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CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306974 When 22.4 L of \({{\rm{H}}_{\rm{2}}}{\rm{(g)}}\) is mixed with 11.2 L of \({\rm{C}}{{\rm{l}}_{\rm{2}}}{\rm{(g)}}\), each at STP, the moles of \({\rm{HCl}}\,{\rm{(g)}}\) formed is equal to

1 1 mole of HCl (g)

2 2 moles of HCl (g)

3 0.5 mole of HCl (g)

4 1.5 mole of HCl (g)

Explanation:

The given problem is related to the concept of stoichiometry of chemical equations. Thus, we have to convert the given volume into their moles and then identify the limiting reagent [possessing minimum number of moles and gets completely used up in the reaction]. The limiting reagent gives the moles of product formed in the reaction.
\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{\rm{H}}_{\rm{2}}}{\rm{(g)}}\,\,{\rm{ + }}\,\,{\rm{C}}{{\rm{l}}_{\rm{2}}}{\rm{(g)}} \to {\rm{2HCl(g)}}\)
\({\rm{Initial vol}}{\rm{. 22}}{\rm{.4 L }}\,\,\,{\rm{11}}{\rm{.2 L }}\,\,\,{\rm{2 mol}}\)
\({\mkern 1mu} {\mkern 1mu} \because \,{\text{22}}{\text{.4L}}\) volume at STP is occupied by \({\rm{C}}{{\rm{l}}_{\rm{2}}}{\rm{ = 1}}{\mkern 1mu} {\mkern 1mu} {\rm{mole}}\)
\(\therefore {\rm{1102}}{\mkern 1mu} {\mkern 1mu} {\rm{L}}\) volume will be occupied by, \({\rm{C}}{{\rm{l}}_{\rm{2}}}{\rm{ = }}\frac{{{\rm{1 \times 11}}{\rm{.2}}}}{{{\rm{22}}{\rm{.4}}}}{\rm{mole = 0}}{\rm{.5mol}}\)
Thus, \({{\rm{H}}_{\rm{2}}}{\rm{(g) + C}}{{\rm{l}}_{\rm{2}}}{\rm{(g)}} \to {\rm{2HC}}{{\rm{l}}_{\rm{2}}}{\rm{(g)}}\)
\({\rm{1}}{\mkern 1mu} {\mkern 1mu} {\rm{mol}}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\rm{0}}{\rm{.5}}{\mkern 1mu} {\mkern 1mu} {\rm{mol}}\)
Since, \({\rm{C}}{{\rm{l}}_{\rm{2}}}\) possesses minimum number of moles, thus it is the limiting reagent
As per equation,
\({\rm{1}}\,\,{\rm{mol}}\,\,{\rm{C}}{{\rm{l}}_{\rm{2}}}\,\, \equiv \,\,{\rm{2}}\,\,{\rm{mol}}\,\,{\rm{HCl}}\)
\(\therefore \,\,{\rm{0}}{\rm{.5}}{\mkern 1mu} {\mkern 1mu} {\rm{mol}}{\mkern 1mu} {\mkern 1mu} {\rm{C}}{{\rm{l}}_{\rm{2}}}{\rm{ = 2 \times 0}}{\rm{.5mol}}{\mkern 1mu} {\mkern 1mu} {\rm{HCl = 1}}{\rm{.0}}{\mkern 1mu} {\mkern 1mu} {\rm{mol}}{\mkern 1mu} {\mkern 1mu} {\rm{HCl}}\)
Hence 1.0 mole of HCl (g) is produced by 0.5 mole of \({\rm{C}}{{\rm{l}}_{\rm{2}}}{\mkern 1mu} {\mkern 1mu} {\rm{[11}}{\rm{.2L]}}\)

