mathrm{H}=13.33 \%\) and \(\mathrm{N}=46.67 \%\). Its empirical formula is
1 \(\mathrm{C}_2 \mathrm{H}_2 \mathrm{~N}\)
2 \(\mathrm{C}_3 \mathrm{H}_7 \mathrm{~N}\)
3 \(\mathrm{CH}_4 \mathrm{~N}\)
4 CHN
Explanation:
(c)
\begin{tabular}{|c|c|c|c|c|}
\hline Element & \(\%\) & \begin{tabular}{c}
No.of \\
moles
\end{tabular} & \begin{tabular}{c}
Molar \\
ratio
\end{tabular} & \begin{tabular}{c}
Simplest \\
ratio
\end{tabular} \\
\hline C & 40 & \begin{tabular}{c}
\(40 / 12\) \\
\(=3.33\)
\end{tabular} & \(\frac{3.33}{3.33}=1\) & 1 \\
\hline H & 13.33 & \begin{tabular}{l}
\(13.33 / 1\) \\
\(=13.33\)
\end{tabular} & \(\frac{13.33}{3.33}=4\) & 4 \\
\hline N & 46.67 & \begin{tabular}{l}
\(46.67 / 14\) \\
\(=3.33\)
\end{tabular} & \(\frac{3.33}{3.33}=1\) & 1 \\
\hline
\end{tabular}
Thus, the empirical formula is \(\mathrm{CH}_4 \mathrm{~N}\).
Karnataka CET 2016
CHEMISTRY(KCET)
285216
The number of oxygen atoms in 4.4 g of\(\mathrm{CO}_2\) is
1 \(1.2 \times 10^{23}\)
2 \(6 \times 10^{22}\)
3 \(6 \times 10^{23}\)
4 \(12 \times 10^{23}\)
Explanation:
(a) 44 g of\(\mathrm{CO}_2\) contain
\(=6.023 \times 10^{23}\) molecules of \(\mathrm{CO}_2\)
4.4 g of \(\mathrm{CO}_2\) will contain \(=\frac{6.023 \times 10^{23}}{44} \times 4.4\)
\(=6.023 \times 10^{22}\) molecules of \(\mathrm{CO}_2\)
1 molecule of \(\mathrm{CO}_2\) contains two oxygen atoms.
\(6.023 \times 10^{22}\) molecules of \(\mathrm{CO}_2\) will contain
\(2 \times 6.023 \times 10^{22}=1.2 \times 10^{23}\) oxygen atoms
285213
\(\quad 1.0 \mathrm{~g}\) of Mg is burnt with 0.28 g of \(\mathrm{O}_2\) in a closed vessel. Which reactant is left in excess and how much?
1 \(\mathrm{Mg}, 5.8 \mathrm{~g}\)
2 \(\mathrm{Mg}, 0.58 \mathrm{~g}\)
3 \(\mathrm{O}_2, 0.24 \mathrm{~g}\)
4 \(\mathrm{O}_2, 2.4 \mathrm{~g}\)
Explanation:
(b) Balanced reaction can be given as
\(2 \mathrm{Mg}+\mathrm{O}_2 \rightarrow 2 \mathrm{MgO}\)
\(24 \times 2=48 \quad 32\)
32 g of \(\mathrm{O}_2\) is required to burn 48 g Mg
So, \(0.28 \mathrm{~g} \mathrm{O}_2\) will be required for
\(\frac{48}{32} \times 0.28 \mathrm{~g} \mathrm{Mg}\)
\(=0.42 \mathrm{~g}\) of Mg
Thus, Mg will remain in excess
\(=1-0.42=0.58 \mathrm{~g}\)
Karnataka CET 2018
CHEMISTRY(KCET)
285214
If\(3.01 \times 10^{20}\) molecules are removed from 98 mg of \(\mathrm{H}_2 \mathrm{SO}_4\), then number of moles of \(\mathrm{H}_2 \mathrm{SO}_4\) left are
