285208
0.48 g of an. organic compound on complete combustion produced 0.22 g of\(\mathrm{CO}_2\). The percentage of C in the given organic compound is :
285209
A pure compound contains 2.4 g of C ,\(1.2 \times 10^{23}\) atoms of \(\mathrm{H}, 0.2\) moles of oxygen atoms. Its empirical formula is
1 \(\mathrm{C}_2 \mathrm{HO}\)
2 \(\mathrm{C}_2 \mathrm{H}_2 \mathrm{O}_2\)
3 \(\mathrm{CH}_2 \mathrm{O}\)
4 CHO
Explanation:
(d) 2.4 g of\(\mathrm{C}=\frac{2.4}{12}=0.2 \mathrm{~mol}\)
\(1.2 \times 10^{23}\) atoms of \(\mathrm{H}=\frac{1.2 \times 10^{23}}{6 \times 10^{23}}=0.2 \mathrm{~mol}\)
\(0.2\) mol of oxygen atoms
\(\therefore\) Simplest ratio \(=\mathrm{C}: \mathrm{H}: \mathrm{O}\)
\(=0.2: 0.2: 0.2=1: 1: 1\)
Empirical formula \(=\mathrm{CHO}\)
Karnataka CET 2021
CHEMISTRY(KCET)
285210
A gas mixture contains\(25 \%\) He and \(75 \% \mathrm{CH}_4\) by volume at a given temperature and pressure. The percentage by mass of methane in the mixture is approximately \(\qquad\)
1 \(25 \%\)
2 \(92 \%\)
3 \(8 \%\)
4 \(75 \%\)
Explanation:
(b) 75 mL of\(\mathrm{CH}_4+25 \mathrm{~mL}\) of \(\mathrm{He}=100 \mathrm{~mL}\) of gas
16 g of \(\mathrm{CH}_4\) has volume 22400 mL (At STP)
Mass of \(\mathrm{CH}_4\) by volume
\(=\frac{75 \times 16}{22400}=0.05357\)
Mass of He by volume
\(=\frac{25 \times 4}{22400}=0.004464\)
Mass percentage of \(\mathrm{CH}_4\)
\(=\frac{0.05357}{0.05357+0.004464} \times 100 \approx 92 \%\)
Karnataka CET 2020
CHEMISTRY(KCET)
285211
0.4 g of dihydrogen is made to react with 7.1 g of dichlorine to form hydrogen chloride. The volume of hydrogen chloride formed at 273 K and 1 bar pressure is
1 4.54 L
2 90.8 L
3 45.4 L
4 9.08 L
Explanation:
(a)\(\mathrm{H}_2(\mathrm{~g})+\mathrm{Cl}_2(\mathrm{~g}) \rightarrow 2 \mathrm{HCl}(\mathrm{g})\)
2 g of \(\mathrm{H}_2\) reacts with 71 g of \(\mathrm{Cl}_2\).
Mass of \(\mathrm{H}_2\) required to react with 7.1 g of \(\mathrm{Cl}_2\)
\(=\frac{2 \times 7.1}{71}=0.2 \mathrm{~g}\)
\(\therefore \mathrm{Cl}_2\) is the limiting reagent.
71 g of \(\mathrm{Cl}_2\) produces \(2 \times 22.7 \mathrm{~L}\) of \(\mathrm{HCl}(\mathrm{g})\)
\(\therefore\) Volume of \(\mathrm{HCl}(\mathrm{g})\) produce
\(=\frac{2 \times 22.7 \times 7.1}{71}=4.54 \mathrm{~L}\)
Karnataka CET 2020
CHEMISTRY(KCET)
285212
The mass of AgCl precipitated when a solution containing 11.70 g of NaCl is added to a solution containing 3.4 g of\(\mathrm{AgNO}_3\) is [Atomic mass of \(\mathrm{Ag}=\) 108, Atomic mass of \(\mathrm{Na}=23\) ]
1 5.74 g
2 1.17 g
3 2.87 g
4 6.8 g
Explanation:
(c)\(\mathrm{NaCl}+\mathrm{AgNO}_3 \rightarrow \mathrm{AgCl}+\mathrm{NaNO}_3\)
No. of moles : \(\frac{11.70}{58.5} \frac{3.4}{170}\)
\(=0.2=0.02\)
Here, \(\mathrm{AgNO}_3\) is limiting reagent.
