160629
Two element \(X\) and \(Y\) combined to form a molecule of \(X_2 Y\) type. If mass of 0.1 mole of \(X\) weighted 2.3 gm and weight of \(3.01 \times 10^{20}\) atom of \(Y\) is \(\mathbf{0 . 0 0 8} \mathbf{g m}\). Molar mass \(X_2 Y\) will be :
1 62
2 31
3 55
4 39
Explanation:
MM of \(x=\frac{2.3}{0.1}\) and MM of \(y=\frac{0.008 \times 6.021 \times 10^{23}}{3.01 \times 10^{20}}\)
NCERT 15
1 RBTS PAPER
160630
On combustion of \(25 \mathrm{ml}\) liquid \(\mathrm{CH}_3 \mathrm{COOH}\) whose specific gravity is 1.2 , gives all gaseous product, voluemof product at STP will be :
1 11.2 lit
2 22.4 lit
3 44.8 lit
4 33.6 lit
Explanation:
\(\mathrm{CH}_3 \mathrm{COOH} \rightarrow 2 \mathrm{CO}+2 \mathrm{H}_2 \mathrm{O}\)No. of moles of \(\mathrm{CH}_3 \mathrm{COOH}=\frac{25 \times 1.2}{60}\)
NCERT 15
1 RBTS PAPER
160648
According to the Bohr theory, which of the following transitions in the hydrogen atom will give rise to the least energetic photon :
1 \(n=6\) to \(n=1\)
2 \(n=5\) to \(n=4\)
3 \(n=6\) to \(n=5\)
4 \(n=5\) to \(n=3\)
Explanation:
\(\Delta \mathrm{E} \propto\left[\frac{1}{\mathrm{~m}_1^2}-\frac{1}{\mathrm{n}_2^2}\right]\), where \(\mathrm{n}_2>\mathrm{n}_1\) \(\therefore n=6\) to \(n=5\) will give least energetic photon.
AIPMT Mains 2011
1 RBTS PAPER
160615
Calculate the mass percentage of sodium element and sulphur element respectively in sodium sulphate :
1 \(34.39 \%, 20.53 \%\)
2 \(32.39 \%, 22.53 \%\)
3 \(38.39 \%, 25.53 \%\)
4 \(23.39 \%, 32.53 \%\)
Explanation:
\(\begin{aligned} & \text { Mass } \% \text { of sodium }=\frac{2 \times 23}{142} \times 100=32.39 \% \\ & \text { Mass } \% \text { of sulphur }=\frac{32}{142} \times 100=22.53 \% \end{aligned}\)
NCERT-15
1 RBTS PAPER
160616
10 mole each of \(\mathrm{KI}\) and \(\mathrm{Hgl}_2\) are mixed and react according to \(\mathrm{KI}+\mathrm{Hgl}_2 \rightarrow \mathrm{K}_2\left[\mathrm{Hgl}_4\right]\) The moles of product formed during the reaction will be:
1 2.5
2 5
3 7.5
4 0.50
Explanation:
\(2 \mathrm{KI}+\mathrm{HgI}_2 \rightarrow \mathrm{K}_2\left[\mathrm{Hgl}_4\right]\) 2 mol KI formed 1 mole \(\mathrm{K}_2\left[\mathrm{Hgl}_4\right]\) 1 mole KI formed \(1 / 2\) mole \(\mathrm{K}_2\left[\mathrm{Hgl}_4\right]\) 10 mole KI formed \(\frac{1}{2} \times 10\) mole \(\mathrm{K}_2\left[\mathrm{HgI}_4\right]\) \(=5\) mole \(\mathrm{K}_2\left[\mathrm{Hgl}_4\right]\)
160629
Two element \(X\) and \(Y\) combined to form a molecule of \(X_2 Y\) type. If mass of 0.1 mole of \(X\) weighted 2.3 gm and weight of \(3.01 \times 10^{20}\) atom of \(Y\) is \(\mathbf{0 . 0 0 8} \mathbf{g m}\). Molar mass \(X_2 Y\) will be :
1 62
2 31
3 55
4 39
Explanation:
MM of \(x=\frac{2.3}{0.1}\) and MM of \(y=\frac{0.008 \times 6.021 \times 10^{23}}{3.01 \times 10^{20}}\)
NCERT 15
1 RBTS PAPER
160630
On combustion of \(25 \mathrm{ml}\) liquid \(\mathrm{CH}_3 \mathrm{COOH}\) whose specific gravity is 1.2 , gives all gaseous product, voluemof product at STP will be :
1 11.2 lit
2 22.4 lit
3 44.8 lit
4 33.6 lit
Explanation:
\(\mathrm{CH}_3 \mathrm{COOH} \rightarrow 2 \mathrm{CO}+2 \mathrm{H}_2 \mathrm{O}\)No. of moles of \(\mathrm{CH}_3 \mathrm{COOH}=\frac{25 \times 1.2}{60}\)
NCERT 15
1 RBTS PAPER
160648
According to the Bohr theory, which of the following transitions in the hydrogen atom will give rise to the least energetic photon :
1 \(n=6\) to \(n=1\)
2 \(n=5\) to \(n=4\)
3 \(n=6\) to \(n=5\)
4 \(n=5\) to \(n=3\)
Explanation:
\(\Delta \mathrm{E} \propto\left[\frac{1}{\mathrm{~m}_1^2}-\frac{1}{\mathrm{n}_2^2}\right]\), where \(\mathrm{n}_2>\mathrm{n}_1\) \(\therefore n=6\) to \(n=5\) will give least energetic photon.