1 \(0.5 \times 10^{-3} \mathrm{~mol}\)
2 \(0.1 \times 10^{-3} \mathrm{~mol}\)
3 \(9.95 \times 10^{-2} \mathrm{~mol}\)
4 \(1.66 \times 10^{-3} \mathrm{~mol}\)
Explanation:
(a) No. of moles of\(\mathrm{H}_2 \mathrm{SO}_4\) in 98 mg
\(=\frac{98 \times 10^{-3}}{98}=1 \times 10^{-3}\)
No. of moles of \(3.01 \times 10^{20}\) molecules of \(\mathrm{H}_2 \mathrm{SO}_4\)
\(=\frac{3.01 \times 10^{20}}{6.02 \times 10^{23}}=\frac{1}{2} \times 10^{-3}\)
Moles of \(\mathrm{H}_2 \mathrm{SO}_4\) left
\(=\left(1 \times 10^{-3}-0.5 \times 10^{-3}\right)\)
\(=0.5 \times 10^{-3}\)
Karnataka CET 2017
CHEMISTRY(KCET)
285215
An organic compound contains\(\mathrm{C}=40 \%\), \(\mathrm{H}=13.33 \%\) and \(\mathrm{N}=46.67 \%\). Its empirical formula is
1 \(\mathrm{C}_2 \mathrm{H}_2 \mathrm{~N}\)
2 \(\mathrm{C}_3 \mathrm{H}_7 \mathrm{~N}\)
3 \(\mathrm{CH}_4 \mathrm{~N}\)
4 CHN
Explanation:
(c)
\begin{tabular}{|c|c|c|c|c|}
\hline Element & \(\%\) & \begin{tabular}{c}
No.of \\
moles
\end{tabular} & \begin{tabular}{c}
Molar \\
ratio
\end{tabular} & \begin{tabular}{c}
Simplest \\
ratio
\end{tabular} \\
\hline C & 40 & \begin{tabular}{c}
\(40 / 12\) \\
\(=3.33\)
\end{tabular} & \(\frac{3.33}{3.33}=1\) & 1 \\
\hline H & 13.33 & \begin{tabular}{l}
\(13.33 / 1\) \\
\(=13.33\)
\end{tabular} & \(\frac{13.33}{3.33}=4\) & 4 \\
\hline N & 46.67 & \begin{tabular}{l}
\(46.67 / 14\) \\
\(=3.33\)
\end{tabular} & \(\frac{3.33}{3.33}=1\) & 1 \\
\hline
\end{tabular}
Thus, the empirical formula is \(\mathrm{CH}_4 \mathrm{~N}\).
Karnataka CET 2016
CHEMISTRY(KCET)
285216
The number of oxygen atoms in 4.4 g of\(\mathrm{CO}_2\) is
1 \(1.2 \times 10^{23}\)
2 \(6 \times 10^{22}\)
3 \(6 \times 10^{23}\)
4 \(12 \times 10^{23}\)
Explanation:
(a) 44 g of\(\mathrm{CO}_2\) contain
\(=6.023 \times 10^{23}\) molecules of \(\mathrm{CO}_2\)
4.4 g of \(\mathrm{CO}_2\) will contain \(=\frac{6.023 \times 10^{23}}{44} \times 4.4\)
\(=6.023 \times 10^{22}\) molecules of \(\mathrm{CO}_2\)
1 molecule of \(\mathrm{CO}_2\) contains two oxygen atoms.
\(6.023 \times 10^{22}\) molecules of \(\mathrm{CO}_2\) will contain
\(2 \times 6.023 \times 10^{22}=1.2 \times 10^{23}\) oxygen atoms
285213
\(\quad 1.0 \mathrm{~g}\) of Mg is burnt with 0.28 g of \(\mathrm{O}_2\) in a closed vessel. Which reactant is left in excess and how much?
