NEET Test Series from KOTA - 10 Papers In MS WORD
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CHEMISTRY(KCET)
285217
What amount of dioxygen (in gram) contains\(1.8 \times\) \(10^{22}\) molecules?
1 0.960
2 96.0
3 0.0960
4 9.60
Explanation:
(a)\(6.022 \times 10^{23}\) molecules are present in 32 g of \(\mathrm{O}_2\)
\(\therefore \quad 1.8 \times 10^{22}\) molecules will be present in
\(\frac{32}{6.022 \times 10^{23}} \times 1.8 \times 10^{22} \mathrm{~g}=0.960{\mathrm{~g} \text { of } \mathrm{O}_2}^2\)
Karnataka CET 2015
CHEMISTRY(KCET)
285218
0.30 g of an organic compound containing\(\mathrm{C}, \mathrm{H}\) and O on combustion yields \(0.44 \mathrm{~g} \mathrm{CO}_2\) and \(0.18 \mathrm{~g} \mathrm{H}_2 \mathrm{O}\). If one mol of compound weighs 60 , then molecular formula of the compound is
1 \(\mathrm{C}_3 \mathrm{H}_8 \mathrm{O}\)
2 \(\mathrm{C}_2 \mathrm{H}_4 \mathrm{O}_2\)
3 \(\mathrm{CH}_2 \mathrm{O}\)
4 \(\mathrm{C}_4 \mathrm{H}_6 \mathrm{O}\)
Explanation:
(b) Percentage of
\(\mathrm{C}=\frac{12}{44} \times \frac{0.44}{0.30} \times 100=40 \%\)
Percentage of \(\mathrm{H}=\frac{2}{18} \times \frac{0.18}{0.30} \times 100=6.6 \%\)
Percentage of \(O=100-(40+6.6)=53.4 \%\)
\begin{tabular}{|c|c|l|l|}
\hline Element & \(\%\) & \begin{tabular}{l}
Molar \\
ratio
\end{tabular} & \begin{tabular}{l}
Simplest \\
ratio
\end{tabular} \\
\hline C & 40 & \(\frac{40}{12}=3.3\) & \(\frac{3.3}{3.3}=1\)
\end{tabular}
Hence, empirical formula \(=\mathrm{CH}_2 \mathrm{O}\)
\(\mathrm{n}=\frac{\text { Molecular mass }}{\text { Empirical formula mass }}=\frac{60}{30}=2\)
\(\Rightarrow\) Molecular formula of the compound
\(=\left(\mathrm{CH}_2 \mathrm{O}\right)_2=\mathrm{C}_2 \mathrm{H}_4 \mathrm{O}_2\)
Karnataka CET 2015
CHEMISTRY(KCET)
285219
\(25 \mathrm{~cm}^3\) of oxalic acid completely neutralised 0.064 g of sodium hydroxide. Molarity of the oxalic acid solution is
1 0.064
2 0.045
3 0.015
4 0.032
Explanation:
(d)\((\mathrm{COOH})_2+2 \mathrm{NaOH} \rightarrow(\mathrm{COONa})_2+2 \mathrm{H}_2 \mathrm{O}\)
No. of moles in 0.064 g of NaOH
\(=\frac{0.064}{40}=0.0016\)
No. of moles of oxalic acid
\(=\frac{0.0016}{2}=8 \times 10^{-4}\)
Volume of solution (in L) \(=\frac{25}{1000}\)
Hence, molarity
\(=\frac{\text { No. of moles ofsolute }}{\text { Volume of solution (in L) }}\)
\(=8 \times 10^{-4} \times \frac{1000}{25}=0.032 \mathrm{M}\)
Karnataka CET 2014
CHEMISTRY(KCET)
285220
10 g of a mixture of BaO and CaO requires\(100 \mathrm{~cm}^3\) of 2.5 M HCl to react completely. The percentage of calcium oxide in the mixture is approximately
(Given : molar mass of \(\mathrm{BaO}=153\) )
1 52.6
2 55.1
3 44.9
4 47.4
Explanation:
(a)\(\mathrm{BaO}+2 \mathrm{HCl} \rightarrow \mathrm{BaCl}_2+\mathrm{H}_2 \mathrm{O}\)
\(\mathrm{CaO}+2 \mathrm{HCl} \rightarrow \mathrm{CaCl}_2+\mathrm{H}_2 \mathrm{O}\)
Reactions show that 2 moles of mix. of BaO and CaO requires 4 moles of HCl to react completely,
Hence, no. of moles of HCl
\(=\frac{2.5 \times 100}{1000}=0.25 \mathrm{moles}\)
Now, initial moles \(=4\) moles \((\mathrm{HCl})\)
\(+2 \text { moles }(\mathrm{BaO}+\mathrm{CaO})\)
Final moles of \(\mathrm{HCl}=0.25\) moles
Final no. of moles of mixture of BaO and CaO
\(=\frac{0.25 \times 2}{4}=0.125 \mathrm{moles}\)
Let the mass of CaO be \(\mathrm{x} g\) and BaO is \((10-x) g\)
Then, \(\frac{\mathrm{x}}{56}+\frac{10-\mathrm{x}}{153}=0.125\)
[Mol. mass of \(\mathrm{CaO}=56, \mathrm{BaO}=153\) ]
\(153 \mathrm{x}+56(10-\mathrm{x})=0.125 \times 56 \times 153\)
\(153 \mathrm{x}+560-56 \mathrm{x}=0.125 \times 56 \times 153\)
\(97 \mathrm{x}=1071-560\)
\(\Rightarrow 97 \mathrm{x}=511\)
\(\Rightarrow \mathrm{x}=5.268\)
\(\therefore \%\) of \(\mathrm{CaO}=\frac{5.268}{10} \times 100=52.68 \%\)
285217
What amount of dioxygen (in gram) contains\(1.8 \times\) \(10^{22}\) molecules?