AIPMT Mains 2011
1 RBTS PAPER
160615
Calculate the mass percentage of sodium element and sulphur element respectively in sodium sulphate :
1 \(34.39 \%, 20.53 \%\)
2 \(32.39 \%, 22.53 \%\)
3 \(38.39 \%, 25.53 \%\)
4 \(23.39 \%, 32.53 \%\)
Explanation:
\(\begin{aligned} & \text { Mass } \% \text { of sodium }=\frac{2 \times 23}{142} \times 100=32.39 \% \\ & \text { Mass } \% \text { of sulphur }=\frac{32}{142} \times 100=22.53 \% \end{aligned}\)
NCERT-15
1 RBTS PAPER
160616
10 mole each of \(\mathrm{KI}\) and \(\mathrm{Hgl}_2\) are mixed and react according to \(\mathrm{KI}+\mathrm{Hgl}_2 \rightarrow \mathrm{K}_2\left[\mathrm{Hgl}_4\right]\) The moles of product formed during the reaction will be:
1 2.5
2 5
3 7.5
4 0.50
Explanation:
\(2 \mathrm{KI}+\mathrm{HgI}_2 \rightarrow \mathrm{K}_2\left[\mathrm{Hgl}_4\right]\) 2 mol KI formed 1 mole \(\mathrm{K}_2\left[\mathrm{Hgl}_4\right]\) 1 mole KI formed \(1 / 2\) mole \(\mathrm{K}_2\left[\mathrm{Hgl}_4\right]\) 10 mole KI formed \(\frac{1}{2} \times 10\) mole \(\mathrm{K}_2\left[\mathrm{HgI}_4\right]\) \(=5\) mole \(\mathrm{K}_2\left[\mathrm{Hgl}_4\right]\)
160629
Two element \(X\) and \(Y\) combined to form a molecule of \(X_2 Y\) type. If mass of 0.1 mole of \(X\) weighted 2.3 gm and weight of \(3.01 \times 10^{20}\) atom of \(Y\) is \(\mathbf{0 . 0 0 8} \mathbf{g m}\). Molar mass \(X_2 Y\) will be :
1 62
2 31
3 55
4 39
Explanation:
MM of \(x=\frac{2.3}{0.1}\) and MM of \(y=\frac{0.008 \times 6.021 \times 10^{23}}{3.01 \times 10^{20}}\)
NCERT 15
1 RBTS PAPER
160630
On combustion of \(25 \mathrm{ml}\) liquid \(\mathrm{CH}_3 \mathrm{COOH}\) whose specific gravity is 1.2 , gives all gaseous product, voluemof product at STP will be :
1 11.2 lit
2 22.4 lit
3 44.8 lit
4 33.6 lit
Explanation:
\(\mathrm{CH}_3 \mathrm{COOH} \rightarrow 2 \mathrm{CO}+2 \mathrm{H}_2 \mathrm{O}\)No. of moles of \(\mathrm{CH}_3 \mathrm{COOH}=\frac{25 \times 1.2}{60}\)
NCERT 15
1 RBTS PAPER
160648
According to the Bohr theory, which of the following transitions in the hydrogen atom will give rise to the least energetic photon :
1 \(n=6\) to \(n=1\)
2 \(n=5\) to \(n=4\)
3 \(n=6\) to \(n=5\)
4 \(n=5\) to \(n=3\)
Explanation:
\(\Delta \mathrm{E} \propto\left[\frac{1}{\mathrm{~m}_1^2}-\frac{1}{\mathrm{n}_2^2}\right]\), where \(\mathrm{n}_2>\mathrm{n}_1\) \(\therefore n=6\) to \(n=5\) will give least energetic photon.