1 \(\mathrm{Mg}, 5.8 \mathrm{~g}\)
2 \(\mathrm{Mg}, 0.58 \mathrm{~g}\)
3 \(\mathrm{O}_2, 0.24 \mathrm{~g}\)
4 \(\mathrm{O}_2, 2.4 \mathrm{~g}\)
Explanation:
(b) Balanced reaction can be given as
\(2 \mathrm{Mg}+\mathrm{O}_2 \rightarrow 2 \mathrm{MgO}\)
\(24 \times 2=48 \quad 32\)
32 g of \(\mathrm{O}_2\) is required to burn 48 g Mg
So, \(0.28 \mathrm{~g} \mathrm{O}_2\) will be required for
\(\frac{48}{32} \times 0.28 \mathrm{~g} \mathrm{Mg}\)
\(=0.42 \mathrm{~g}\) of Mg
Thus, Mg will remain in excess
\(=1-0.42=0.58 \mathrm{~g}\)
Karnataka CET 2018
CHEMISTRY(KCET)
285214
If\(3.01 \times 10^{20}\) molecules are removed from 98 mg of \(\mathrm{H}_2 \mathrm{SO}_4\), then number of moles of \(\mathrm{H}_2 \mathrm{SO}_4\) left are
1 \(0.5 \times 10^{-3} \mathrm{~mol}\)
2 \(0.1 \times 10^{-3} \mathrm{~mol}\)
3 \(9.95 \times 10^{-2} \mathrm{~mol}\)
4 \(1.66 \times 10^{-3} \mathrm{~mol}\)
Explanation:
(a) No. of moles of\(\mathrm{H}_2 \mathrm{SO}_4\) in 98 mg
\(=\frac{98 \times 10^{-3}}{98}=1 \times 10^{-3}\)
No. of moles of \(3.01 \times 10^{20}\) molecules of \(\mathrm{H}_2 \mathrm{SO}_4\)
\(=\frac{3.01 \times 10^{20}}{6.02 \times 10^{23}}=\frac{1}{2} \times 10^{-3}\)
Moles of \(\mathrm{H}_2 \mathrm{SO}_4\) left
\(=\left(1 \times 10^{-3}-0.5 \times 10^{-3}\right)\)
\(=0.5 \times 10^{-3}\)
Karnataka CET 2017
CHEMISTRY(KCET)
285215
An organic compound contains\(\mathrm{C}=40 \%\), \(\mathrm{H}=13.33 \%\) and \(\mathrm{N}=46.67 \%\). Its empirical formula is
1 \(\mathrm{C}_2 \mathrm{H}_2 \mathrm{~N}\)
2 \(\mathrm{C}_3 \mathrm{H}_7 \mathrm{~N}\)
3 \(\mathrm{CH}_4 \mathrm{~N}\)
4 CHN
Explanation:
(c)
\begin{tabular}{|c|c|c|c|c|}
\hline Element & \(\%\) & \begin{tabular}{c}
No.of \\
moles
\end{tabular} & \begin{tabular}{c}
Molar \\
ratio
\end{tabular} & \begin{tabular}{c}
Simplest \\
ratio
\end{tabular} \\
\hline C & 40 & \begin{tabular}{c}
\(40 / 12\) \\
\(=3.33\)
\end{tabular} & \(\frac{3.33}{3.33}=1\) & 1 \\
\hline H & 13.33 & \begin{tabular}{l}
\(13.33 / 1\) \\
\(=13.33\)
\end{tabular} & \(\frac{13.33}{3.33}=4\) & 4 \\
\hline N & 46.67 & \begin{tabular}{l}
\(46.67 / 14\) \\
\(=3.33\)
\end{tabular} & \(\frac{3.33}{3.33}=1\) & 1 \\
\hline
\end{tabular}
Thus, the empirical formula is \(\mathrm{CH}_4 \mathrm{~N}\).
Karnataka CET 2016
CHEMISTRY(KCET)
285216
The number of oxygen atoms in 4.4 g of\(\mathrm{CO}_2\) is
1 \(1.2 \times 10^{23}\)
2 \(6 \times 10^{22}\)
3 \(6 \times 10^{23}\)
4 \(12 \times 10^{23}\)
Explanation:
(a) 44 g of\(\mathrm{CO}_2\) contain
\(=6.023 \times 10^{23}\) molecules of \(\mathrm{CO}_2\)
4.4 g of \(\mathrm{CO}_2\) will contain \(=\frac{6.023 \times 10^{23}}{44} \times 4.4\)
\(=6.023 \times 10^{22}\) molecules of \(\mathrm{CO}_2\)
1 molecule of \(\mathrm{CO}_2\) contains two oxygen atoms.
\(6.023 \times 10^{22}\) molecules of \(\mathrm{CO}_2\) will contain
\(2 \times 6.023 \times 10^{22}=1.2 \times 10^{23}\) oxygen atoms
285213
\(\quad 1.0 \mathrm{~g}\) of Mg is burnt with 0.28 g of \(\mathrm{O}_2\) in a closed vessel. Which reactant is left in excess and how much?