\(\therefore\) Mass of AgCl precipitated
\(=0.02 \times 143.5=2.87 \mathrm{~g}\)
285208
0.48 g of an. organic compound on complete combustion produced 0.22 g of\(\mathrm{CO}_2\). The percentage of C in the given organic compound is :
285209
A pure compound contains 2.4 g of C ,\(1.2 \times 10^{23}\) atoms of \(\mathrm{H}, 0.2\) moles of oxygen atoms. Its empirical formula is
1 \(\mathrm{C}_2 \mathrm{HO}\)
2 \(\mathrm{C}_2 \mathrm{H}_2 \mathrm{O}_2\)
3 \(\mathrm{CH}_2 \mathrm{O}\)
4 CHO
Explanation:
(d) 2.4 g of\(\mathrm{C}=\frac{2.4}{12}=0.2 \mathrm{~mol}\)
\(1.2 \times 10^{23}\) atoms of \(\mathrm{H}=\frac{1.2 \times 10^{23}}{6 \times 10^{23}}=0.2 \mathrm{~mol}\)
\(0.2\) mol of oxygen atoms
\(\therefore\) Simplest ratio \(=\mathrm{C}: \mathrm{H}: \mathrm{O}\)
\(=0.2: 0.2: 0.2=1: 1: 1\)
Empirical formula \(=\mathrm{CHO}\)
Karnataka CET 2021
CHEMISTRY(KCET)
285210
A gas mixture contains\(25 \%\) He and \(75 \% \mathrm{CH}_4\) by volume at a given temperature and pressure. The percentage by mass of methane in the mixture is approximately \(\qquad\)
1 \(25 \%\)
2 \(92 \%\)
3 \(8 \%\)
4 \(75 \%\)
Explanation:
(b) 75 mL of\(\mathrm{CH}_4+25 \mathrm{~mL}\) of \(\mathrm{He}=100 \mathrm{~mL}\) of gas
16 g of \(\mathrm{CH}_4\) has volume 22400 mL (At STP)
Mass of \(\mathrm{CH}_4\) by volume
\(=\frac{75 \times 16}{22400}=0.05357\)
Mass of He by volume
\(=\frac{25 \times 4}{22400}=0.004464\)
Mass percentage of \(\mathrm{CH}_4\)
\(=\frac{0.05357}{0.05357+0.004464} \times 100 \approx 92 \%\)
Karnataka CET 2020
CHEMISTRY(KCET)
285211
0.4 g of dihydrogen is made to react with 7.1 g of dichlorine to form hydrogen chloride. The volume of hydrogen chloride formed at 273 K and 1 bar pressure is
1 4.54 L
2 90.8 L
3 45.4 L
4 9.08 L
Explanation:
(a)\(\mathrm{H}_2(\mathrm{~g})+\mathrm{Cl}_2(\mathrm{~g}) \rightarrow 2 \mathrm{HCl}(\mathrm{g})\)
2 g of \(\mathrm{H}_2\) reacts with 71 g of \(\mathrm{Cl}_2\).
Mass of \(\mathrm{H}_2\) required to react with 7.1 g of \(\mathrm{Cl}_2\)
\(=\frac{2 \times 7.1}{71}=0.2 \mathrm{~g}\)
\(\therefore \mathrm{Cl}_2\) is the limiting reagent.
71 g of \(\mathrm{Cl}_2\) produces \(2 \times 22.7 \mathrm{~L}\) of \(\mathrm{HCl}(\mathrm{g})\)
\(\therefore\) Volume of \(\mathrm{HCl}(\mathrm{g})\) produce
\(=\frac{2 \times 22.7 \times 7.1}{71}=4.54 \mathrm{~L}\)
Karnataka CET 2020
CHEMISTRY(KCET)
285212
The mass of AgCl precipitated when a solution containing 11.70 g of NaCl is added to a solution containing 3.4 g of\(\mathrm{AgNO}_3\) is [Atomic mass of \(\mathrm{Ag}=\) 108, Atomic mass of \(\mathrm{Na}=23\) ]
1 5.74 g
2 1.17 g
3 2.87 g
4 6.8 g
Explanation:
(c)\(\mathrm{NaCl}+\mathrm{AgNO}_3 \rightarrow \mathrm{AgCl}+\mathrm{NaNO}_3\)
No. of moles : \(\frac{11.70}{58.5} \frac{3.4}{170}\)
\(=0.2=0.02\)
Here, \(\mathrm{AgNO}_3\) is limiting reagent.