AIPMT Mains 2011
1 RBTS PAPER
160615
Calculate the mass percentage of sodium element and sulphur element respectively in sodium sulphate :
1 \(34.39 \%, 20.53 \%\)
2 \(32.39 \%, 22.53 \%\)
3 \(38.39 \%, 25.53 \%\)
4 \(23.39 \%, 32.53 \%\)
Explanation:
\(\begin{aligned} & \text { Mass } \% \text { of sodium }=\frac{2 \times 23}{142} \times 100=32.39 \% \\ & \text { Mass } \% \text { of sulphur }=\frac{32}{142} \times 100=22.53 \% \end{aligned}\)
NCERT-15
1 RBTS PAPER
160616
10 mole each of \(\mathrm{KI}\) and \(\mathrm{Hgl}_2\) are mixed and react according to \(\mathrm{KI}+\mathrm{Hgl}_2 \rightarrow \mathrm{K}_2\left[\mathrm{Hgl}_4\right]\) The moles of product formed during the reaction will be:
1 2.5
2 5
3 7.5
4 0.50
Explanation:
\(2 \mathrm{KI}+\mathrm{HgI}_2 \rightarrow \mathrm{K}_2\left[\mathrm{Hgl}_4\right]\) 2 mol KI formed 1 mole \(\mathrm{K}_2\left[\mathrm{Hgl}_4\right]\) 1 mole KI formed \(1 / 2\) mole \(\mathrm{K}_2\left[\mathrm{Hgl}_4\right]\) 10 mole KI formed \(\frac{1}{2} \times 10\) mole \(\mathrm{K}_2\left[\mathrm{HgI}_4\right]\) \(=5\) mole \(\mathrm{K}_2\left[\mathrm{Hgl}_4\right]\)
160629
Two element \(X\) and \(Y\) combined to form a molecule of \(X_2 Y\) type. If mass of 0.1 mole of \(X\) weighted 2.3 gm and weight of \(3.01 \times 10^{20}\) atom of \(Y\) is \(\mathbf{0 . 0 0 8} \mathbf{g m}\). Molar mass \(X_2 Y\) will be :
1 62
2 31
3 55
4 39
Explanation:
MM of \(x=\frac{2.3}{0.1}\) and MM of \(y=\frac{0.008 \times 6.021 \times 10^{23}}{3.01 \times 10^{20}}\)
NCERT 15
1 RBTS PAPER
160630
On combustion of \(25 \mathrm{ml}\) liquid \(\mathrm{CH}_3 \mathrm{COOH}\) whose specific gravity is 1.2 , gives all gaseous product, voluemof product at STP will be :
1 11.2 lit
2 22.4 lit
3 44.8 lit
4 33.6 lit
Explanation:
\(\mathrm{CH}_3 \mathrm{COOH} \rightarrow 2 \mathrm{CO}+2 \mathrm{H}_2 \mathrm{O}\)No. of moles of \(\mathrm{CH}_3 \mathrm{COOH}=\frac{25 \times 1.2}{60}\)
NCERT 15
1 RBTS PAPER
160648
According to the Bohr theory, which of the following transitions in the hydrogen atom will give rise to the least energetic photon :
1 \(n=6\) to \(n=1\)
2 \(n=5\) to \(n=4\)
3 \(n=6\) to \(n=5\)
4 \(n=5\) to \(n=3\)
Explanation:
\(\Delta \mathrm{E} \propto\left[\frac{1}{\mathrm{~m}_1^2}-\frac{1}{\mathrm{n}_2^2}\right]\), where \(\mathrm{n}_2>\mathrm{n}_1\) \(\therefore n=6\) to \(n=5\) will give least energetic photon.