1 0.960
2 96.0
3 0.0960
4 9.60
Explanation:
(a)\(6.022 \times 10^{23}\) molecules are present in 32 g of \(\mathrm{O}_2\)
\(\therefore \quad 1.8 \times 10^{22}\) molecules will be present in
\(\frac{32}{6.022 \times 10^{23}} \times 1.8 \times 10^{22} \mathrm{~g}=0.960{\mathrm{~g} \text { of } \mathrm{O}_2}^2\)
Karnataka CET 2015
CHEMISTRY(KCET)
285218
0.30 g of an organic compound containing\(\mathrm{C}, \mathrm{H}\) and O on combustion yields \(0.44 \mathrm{~g} \mathrm{CO}_2\) and \(0.18 \mathrm{~g} \mathrm{H}_2 \mathrm{O}\). If one mol of compound weighs 60 , then molecular formula of the compound is
1 \(\mathrm{C}_3 \mathrm{H}_8 \mathrm{O}\)
2 \(\mathrm{C}_2 \mathrm{H}_4 \mathrm{O}_2\)
3 \(\mathrm{CH}_2 \mathrm{O}\)
4 \(\mathrm{C}_4 \mathrm{H}_6 \mathrm{O}\)
Explanation:
(b) Percentage of
\(\mathrm{C}=\frac{12}{44} \times \frac{0.44}{0.30} \times 100=40 \%\)
Percentage of \(\mathrm{H}=\frac{2}{18} \times \frac{0.18}{0.30} \times 100=6.6 \%\)
Percentage of \(O=100-(40+6.6)=53.4 \%\)
\begin{tabular}{|c|c|l|l|}
\hline Element & \(\%\) & \begin{tabular}{l}
Molar \\
ratio
\end{tabular} & \begin{tabular}{l}
Simplest \\
ratio
\end{tabular} \\
\hline C & 40 & \(\frac{40}{12}=3.3\) & \(\frac{3.3}{3.3}=1\)
\end{tabular}
Hence, empirical formula \(=\mathrm{CH}_2 \mathrm{O}\)
\(\mathrm{n}=\frac{\text { Molecular mass }}{\text { Empirical formula mass }}=\frac{60}{30}=2\)
\(\Rightarrow\) Molecular formula of the compound
\(=\left(\mathrm{CH}_2 \mathrm{O}\right)_2=\mathrm{C}_2 \mathrm{H}_4 \mathrm{O}_2\)
Karnataka CET 2015
CHEMISTRY(KCET)
285219
\(25 \mathrm{~cm}^3\) of oxalic acid completely neutralised 0.064 g of sodium hydroxide. Molarity of the oxalic acid solution is
1 0.064
2 0.045
3 0.015
4 0.032
Explanation:
(d)\((\mathrm{COOH})_2+2 \mathrm{NaOH} \rightarrow(\mathrm{COONa})_2+2 \mathrm{H}_2 \mathrm{O}\)
No. of moles in 0.064 g of NaOH
\(=\frac{0.064}{40}=0.0016\)
No. of moles of oxalic acid
\(=\frac{0.0016}{2}=8 \times 10^{-4}\)
Volume of solution (in L) \(=\frac{25}{1000}\)
Hence, molarity
\(=\frac{\text { No. of moles ofsolute }}{\text { Volume of solution (in L) }}\)
\(=8 \times 10^{-4} \times \frac{1000}{25}=0.032 \mathrm{M}\)
Karnataka CET 2014
CHEMISTRY(KCET)
285220
10 g of a mixture of BaO and CaO requires\(100 \mathrm{~cm}^3\) of 2.5 M HCl to react completely. The percentage of calcium oxide in the mixture is approximately
(Given : molar mass of \(\mathrm{BaO}=153\) )
1 52.6
2 55.1