\(\therefore\) Mass of AgCl precipitated
\(=0.02 \times 143.5=2.87 \mathrm{~g}\)
285208
0.48 g of an. organic compound on complete combustion produced 0.22 g of\(\mathrm{CO}_2\). The percentage of C in the given organic compound is :
285209
A pure compound contains 2.4 g of C ,\(1.2 \times 10^{23}\) atoms of \(\mathrm{H}, 0.2\) moles of oxygen atoms. Its empirical formula is
1 \(\mathrm{C}_2 \mathrm{HO}\)
2 \(\mathrm{C}_2 \mathrm{H}_2 \mathrm{O}_2\)
3 \(\mathrm{CH}_2 \mathrm{O}\)
4 CHO
Explanation:
(d) 2.4 g of\(\mathrm{C}=\frac{2.4}{12}=0.2 \mathrm{~mol}\)
\(1.2 \times 10^{23}\) atoms of \(\mathrm{H}=\frac{1.2 \times 10^{23}}{6 \times 10^{23}}=0.2 \mathrm{~mol}\)
\(0.2\) mol of oxygen atoms
\(\therefore\) Simplest ratio \(=\mathrm{C}: \mathrm{H}: \mathrm{O}\)
\(=0.2: 0.2: 0.2=1: 1: 1\)
Empirical formula \(=\mathrm{CHO}\)
Karnataka CET 2021
CHEMISTRY(KCET)
285210
A gas mixture contains\(25 \%\) He and \(75 \% \mathrm{CH}_4\) by volume at a given temperature and pressure. The percentage by mass of methane in the mixture is approximately \(\qquad\)
1 \(25 \%\)
2 \(92 \%\)
3 \(8 \%\)
4 \(75 \%\)
Explanation:
(b) 75 mL of\(\mathrm{CH}_4+25 \mathrm{~mL}\) of \(\mathrm{He}=100 \mathrm{~mL}\) of gas
16 g of \(\mathrm{CH}_4\) has volume 22400 mL (At STP)
Mass of \(\mathrm{CH}_4\) by volume
\(=\frac{75 \times 16}{22400}=0.05357\)
Mass of He by volume
\(=\frac{25 \times 4}{22400}=0.004464\)
Mass percentage of \(\mathrm{CH}_4\)
\(=\frac{0.05357}{0.05357+0.004464} \times 100 \approx 92 \%\)
Karnataka CET 2020
CHEMISTRY(KCET)
285211
0.4 g of dihydrogen is made to react with 7.1 g of dichlorine to form hydrogen chloride. The volume of hydrogen chloride formed at 273 K and 1 bar pressure is
1 4.54 L
2 90.8 L
3 45.4 L
4 9.08 L
Explanation:
(a)\(\mathrm{H}_2(\mathrm{~g})+\mathrm{Cl}_2(\mathrm{~g}) \rightarrow 2 \mathrm{HCl}(\mathrm{g})\)
2 g of \(\mathrm{H}_2\) reacts with 71 g of \(\mathrm{Cl}_2\).
Mass of \(\mathrm{H}_2\) required to react with 7.1 g of \(\mathrm{Cl}_2\)
\(=\frac{2 \times 7.1}{71}=0.2 \mathrm{~g}\)
\(\therefore \mathrm{Cl}_2\) is the limiting reagent.
71 g of \(\mathrm{Cl}_2\) produces \(2 \times 22.7 \mathrm{~L}\) of \(\mathrm{HCl}(\mathrm{g})\)
\(\therefore\) Volume of \(\mathrm{HCl}(\mathrm{g})\) produce
\(=\frac{2 \times 22.7 \times 7.1}{71}=4.54 \mathrm{~L}\)
Karnataka CET 2020
CHEMISTRY(KCET)
285212
The mass of AgCl precipitated when a solution containing 11.70 g of NaCl is added to a solution containing 3.4 g of\(\mathrm{AgNO}_3\) is [Atomic mass of \(\mathrm{Ag}=\) 108, Atomic mass of \(\mathrm{Na}=23\) ]
1 5.74 g
2 1.17 g
3 2.87 g
4 6.8 g
Explanation:
(c)\(\mathrm{NaCl}+\mathrm{AgNO}_3 \rightarrow \mathrm{AgCl}+\mathrm{NaNO}_3\)
No. of moles : \(\frac{11.70}{58.5} \frac{3.4}{170}\)
\(=0.2=0.02\)
Here, \(\mathrm{AgNO}_3\) is limiting reagent.