AIPMT Mains 2011
1 RBTS PAPER
160615
Calculate the mass percentage of sodium element and sulphur element respectively in sodium sulphate :
1 \(34.39 \%, 20.53 \%\)
2 \(32.39 \%, 22.53 \%\)
3 \(38.39 \%, 25.53 \%\)
4 \(23.39 \%, 32.53 \%\)
Explanation:
\(\begin{aligned} & \text { Mass } \% \text { of sodium }=\frac{2 \times 23}{142} \times 100=32.39 \% \\ & \text { Mass } \% \text { of sulphur }=\frac{32}{142} \times 100=22.53 \% \end{aligned}\)
NCERT-15
1 RBTS PAPER
160616
10 mole each of \(\mathrm{KI}\) and \(\mathrm{Hgl}_2\) are mixed and react according to \(\mathrm{KI}+\mathrm{Hgl}_2 \rightarrow \mathrm{K}_2\left[\mathrm{Hgl}_4\right]\) The moles of product formed during the reaction will be:
1 2.5
2 5
3 7.5
4 0.50
Explanation:
\(2 \mathrm{KI}+\mathrm{HgI}_2 \rightarrow \mathrm{K}_2\left[\mathrm{Hgl}_4\right]\) 2 mol KI formed 1 mole \(\mathrm{K}_2\left[\mathrm{Hgl}_4\right]\) 1 mole KI formed \(1 / 2\) mole \(\mathrm{K}_2\left[\mathrm{Hgl}_4\right]\) 10 mole KI formed \(\frac{1}{2} \times 10\) mole \(\mathrm{K}_2\left[\mathrm{HgI}_4\right]\) \(=5\) mole \(\mathrm{K}_2\left[\mathrm{Hgl}_4\right]\)
160629
Two element \(X\) and \(Y\) combined to form a molecule of \(X_2 Y\) type. If mass of 0.1 mole of \(X\) weighted 2.3 gm and weight of \(3.01 \times 10^{20}\) atom of \(Y\) is \(\mathbf{0 . 0 0 8} \mathbf{g m}\). Molar mass \(X_2 Y\) will be :
1 62
2 31
3 55
4 39
Explanation:
MM of \(x=\frac{2.3}{0.1}\) and MM of \(y=\frac{0.008 \times 6.021 \times 10^{23}}{3.01 \times 10^{20}}\)
NCERT 15
1 RBTS PAPER
160630
On combustion of \(25 \mathrm{ml}\) liquid \(\mathrm{CH}_3 \mathrm{COOH}\) whose specific gravity is 1.2 , gives all gaseous product, voluemof product at STP will be :
1 11.2 lit
2 22.4 lit
3 44.8 lit
4 33.6 lit
Explanation:
\(\mathrm{CH}_3 \mathrm{COOH} \rightarrow 2 \mathrm{CO}+2 \mathrm{H}_2 \mathrm{O}\)No. of moles of \(\mathrm{CH}_3 \mathrm{COOH}=\frac{25 \times 1.2}{60}\)
NCERT 15
1 RBTS PAPER
160648
According to the Bohr theory, which of the following transitions in the hydrogen atom will give rise to the least energetic photon :
1 \(n=6\) to \(n=1\)
2 \(n=5\) to \(n=4\)
3 \(n=6\) to \(n=5\)
4 \(n=5\) to \(n=3\)
Explanation:
\(\Delta \mathrm{E} \propto\left[\frac{1}{\mathrm{~m}_1^2}-\frac{1}{\mathrm{n}_2^2}\right]\), where \(\mathrm{n}_2>\mathrm{n}_1\) \(\therefore n=6\) to \(n=5\) will give least energetic photon.
AIPMT Mains 2011
1 RBTS PAPER
160615
Calculate the mass percentage of sodium element and sulphur element respectively in sodium sulphate :
1 \(34.39 \%, 20.53 \%\)
2 \(32.39 \%, 22.53 \%\)
3 \(38.39 \%, 25.53 \%\)
4 \(23.39 \%, 32.53 \%\)
Explanation:
\(\begin{aligned} & \text { Mass } \% \text { of sodium }=\frac{2 \times 23}{142} \times 100=32.39 \% \\ & \text { Mass } \% \text { of sulphur }=\frac{32}{142} \times 100=22.53 \% \end{aligned}\)
NCERT-15
1 RBTS PAPER
160616
10 mole each of \(\mathrm{KI}\) and \(\mathrm{Hgl}_2\) are mixed and react according to \(\mathrm{KI}+\mathrm{Hgl}_2 \rightarrow \mathrm{K}_2\left[\mathrm{Hgl}_4\right]\) The moles of product formed during the reaction will be:
1 2.5
2 5
3 7.5
4 0.50
Explanation:
\(2 \mathrm{KI}+\mathrm{HgI}_2 \rightarrow \mathrm{K}_2\left[\mathrm{Hgl}_4\right]\) 2 mol KI formed 1 mole \(\mathrm{K}_2\left[\mathrm{Hgl}_4\right]\) 1 mole KI formed \(1 / 2\) mole \(\mathrm{K}_2\left[\mathrm{Hgl}_4\right]\) 10 mole KI formed \(\frac{1}{2} \times 10\) mole \(\mathrm{K}_2\left[\mathrm{HgI}_4\right]\) \(=5\) mole \(\mathrm{K}_2\left[\mathrm{Hgl}_4\right]\)