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306975 What mass of 95% pure \({\rm{CaC}}{{\rm{O}}_{\rm{3}}}\) will be required to neutralise 50 mL of 0.5 M HCl solution according to the following reaction?
\({\rm{CaC}}{{\rm{O}}_{{\rm{3(s)}}}}{\rm{ + 2HC}}{{\rm{l}}_{{\rm{(aq)}}}} \to {\rm{CaC}}{{\rm{L}}_{{\rm{2(g)}}}}{\rm{ + }}{{\rm{H}}_{\rm{2}}}{{\rm{O}}_{{\rm{(l)}}}}\)
[Calculate upto second place of decimal point]

1 1.32 g

2 3.65 g

3 9.50 g

4 1.25 g

NEET - 2022

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306976 The volume of \(\mathrm{Cl}_{2}\) at STP obtained on reacting \(4.35 \mathrm{~g} \,\mathrm{MnO}_{2}\) with conc. \(\mathrm{HCl}\) (At. wt. of \(\mathrm{Mn}=55\) )

1 4.48 litre

2 2.24 litres

3 1.12 litre

4 0.56 litre

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306977 A sample of a hydrate of barium chloride weighing 61 g was heated until all the water of hydration is removed. The dried sample weighed 52 g. The formula of the hydrated salt is:
(atomic mass : Ba = 137 amu , Cl = 35.5 amu)

1 \({\rm{BaC}}{{\rm{l}}_{\rm{2}}}{\rm{.4}}{{\rm{H}}_{\rm{2}}}{\rm{O}}\)

2 \({\rm{BaC}}{{\rm{l}}_{\rm{2}}}{\rm{.3}}{{\rm{H}}_{\rm{2}}}{\rm{O}}\)

3 \({\rm{BaC}}{{\rm{l}}_{\rm{2}}}{\rm{.}}{{\rm{H}}_{\rm{2}}}{\rm{O}}\)

4 \({\rm{BaC}}{{\rm{l}}_{\rm{2}}}{\rm{.2}}{{\rm{H}}_{\rm{2}}}{\rm{O}}\)

JEE - 2015

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306974 When 22.4 L of \({{\rm{H}}_{\rm{2}}}{\rm{(g)}}\) is mixed with 11.2 L of \({\rm{C}}{{\rm{l}}_{\rm{2}}}{\rm{(g)}}\), each at STP, the moles of \({\rm{HCl}}\,{\rm{(g)}}\) formed is equal to

1 1 mole of HCl (g)

2 2 moles of HCl (g)

3 0.5 mole of HCl (g)

4 1.5 mole of HCl (g)

Explanation:

The given problem is related to the concept of stoichiometry of chemical equations. Thus, we have to convert the given volume into their moles and then identify the limiting reagent [possessing minimum number of moles and gets completely used up in the reaction]. The limiting reagent gives the moles of product formed in the reaction.
\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{\rm{H}}_{\rm{2}}}{\rm{(g)}}\,\,{\rm{ + }}\,\,{\rm{C}}{{\rm{l}}_{\rm{2}}}{\rm{(g)}} \to {\rm{2HCl(g)}}\)
\({\rm{Initial vol}}{\rm{. 22}}{\rm{.4 L }}\,\,\,{\rm{11}}{\rm{.2 L }}\,\,\,{\rm{2 mol}}\)
\({\mkern 1mu} {\mkern 1mu} \because \,{\text{22}}{\text{.4L}}\) volume at STP is occupied by \({\rm{C}}{{\rm{l}}_{\rm{2}}}{\rm{ = 1}}{\mkern 1mu} {\mkern 1mu} {\rm{mole}}\)
\(\therefore {\rm{1102}}{\mkern 1mu} {\mkern 1mu} {\rm{L}}\) volume will be occupied by, \({\rm{C}}{{\rm{l}}_{\rm{2}}}{\rm{ = }}\frac{{{\rm{1 \times 11}}{\rm{.2}}}}{{{\rm{22}}{\rm{.4}}}}{\rm{mole = 0}}{\rm{.5mol}}\)
Thus, \({{\rm{H}}_{\rm{2}}}{\rm{(g) + C}}{{\rm{l}}_{\rm{2}}}{\rm{(g)}} \to {\rm{2HC}}{{\rm{l}}_{\rm{2}}}{\rm{(g)}}\)
\({\rm{1}}{\mkern 1mu} {\mkern 1mu} {\rm{mol}}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\rm{0}}{\rm{.5}}{\mkern 1mu} {\mkern 1mu} {\rm{mol}}\)
Since, \({\rm{C}}{{\rm{l}}_{\rm{2}}}\) possesses minimum number of moles, thus it is the limiting reagent
As per equation,
\({\rm{1}}\,\,{\rm{mol}}\,\,{\rm{C}}{{\rm{l}}_{\rm{2}}}\,\, \equiv \,\,{\rm{2}}\,\,{\rm{mol}}\,\,{\rm{HCl}}\)
\(\therefore \,\,{\rm{0}}{\rm{.5}}{\mkern 1mu} {\mkern 1mu} {\rm{mol}}{\mkern 1mu} {\mkern 1mu} {\rm{C}}{{\rm{l}}_{\rm{2}}}{\rm{ = 2 \times 0}}{\rm{.5mol}}{\mkern 1mu} {\mkern 1mu} {\rm{HCl = 1}}{\rm{.0}}{\mkern 1mu} {\mkern 1mu} {\rm{mol}}{\mkern 1mu} {\mkern 1mu} {\rm{HCl}}\)
Hence 1.0 mole of HCl (g) is produced by 0.5 mole of \({\rm{C}}{{\rm{l}}_{\rm{2}}}{\mkern 1mu} {\mkern 1mu} {\rm{[11}}{\rm{.2L]}}\)

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306975 What mass of 95% pure \({\rm{CaC}}{{\rm{O}}_{\rm{3}}}\) will be required to neutralise 50 mL of 0.5 M HCl solution according to the following reaction?
\({\rm{CaC}}{{\rm{O}}_{{\rm{3(s)}}}}{\rm{ + 2HC}}{{\rm{l}}_{{\rm{(aq)}}}} \to {\rm{CaC}}{{\rm{L}}_{{\rm{2(g)}}}}{\rm{ + }}{{\rm{H}}_{\rm{2}}}{{\rm{O}}_{{\rm{(l)}}}}\)
[Calculate upto second place of decimal point]

1 1.32 g

2 3.65 g

3 9.50 g

4 1.25 g

NEET - 2022

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306976 The volume of \(\mathrm{Cl}_{2}\) at STP obtained on reacting \(4.35 \mathrm{~g} \,\mathrm{MnO}_{2}\) with conc. \(\mathrm{HCl}\) (At. wt. of \(\mathrm{Mn}=55\) )

1 4.48 litre

2 2.24 litres

3 1.12 litre

4 0.56 litre

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306977 A sample of a hydrate of barium chloride weighing 61 g was heated until all the water of hydration is removed. The dried sample weighed 52 g. The formula of the hydrated salt is:
(atomic mass : Ba = 137 amu , Cl = 35.5 amu)

1 \({\rm{BaC}}{{\rm{l}}_{\rm{2}}}{\rm{.4}}{{\rm{H}}_{\rm{2}}}{\rm{O}}\)

2 \({\rm{BaC}}{{\rm{l}}_{\rm{2}}}{\rm{.3}}{{\rm{H}}_{\rm{2}}}{\rm{O}}\)

3 \({\rm{BaC}}{{\rm{l}}_{\rm{2}}}{\rm{.}}{{\rm{H}}_{\rm{2}}}{\rm{O}}\)

4 \({\rm{BaC}}{{\rm{l}}_{\rm{2}}}{\rm{.2}}{{\rm{H}}_{\rm{2}}}{\rm{O}}\)