\(\therefore\) Mass of AgCl precipitated
\(=0.02 \times 143.5=2.87 \mathrm{~g}\)
285208
0.48 g of an. organic compound on complete combustion produced 0.22 g of\(\mathrm{CO}_2\). The percentage of C in the given organic compound is :
285209
A pure compound contains 2.4 g of C ,\(1.2 \times 10^{23}\) atoms of \(\mathrm{H}, 0.2\) moles of oxygen atoms. Its empirical formula is
1 \(\mathrm{C}_2 \mathrm{HO}\)
2 \(\mathrm{C}_2 \mathrm{H}_2 \mathrm{O}_2\)
3 \(\mathrm{CH}_2 \mathrm{O}\)
4 CHO
Explanation:
(d) 2.4 g of\(\mathrm{C}=\frac{2.4}{12}=0.2 \mathrm{~mol}\)
\(1.2 \times 10^{23}\) atoms of \(\mathrm{H}=\frac{1.2 \times 10^{23}}{6 \times 10^{23}}=0.2 \mathrm{~mol}\)
\(0.2\) mol of oxygen atoms
\(\therefore\) Simplest ratio \(=\mathrm{C}: \mathrm{H}: \mathrm{O}\)
\(=0.2: 0.2: 0.2=1: 1: 1\)
Empirical formula \(=\mathrm{CHO}\)
Karnataka CET 2021
CHEMISTRY(KCET)
285210
A gas mixture contains\(25 \%\) He and \(75 \% \mathrm{CH}_4\) by volume at a given temperature and pressure. The percentage by mass of methane in the mixture is approximately \(\qquad\)
1 \(25 \%\)
2 \(92 \%\)
3 \(8 \%\)
4 \(75 \%\)
Explanation:
(b) 75 mL of\(\mathrm{CH}_4+25 \mathrm{~mL}\) of \(\mathrm{He}=100 \mathrm{~mL}\) of gas
16 g of \(\mathrm{CH}_4\) has volume 22400 mL (At STP)
Mass of \(\mathrm{CH}_4\) by volume
\(=\frac{75 \times 16}{22400}=0.05357\)
Mass of He by volume
\(=\frac{25 \times 4}{22400}=0.004464\)
Mass percentage of \(\mathrm{CH}_4\)
\(=\frac{0.05357}{0.05357+0.004464} \times 100 \approx 92 \%\)
Karnataka CET 2020
CHEMISTRY(KCET)
285211
0.4 g of dihydrogen is made to react with 7.1 g of dichlorine to form hydrogen chloride. The volume of hydrogen chloride formed at 273 K and 1 bar pressure is
1 4.54 L
2 90.8 L
3 45.4 L
4 9.08 L
Explanation:
(a)\(\mathrm{H}_2(\mathrm{~g})+\mathrm{Cl}_2(\mathrm{~g}) \rightarrow 2 \mathrm{HCl}(\mathrm{g})\)
2 g of \(\mathrm{H}_2\) reacts with 71 g of \(\mathrm{Cl}_2\).
Mass of \(\mathrm{H}_2\) required to react with 7.1 g of \(\mathrm{Cl}_2\)
\(=\frac{2 \times 7.1}{71}=0.2 \mathrm{~g}\)
\(\therefore \mathrm{Cl}_2\) is the limiting reagent.
71 g of \(\mathrm{Cl}_2\) produces \(2 \times 22.7 \mathrm{~L}\) of \(\mathrm{HCl}(\mathrm{g})\)
\(\therefore\) Volume of \(\mathrm{HCl}(\mathrm{g})\) produce
\(=\frac{2 \times 22.7 \times 7.1}{71}=4.54 \mathrm{~L}\)
Karnataka CET 2020
CHEMISTRY(KCET)
285212
The mass of AgCl precipitated when a solution containing 11.70 g of NaCl is added to a solution containing 3.4 g of\(\mathrm{AgNO}_3\) is [Atomic mass of \(\mathrm{Ag}=\) 108, Atomic mass of \(\mathrm{Na}=23\) ]
1 5.74 g
2 1.17 g
3 2.87 g
4 6.8 g
Explanation:
(c)\(\mathrm{NaCl}+\mathrm{AgNO}_3 \rightarrow \mathrm{AgCl}+\mathrm{NaNO}_3\)
No. of moles : \(\frac{11.70}{58.5} \frac{3.4}{170}\)
\(=0.2=0.02\)
Here, \(\mathrm{AgNO}_3\) is limiting reagent.