3 44.9
4 47.4
Explanation:
(a)\(\mathrm{BaO}+2 \mathrm{HCl} \rightarrow \mathrm{BaCl}_2+\mathrm{H}_2 \mathrm{O}\)
\(\mathrm{CaO}+2 \mathrm{HCl} \rightarrow \mathrm{CaCl}_2+\mathrm{H}_2 \mathrm{O}\)
Reactions show that 2 moles of mix. of BaO and CaO requires 4 moles of HCl to react completely,
Hence, no. of moles of HCl
\(=\frac{2.5 \times 100}{1000}=0.25 \mathrm{moles}\)
Now, initial moles \(=4\) moles \((\mathrm{HCl})\)
\(+2 \text { moles }(\mathrm{BaO}+\mathrm{CaO})\)
Final moles of \(\mathrm{HCl}=0.25\) moles
Final no. of moles of mixture of BaO and CaO
\(=\frac{0.25 \times 2}{4}=0.125 \mathrm{moles}\)
Let the mass of CaO be \(\mathrm{x} g\) and BaO is \((10-x) g\)
Then, \(\frac{\mathrm{x}}{56}+\frac{10-\mathrm{x}}{153}=0.125\)
[Mol. mass of \(\mathrm{CaO}=56, \mathrm{BaO}=153\) ]
\(153 \mathrm{x}+56(10-\mathrm{x})=0.125 \times 56 \times 153\)
\(153 \mathrm{x}+560-56 \mathrm{x}=0.125 \times 56 \times 153\)
\(97 \mathrm{x}=1071-560\)
\(\Rightarrow 97 \mathrm{x}=511\)
\(\Rightarrow \mathrm{x}=5.268\)
\(\therefore \%\) of \(\mathrm{CaO}=\frac{5.268}{10} \times 100=52.68 \%\)
285217
What amount of dioxygen (in gram) contains\(1.8 \times\) \(10^{22}\) molecules?
1 0.960
2 96.0
3 0.0960
4 9.60
Explanation:
(a)\(6.022 \times 10^{23}\) molecules are present in 32 g of \(\mathrm{O}_2\)
\(\therefore \quad 1.8 \times 10^{22}\) molecules will be present in
\(\frac{32}{6.022 \times 10^{23}} \times 1.8 \times 10^{22} \mathrm{~g}=0.960{\mathrm{~g} \text { of } \mathrm{O}_2}^2\)
Karnataka CET 2015
CHEMISTRY(KCET)
285218
0.30 g of an organic compound containing\(\mathrm{C}, \mathrm{H}\) and O on combustion yields \(0.44 \mathrm{~g} \mathrm{CO}_2\) and \(0.18 \mathrm{~g} \mathrm{H}_2 \mathrm{O}\). If one mol of compound weighs 60 , then molecular formula of the compound is
1 \(\mathrm{C}_3 \mathrm{H}_8 \mathrm{O}\)
2 \(\mathrm{C}_2 \mathrm{H}_4 \mathrm{O}_2\)
3 \(\mathrm{CH}_2 \mathrm{O}\)
4 \(\mathrm{C}_4 \mathrm{H}_6 \mathrm{O}\)
Explanation:
(b) Percentage of
\(\mathrm{C}=\frac{12}{44} \times \frac{0.44}{0.30} \times 100=40 \%\)
Percentage of \(\mathrm{H}=\frac{2}{18} \times \frac{0.18}{0.30} \times 100=6.6 \%\)
Percentage of \(O=100-(40+6.6)=53.4 \%\)
\begin{tabular}{|c|c|l|l|}
\hline Element & \(\%\) & \begin{tabular}{l}
Molar \\
ratio
\end{tabular} & \begin{tabular}{l}
Simplest \\
ratio
\end{tabular} \\
\hline C & 40 & \(\frac{40}{12}=3.3\) & \(\frac{3.3}{3.3}=1\)
\end{tabular}