JEE - 2015

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306974 When 22.4 L of \({{\rm{H}}_{\rm{2}}}{\rm{(g)}}\) is mixed with 11.2 L of \({\rm{C}}{{\rm{l}}_{\rm{2}}}{\rm{(g)}}\), each at STP, the moles of \({\rm{HCl}}\,{\rm{(g)}}\) formed is equal to

1 1 mole of HCl (g)

2 2 moles of HCl (g)

3 0.5 mole of HCl (g)

4 1.5 mole of HCl (g)

Explanation:

The given problem is related to the concept of stoichiometry of chemical equations. Thus, we have to convert the given volume into their moles and then identify the limiting reagent [possessing minimum number of moles and gets completely used up in the reaction]. The limiting reagent gives the moles of product formed in the reaction.
\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{\rm{H}}_{\rm{2}}}{\rm{(g)}}\,\,{\rm{ + }}\,\,{\rm{C}}{{\rm{l}}_{\rm{2}}}{\rm{(g)}} \to {\rm{2HCl(g)}}\)
\({\rm{Initial vol}}{\rm{. 22}}{\rm{.4 L }}\,\,\,{\rm{11}}{\rm{.2 L }}\,\,\,{\rm{2 mol}}\)
\({\mkern 1mu} {\mkern 1mu} \because \,{\text{22}}{\text{.4L}}\) volume at STP is occupied by \({\rm{C}}{{\rm{l}}_{\rm{2}}}{\rm{ = 1}}{\mkern 1mu} {\mkern 1mu} {\rm{mole}}\)
\(\therefore {\rm{1102}}{\mkern 1mu} {\mkern 1mu} {\rm{L}}\) volume will be occupied by, \({\rm{C}}{{\rm{l}}_{\rm{2}}}{\rm{ = }}\frac{{{\rm{1 \times 11}}{\rm{.2}}}}{{{\rm{22}}{\rm{.4}}}}{\rm{mole = 0}}{\rm{.5mol}}\)
Thus, \({{\rm{H}}_{\rm{2}}}{\rm{(g) + C}}{{\rm{l}}_{\rm{2}}}{\rm{(g)}} \to {\rm{2HC}}{{\rm{l}}_{\rm{2}}}{\rm{(g)}}\)
\({\rm{1}}{\mkern 1mu} {\mkern 1mu} {\rm{mol}}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\rm{0}}{\rm{.5}}{\mkern 1mu} {\mkern 1mu} {\rm{mol}}\)
Since, \({\rm{C}}{{\rm{l}}_{\rm{2}}}\) possesses minimum number of moles, thus it is the limiting reagent
As per equation,
\({\rm{1}}\,\,{\rm{mol}}\,\,{\rm{C}}{{\rm{l}}_{\rm{2}}}\,\, \equiv \,\,{\rm{2}}\,\,{\rm{mol}}\,\,{\rm{HCl}}\)
\(\therefore \,\,{\rm{0}}{\rm{.5}}{\mkern 1mu} {\mkern 1mu} {\rm{mol}}{\mkern 1mu} {\mkern 1mu} {\rm{C}}{{\rm{l}}_{\rm{2}}}{\rm{ = 2 \times 0}}{\rm{.5mol}}{\mkern 1mu} {\mkern 1mu} {\rm{HCl = 1}}{\rm{.0}}{\mkern 1mu} {\mkern 1mu} {\rm{mol}}{\mkern 1mu} {\mkern 1mu} {\rm{HCl}}\)
Hence 1.0 mole of HCl (g) is produced by 0.5 mole of \({\rm{C}}{{\rm{l}}_{\rm{2}}}{\mkern 1mu} {\mkern 1mu} {\rm{[11}}{\rm{.2L]}}\)