1 \(\mathrm{Mg}, 5.8 \mathrm{~g}\)
2 \(\mathrm{Mg}, 0.58 \mathrm{~g}\)
3 \(\mathrm{O}_2, 0.24 \mathrm{~g}\)
4 \(\mathrm{O}_2, 2.4 \mathrm{~g}\)
Explanation:
(b) Balanced reaction can be given as
\(2 \mathrm{Mg}+\mathrm{O}_2 \rightarrow 2 \mathrm{MgO}\)
\(24 \times 2=48 \quad 32\)
32 g of \(\mathrm{O}_2\) is required to burn 48 g Mg
So, \(0.28 \mathrm{~g} \mathrm{O}_2\) will be required for
\(\frac{48}{32} \times 0.28 \mathrm{~g} \mathrm{Mg}\)
\(=0.42 \mathrm{~g}\) of Mg
Thus, Mg will remain in excess
\(=1-0.42=0.58 \mathrm{~g}\)
Karnataka CET 2018
CHEMISTRY(KCET)
285214
If\(3.01 \times 10^{20}\) molecules are removed from 98 mg of \(\mathrm{H}_2 \mathrm{SO}_4\), then number of moles of \(\mathrm{H}_2 \mathrm{SO}_4\) left are
1 \(0.5 \times 10^{-3} \mathrm{~mol}\)
2 \(0.1 \times 10^{-3} \mathrm{~mol}\)
3 \(9.95 \times 10^{-2} \mathrm{~mol}\)
4 \(1.66 \times 10^{-3} \mathrm{~mol}\)
Explanation:
(a) No. of moles of\(\mathrm{H}_2 \mathrm{SO}_4\) in 98 mg
\(=\frac{98 \times 10^{-3}}{98}=1 \times 10^{-3}\)
No. of moles of \(3.01 \times 10^{20}\) molecules of \(\mathrm{H}_2 \mathrm{SO}_4\)
\(=\frac{3.01 \times 10^{20}}{6.02 \times 10^{23}}=\frac{1}{2} \times 10^{-3}\)
Moles of \(\mathrm{H}_2 \mathrm{SO}_4\) left
\(=\left(1 \times 10^{-3}-0.5 \times 10^{-3}\right)\)
\(=0.5 \times 10^{-3}\)
Karnataka CET 2017
CHEMISTRY(KCET)
285215
An organic compound contains\(\mathrm{C}=40 \%\), \(\mathrm{H}=13.33 \%\) and \(\mathrm{N}=46.67 \%\). Its empirical formula is
1 \(\mathrm{C}_2 \mathrm{H}_2 \mathrm{~N}\)
2 \(\mathrm{C}_3 \mathrm{H}_7 \mathrm{~N}\)
3 \(\mathrm{CH}_4 \mathrm{~N}\)
4 CHN
Explanation:
(c)
\begin{tabular}{|c|c|c|c|c|}
\hline Element & \(\%\) & \begin{tabular}{c}
No.of \\
moles
\end{tabular} & \begin{tabular}{c}
Molar \\
ratio
\end{tabular} & \begin{tabular}{c}
Simplest \\
ratio
\end{tabular} \\
\hline C & 40 & \begin{tabular}{c}
\(40 / 12\) \\
\(=3.33\)
\end{tabular} & \(\frac{3.33}{3.33}=1\) & 1 \\
\hline H & 13.33 & \begin{tabular}{l}
\(13.33 / 1\) \\
\(=13.33\)
\end{tabular} & \(\frac{13.33}{3.33}=4\) & 4 \\
\hline N & 46.67 & \begin{tabular}{l}
\(46.67 / 14\) \\
\(=3.33\)
\end{tabular} & \(\frac{3.33}{3.33}=1\) & 1 \\
\hline
\end{tabular}
Thus, the empirical formula is \(\mathrm{CH}_4 \mathrm{~N}\).
Karnataka CET 2016
CHEMISTRY(KCET)
285216
The number of oxygen atoms in 4.4 g of\(\mathrm{CO}_2\) is
1 \(1.2 \times 10^{23}\)
2 \(6 \times 10^{22}\)
3 \(6 \times 10^{23}\)
4 \(12 \times 10^{23}\)
Explanation:
(a) 44 g of\(\mathrm{CO}_2\) contain
\(=6.023 \times 10^{23}\) molecules of \(\mathrm{CO}_2\)
4.4 g of \(\mathrm{CO}_2\) will contain \(=\frac{6.023 \times 10^{23}}{44} \times 4.4\)
\(=6.023 \times 10^{22}\) molecules of \(\mathrm{CO}_2\)
1 molecule of \(\mathrm{CO}_2\) contains two oxygen atoms.
\(6.023 \times 10^{22}\) molecules of \(\mathrm{CO}_2\) will contain
\(2 \times 6.023 \times 10^{22}=1.2 \times 10^{23}\) oxygen atoms