\(\therefore\) Mass of AgCl precipitated
\(=0.02 \times 143.5=2.87 \mathrm{~g}\)
285208
0.48 g of an. organic compound on complete combustion produced 0.22 g of\(\mathrm{CO}_2\). The percentage of C in the given organic compound is :
285209
A pure compound contains 2.4 g of C ,\(1.2 \times 10^{23}\) atoms of \(\mathrm{H}, 0.2\) moles of oxygen atoms. Its empirical formula is
1 \(\mathrm{C}_2 \mathrm{HO}\)
2 \(\mathrm{C}_2 \mathrm{H}_2 \mathrm{O}_2\)
3 \(\mathrm{CH}_2 \mathrm{O}\)
4 CHO
Explanation:
(d) 2.4 g of\(\mathrm{C}=\frac{2.4}{12}=0.2 \mathrm{~mol}\)
\(1.2 \times 10^{23}\) atoms of \(\mathrm{H}=\frac{1.2 \times 10^{23}}{6 \times 10^{23}}=0.2 \mathrm{~mol}\)
\(0.2\) mol of oxygen atoms
\(\therefore\) Simplest ratio \(=\mathrm{C}: \mathrm{H}: \mathrm{O}\)
\(=0.2: 0.2: 0.2=1: 1: 1\)
Empirical formula \(=\mathrm{CHO}\)
Karnataka CET 2021
CHEMISTRY(KCET)
285210
A gas mixture contains\(25 \%\) He and \(75 \% \mathrm{CH}_4\) by volume at a given temperature and pressure. The percentage by mass of methane in the mixture is approximately \(\qquad\)
1 \(25 \%\)
2 \(92 \%\)
3 \(8 \%\)
4 \(75 \%\)
Explanation:
(b) 75 mL of\(\mathrm{CH}_4+25 \mathrm{~mL}\) of \(\mathrm{He}=100 \mathrm{~mL}\) of gas
16 g of \(\mathrm{CH}_4\) has volume 22400 mL (At STP)
Mass of \(\mathrm{CH}_4\) by volume
\(=\frac{75 \times 16}{22400}=0.05357\)
Mass of He by volume
\(=\frac{25 \times 4}{22400}=0.004464\)
Mass percentage of \(\mathrm{CH}_4\)
\(=\frac{0.05357}{0.05357+0.004464} \times 100 \approx 92 \%\)
Karnataka CET 2020
CHEMISTRY(KCET)
285211
0.4 g of dihydrogen is made to react with 7.1 g of dichlorine to form hydrogen chloride. The volume of hydrogen chloride formed at 273 K and 1 bar pressure is
1 4.54 L
2 90.8 L
3 45.4 L
4 9.08 L
Explanation:
(a)\(\mathrm{H}_2(\mathrm{~g})+\mathrm{Cl}_2(\mathrm{~g}) \rightarrow 2 \mathrm{HCl}(\mathrm{g})\)
2 g of \(\mathrm{H}_2\) reacts with 71 g of \(\mathrm{Cl}_2\).
Mass of \(\mathrm{H}_2\) required to react with 7.1 g of \(\mathrm{Cl}_2\)
\(=\frac{2 \times 7.1}{71}=0.2 \mathrm{~g}\)
\(\therefore \mathrm{Cl}_2\) is the limiting reagent.
71 g of \(\mathrm{Cl}_2\) produces \(2 \times 22.7 \mathrm{~L}\) of \(\mathrm{HCl}(\mathrm{g})\)
\(\therefore\) Volume of \(\mathrm{HCl}(\mathrm{g})\) produce
\(=\frac{2 \times 22.7 \times 7.1}{71}=4.54 \mathrm{~L}\)
Karnataka CET 2020
CHEMISTRY(KCET)
285212
The mass of AgCl precipitated when a solution containing 11.70 g of NaCl is added to a solution containing 3.4 g of\(\mathrm{AgNO}_3\) is [Atomic mass of \(\mathrm{Ag}=\) 108, Atomic mass of \(\mathrm{Na}=23\) ]
1 5.74 g
2 1.17 g
3 2.87 g
4 6.8 g
Explanation:
(c)\(\mathrm{NaCl}+\mathrm{AgNO}_3 \rightarrow \mathrm{AgCl}+\mathrm{NaNO}_3\)
No. of moles : \(\frac{11.70}{58.5} \frac{3.4}{170}\)
\(=0.2=0.02\)
Here, \(\mathrm{AgNO}_3\) is limiting reagent.
\(\therefore\) Mass of AgCl precipitated
\(=0.02 \times 143.5=2.87 \mathrm{~g}\)