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306975 What mass of 95% pure \({\rm{CaC}}{{\rm{O}}_{\rm{3}}}\) will be required to neutralise 50 mL of 0.5 M HCl solution according to the following reaction?
\({\rm{CaC}}{{\rm{O}}_{{\rm{3(s)}}}}{\rm{ + 2HC}}{{\rm{l}}_{{\rm{(aq)}}}} \to {\rm{CaC}}{{\rm{L}}_{{\rm{2(g)}}}}{\rm{ + }}{{\rm{H}}_{\rm{2}}}{{\rm{O}}_{{\rm{(l)}}}}\)
[Calculate upto second place of decimal point]

1 1.32 g

2 3.65 g

3 9.50 g

4 1.25 g

NEET - 2022

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306976 The volume of \(\mathrm{Cl}_{2}\) at STP obtained on reacting \(4.35 \mathrm{~g} \,\mathrm{MnO}_{2}\) with conc. \(\mathrm{HCl}\) (At. wt. of \(\mathrm{Mn}=55\) )

1 4.48 litre

2 2.24 litres

3 1.12 litre

4 0.56 litre

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306977 A sample of a hydrate of barium chloride weighing 61 g was heated until all the water of hydration is removed. The dried sample weighed 52 g. The formula of the hydrated salt is:
(atomic mass : Ba = 137 amu , Cl = 35.5 amu)

1 \({\rm{BaC}}{{\rm{l}}_{\rm{2}}}{\rm{.4}}{{\rm{H}}_{\rm{2}}}{\rm{O}}\)

2 \({\rm{BaC}}{{\rm{l}}_{\rm{2}}}{\rm{.3}}{{\rm{H}}_{\rm{2}}}{\rm{O}}\)

3 \({\rm{BaC}}{{\rm{l}}_{\rm{2}}}{\rm{.}}{{\rm{H}}_{\rm{2}}}{\rm{O}}\)

4 \({\rm{BaC}}{{\rm{l}}_{\rm{2}}}{\rm{.2}}{{\rm{H}}_{\rm{2}}}{\rm{O}}\)

JEE - 2015

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306974 When 22.4 L of \({{\rm{H}}_{\rm{2}}}{\rm{(g)}}\) is mixed with 11.2 L of \({\rm{C}}{{\rm{l}}_{\rm{2}}}{\rm{(g)}}\), each at STP, the moles of \({\rm{HCl}}\,{\rm{(g)}}\) formed is equal to

1 1 mole of HCl (g)

2 2 moles of HCl (g)

3 0.5 mole of HCl (g)

4 1.5 mole of HCl (g)

Explanation:

The given problem is related to the concept of stoichiometry of chemical equations. Thus, we have to convert the given volume into their moles and then identify the limiting reagent [possessing minimum number of moles and gets completely used up in the reaction]. The limiting reagent gives the moles of product formed in the reaction.
\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{\rm{H}}_{\rm{2}}}{\rm{(g)}}\,\,{\rm{ + }}\,\,{\rm{C}}{{\rm{l}}_{\rm{2}}}{\rm{(g)}} \to {\rm{2HCl(g)}}\)
\({\rm{Initial vol}}{\rm{. 22}}{\rm{.4 L }}\,\,\,{\rm{11}}{\rm{.2 L }}\,\,\,{\rm{2 mol}}\)
\({\mkern 1mu} {\mkern 1mu} \because \,{\text{22}}{\text{.4L}}\) volume at STP is occupied by \({\rm{C}}{{\rm{l}}_{\rm{2}}}{\rm{ = 1}}{\mkern 1mu} {\mkern 1mu} {\rm{mole}}\)
\(\therefore {\rm{1102}}{\mkern 1mu} {\mkern 1mu} {\rm{L}}\) volume will be occupied by, \({\rm{C}}{{\rm{l}}_{\rm{2}}}{\rm{ = }}\frac{{{\rm{1 \times 11}}{\rm{.2}}}}{{{\rm{22}}{\rm{.4}}}}{\rm{mole = 0}}{\rm{.5mol}}\)
Thus, \({{\rm{H}}_{\rm{2}}}{\rm{(g) + C}}{{\rm{l}}_{\rm{2}}}{\rm{(g)}} \to {\rm{2HC}}{{\rm{l}}_{\rm{2}}}{\rm{(g)}}\)
\({\rm{1}}{\mkern 1mu} {\mkern 1mu} {\rm{mol}}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\rm{0}}{\rm{.5}}{\mkern 1mu} {\mkern 1mu} {\rm{mol}}\)
Since, \({\rm{C}}{{\rm{l}}_{\rm{2}}}\) possesses minimum number of moles, thus it is the limiting reagent
As per equation,
\({\rm{1}}\,\,{\rm{mol}}\,\,{\rm{C}}{{\rm{l}}_{\rm{2}}}\,\, \equiv \,\,{\rm{2}}\,\,{\rm{mol}}\,\,{\rm{HCl}}\)
\(\therefore \,\,{\rm{0}}{\rm{.5}}{\mkern 1mu} {\mkern 1mu} {\rm{mol}}{\mkern 1mu} {\mkern 1mu} {\rm{C}}{{\rm{l}}_{\rm{2}}}{\rm{ = 2 \times 0}}{\rm{.5mol}}{\mkern 1mu} {\mkern 1mu} {\rm{HCl = 1}}{\rm{.0}}{\mkern 1mu} {\mkern 1mu} {\rm{mol}}{\mkern 1mu} {\mkern 1mu} {\rm{HCl}}\)
Hence 1.0 mole of HCl (g) is produced by 0.5 mole of \({\rm{C}}{{\rm{l}}_{\rm{2}}}{\mkern 1mu} {\mkern 1mu} {\rm{[11}}{\rm{.2L]}}\)

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306975 What mass of 95% pure \({\rm{CaC}}{{\rm{O}}_{\rm{3}}}\) will be required to neutralise 50 mL of 0.5 M HCl solution according to the following reaction?
\({\rm{CaC}}{{\rm{O}}_{{\rm{3(s)}}}}{\rm{ + 2HC}}{{\rm{l}}_{{\rm{(aq)}}}} \to {\rm{CaC}}{{\rm{L}}_{{\rm{2(g)}}}}{\rm{ + }}{{\rm{H}}_{\rm{2}}}{{\rm{O}}_{{\rm{(l)}}}}\)
[Calculate upto second place of decimal point]

1 1.32 g

2 3.65 g

3 9.50 g

4 1.25 g

NEET - 2022

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306976 The volume of \(\mathrm{Cl}_{2}\) at STP obtained on reacting \(4.35 \mathrm{~g} \,\mathrm{MnO}_{2}\) with conc. \(\mathrm{HCl}\) (At. wt. of \(\mathrm{Mn}=55\) )

1 4.48 litre

2 2.24 litres

3 1.12 litre

4 0.56 litre

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306977 A sample of a hydrate of barium chloride weighing 61 g was heated until all the water of hydration is removed. The dried sample weighed 52 g. The formula of the hydrated salt is:
(atomic mass : Ba = 137 amu , Cl = 35.5 amu)

1 \({\rm{BaC}}{{\rm{l}}_{\rm{2}}}{\rm{.4}}{{\rm{H}}_{\rm{2}}}{\rm{O}}\)

2 \({\rm{BaC}}{{\rm{l}}_{\rm{2}}}{\rm{.3}}{{\rm{H}}_{\rm{2}}}{\rm{O}}\)

3 \({\rm{BaC}}{{\rm{l}}_{\rm{2}}}{\rm{.}}{{\rm{H}}_{\rm{2}}}{\rm{O}}\)

4 \({\rm{BaC}}{{\rm{l}}_{\rm{2}}}{\rm{.2}}{{\rm{H}}_{\rm{2}}}{\rm{O}}\)

JEE - 2015

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