Hence, empirical formula \(=\mathrm{CH}_2 \mathrm{O}\)
\(\mathrm{n}=\frac{\text { Molecular mass }}{\text { Empirical formula mass }}=\frac{60}{30}=2\)
\(\Rightarrow\) Molecular formula of the compound
\(=\left(\mathrm{CH}_2 \mathrm{O}\right)_2=\mathrm{C}_2 \mathrm{H}_4 \mathrm{O}_2\)
Karnataka CET 2015
CHEMISTRY(KCET)
285219
\(25 \mathrm{~cm}^3\) of oxalic acid completely neutralised 0.064 g of sodium hydroxide. Molarity of the oxalic acid solution is
1 0.064
2 0.045
3 0.015
4 0.032
Explanation:
(d)\((\mathrm{COOH})_2+2 \mathrm{NaOH} \rightarrow(\mathrm{COONa})_2+2 \mathrm{H}_2 \mathrm{O}\)
No. of moles in 0.064 g of NaOH
\(=\frac{0.064}{40}=0.0016\)
No. of moles of oxalic acid
\(=\frac{0.0016}{2}=8 \times 10^{-4}\)
Volume of solution (in L) \(=\frac{25}{1000}\)
Hence, molarity
\(=\frac{\text { No. of moles ofsolute }}{\text { Volume of solution (in L) }}\)
\(=8 \times 10^{-4} \times \frac{1000}{25}=0.032 \mathrm{M}\)
Karnataka CET 2014
CHEMISTRY(KCET)
285220
10 g of a mixture of BaO and CaO requires\(100 \mathrm{~cm}^3\) of 2.5 M HCl to react completely. The percentage of calcium oxide in the mixture is approximately
(Given : molar mass of \(\mathrm{BaO}=153\) )
1 52.6
2 55.1
3 44.9
4 47.4
Explanation:
(a)\(\mathrm{BaO}+2 \mathrm{HCl} \rightarrow \mathrm{BaCl}_2+\mathrm{H}_2 \mathrm{O}\)
\(\mathrm{CaO}+2 \mathrm{HCl} \rightarrow \mathrm{CaCl}_2+\mathrm{H}_2 \mathrm{O}\)
Reactions show that 2 moles of mix. of BaO and CaO requires 4 moles of HCl to react completely,
Hence, no. of moles of HCl
\(=\frac{2.5 \times 100}{1000}=0.25 \mathrm{moles}\)
Now, initial moles \(=4\) moles \((\mathrm{HCl})\)
\(+2 \text { moles }(\mathrm{BaO}+\mathrm{CaO})\)
Final moles of \(\mathrm{HCl}=0.25\) moles
Final no. of moles of mixture of BaO and CaO
\(=\frac{0.25 \times 2}{4}=0.125 \mathrm{moles}\)
Let the mass of CaO be \(\mathrm{x} g\) and BaO is \((10-x) g\)
Then, \(\frac{\mathrm{x}}{56}+\frac{10-\mathrm{x}}{153}=0.125\)
[Mol. mass of \(\mathrm{CaO}=56, \mathrm{BaO}=153\) ]
\(153 \mathrm{x}+56(10-\mathrm{x})=0.125 \times 56 \times 153\)
\(153 \mathrm{x}+560-56 \mathrm{x}=0.125 \times 56 \times 153\)
\(97 \mathrm{x}=1071-560\)
\(\Rightarrow 97 \mathrm{x}=511\)
\(\Rightarrow \mathrm{x}=5.268\)
\(\therefore \%\) of \(\mathrm{CaO}=\frac{5.268}{10} \times 100=52.68 \%\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
CHEMISTRY(KCET)
285217
What amount of dioxygen (in gram) contains\(1.8 \times\) \(10^{22}\) molecules?
1 0.960
2 96.0
3 0.0960
4 9.60
Explanation:
(a)\(6.022 \times 10^{23}\) molecules are present in 32 g of \(\mathrm{O}_2\)
\(\therefore \quad 1.8 \times 10^{22}\) molecules will be present in
\(\frac{32}{6.022 \times 10^{23}} \times 1.8 \times 10^{22} \mathrm{~g}=0.960{\mathrm{~g} \text { of } \mathrm{O}_2}^2\)
Karnataka CET 2015
CHEMISTRY(KCET)
285218
0.30 g of an organic compound containing\(\mathrm{C}, \mathrm{H}\) and O on combustion yields \(0.44 \mathrm{~g} \mathrm{CO}_2\) and \(0.18 \mathrm{~g} \mathrm{H}_2 \mathrm{O}\). If one mol of compound weighs 60 , then molecular formula of the compound is
1 \(\mathrm{C}_3 \mathrm{H}_8 \mathrm{O}\)
2 \(\mathrm{C}_2 \mathrm{H}_4 \mathrm{O}_2\)
3 \(\mathrm{CH}_2 \mathrm{O}\)
4 \(\mathrm{C}_4 \mathrm{H}_6 \mathrm{O}\)
Explanation:
(b) Percentage of
\(\mathrm{C}=\frac{12}{44} \times \frac{0.44}{0.30} \times 100=40 \%\)
Percentage of \(\mathrm{H}=\frac{2}{18} \times \frac{0.18}{0.30} \times 100=6.6 \%\)
Percentage of \(O=100-(40+6.6)=53.4 \%\)
\begin{tabular}{|c|c|l|l|}
\hline Element & \(\%\) & \begin{tabular}{l}
Molar \\
ratio
\end{tabular} & \begin{tabular}{l}
Simplest \\
ratio
\end{tabular} \\
\hline C & 40 & \(\frac{40}{12}=3.3\) & \(\frac{3.3}{3.3}=1\)
\end{tabular}
Hence, empirical formula \(=\mathrm{CH}_2 \mathrm{O}\)
\(\mathrm{n}=\frac{\text { Molecular mass }}{\text { Empirical formula mass }}=\frac{60}{30}=2\)
\(\Rightarrow\) Molecular formula of the compound
\(=\left(\mathrm{CH}_2 \mathrm{O}\right)_2=\mathrm{C}_2 \mathrm{H}_4 \mathrm{O}_2\)
Karnataka CET 2015
CHEMISTRY(KCET)
285219
\(25 \mathrm{~cm}^3\) of oxalic acid completely neutralised 0.064 g of sodium hydroxide. Molarity of the oxalic acid solution is
1 0.064
2 0.045
3 0.015
4 0.032
Explanation:
(d)\((\mathrm{COOH})_2+2 \mathrm{NaOH} \rightarrow(\mathrm{COONa})_2+2 \mathrm{H}_2 \mathrm{O}\)
No. of moles in 0.064 g of NaOH
\(=\frac{0.064}{40}=0.0016\)
No. of moles of oxalic acid
\(=\frac{0.0016}{2}=8 \times 10^{-4}\)
Volume of solution (in L) \(=\frac{25}{1000}\)
Hence, molarity
\(=\frac{\text { No. of moles ofsolute }}{\text { Volume of solution (in L) }}\)
\(=8 \times 10^{-4} \times \frac{1000}{25}=0.032 \mathrm{M}\)
Karnataka CET 2014
CHEMISTRY(KCET)
285220
10 g of a mixture of BaO and CaO requires\(100 \mathrm{~cm}^3\) of 2.5 M HCl to react completely. The percentage of calcium oxide in the mixture is approximately
(Given : molar mass of \(\mathrm{BaO}=153\) )
1 52.6
2 55.1
3 44.9
4 47.4
Explanation:
(a)\(\mathrm{BaO}+2 \mathrm{HCl} \rightarrow \mathrm{BaCl}_2+\mathrm{H}_2 \mathrm{O}\)
\(\mathrm{CaO}+2 \mathrm{HCl} \rightarrow \mathrm{CaCl}_2+\mathrm{H}_2 \mathrm{O}\)
Reactions show that 2 moles of mix. of BaO and CaO requires 4 moles of HCl to react completely,
Hence, no. of moles of HCl
\(=\frac{2.5 \times 100}{1000}=0.25 \mathrm{moles}\)
Now, initial moles \(=4\) moles \((\mathrm{HCl})\)
\(+2 \text { moles }(\mathrm{BaO}+\mathrm{CaO})\)
Final moles of \(\mathrm{HCl}=0.25\) moles
Final no. of moles of mixture of BaO and CaO
\(=\frac{0.25 \times 2}{4}=0.125 \mathrm{moles}\)
Let the mass of CaO be \(\mathrm{x} g\) and BaO is \((10-x) g\)
Then, \(\frac{\mathrm{x}}{56}+\frac{10-\mathrm{x}}{153}=0.125\)
[Mol. mass of \(\mathrm{CaO}=56, \mathrm{BaO}=153\) ]
\(153 \mathrm{x}+56(10-\mathrm{x})=0.125 \times 56 \times 153\)
\(153 \mathrm{x}+560-56 \mathrm{x}=0.125 \times 56 \times 153\)
\(97 \mathrm{x}=1071-560\)
\(\Rightarrow 97 \mathrm{x}=511\)
\(\Rightarrow \mathrm{x}=5.268\)
\(\therefore \%\) of \(\mathrm{CaO}=\frac{5.268}{10} \times 100=52.68 \%\)