228888
$20 \mathrm{~mL}$ of acetic acid reacts with $20 \mathrm{~mL}$ of ethyl alcohol to form ethyl acetate. The density of acid and alcohol are $1 \mathrm{~g} / \mathrm{mL}$ and $0.7 \mathrm{~g} / \mathrm{mL}$ respectively. The limiting reagent in this reaction is
1 acetic acid
2 ethyl alcohol
3 acetic acid and ethyl alcohol
4 ester.
Explanation:
Mass of $20 \mathrm{~mL}$ of acetic acid $=20 \times 1=20 \mathrm{~g}$ Mass of $20 \mathrm{~mL}$ of ethyl alcohol $=20 \times 0.7=14 \mathrm{~g}$ $60 \mathrm{~g}$ of acetic acid $=46 \mathrm{~g}$ of ethyl alcohol $20 \mathrm{~g}$ of acetic acid $=15.3 \mathrm{~g}$ of ethyl alcohol Therefore, ethyl alcohol is the limiting reagent.
COMEDK-2015
Some Basic Concepts of Chemistry
228891
If $\mathrm{NaCl}$ is doped with $10^{-3} \mathrm{~mol} \%$ of $\mathrm{SrCl}_2$, the number of cation vacancies will be
1 $6.023 \times 10^{18}$
2 $1 \times 10^{-3}$
3 $6 \times 10^{12}$
4 $6.023 \times 10^{23}$
Explanation:
1 cation of $\mathrm{Sr}^{2+}$ will create 1 cation vacancy in $\mathrm{NaCl}$. Therefore, the number of cation vacancy created in the lattice of $\mathrm{NaCl}$ is equal to the number of divalent $\mathrm{Sr}^{2+}$ ions added. Concentration of cation vacancy on being doped with $10^{-3} \mathrm{~mol} \%$ of $\mathrm{SrCl}_2$. $=10^{-3} \mathrm{~mol} \%=10^{-3} / 100=10^{-5} \mathrm{~mol}$ $=10^{-5} \times 6.023 \times 10^{23}=6.023 \times 10^{18}$ Therefore, Number of cation vacancies $=6.023 \times 10^{18}$
CG PET- 2015
Some Basic Concepts of Chemistry
228894
The number of moles of electrons required to deposit $36 \mathrm{~g}$ of $\mathrm{Al}$ from an aqueous solution of $\mathrm{Al}\left(\mathrm{NO}_3\right)_3$ is (At. wt. of $\mathrm{Al}=27$ )
1 4
2 2
3 3
4 1
Explanation:
Mass of $\mathrm{Al}$ in $1 \mathrm{~mol}$ of $\mathrm{Al}\left(\mathrm{NO}_3\right)_3$ is the $27 \mathrm{~g}$. For 1 mole of $\mathrm{Al}(27 \mathrm{~g}), 3$ mole of electrons are required to deposit $27 \mathrm{~g}$ of $\mathrm{Al}$. As 3 mole of electrons required to deposit $\rightarrow 27 \mathrm{~g}$ of $\mathrm{Al}$ So, for $36 \mathrm{~g} \rightarrow 36 \times \frac{3}{27}=\frac{36}{9}=4 \mathrm{moles}$
AP EAMCET (Engg.) - 2012
Some Basic Concepts of Chemistry
228906
The weight of one molecule of a compound of molecular formula $\mathrm{C}_{60} \mathrm{H}_{122}$ is
1 $1.2 \times 10^{-20} \mathrm{~g}$
2 $5.025 \times 10^{-23} \mathrm{~g}$
3 $1.4 \times 10^{-21} \mathrm{~g}$
4 $6.023 \times 10^{-20} \mathrm{~g}$
Explanation:
The molecular weight of $\mathrm{C}_{60} \mathrm{H}_{122}$ is $842 \mathrm{~g} / \mathrm{mol}$ one mole of $\mathrm{C}_{60} \mathrm{H}_{122}$ weight $6.02 \times 10^{23} \mathrm{~g}$. Weight of one molecule of $\mathrm{C}_{60} \mathrm{H}_{122}$ is $\frac{842}{6.02 \times 10^{23}}=1.4 \times 10^{-21} \mathrm{~g}$
AIIMS-2002
Some Basic Concepts of Chemistry
228885
Pink colour of non-stoichiometric $\mathrm{LiCl}$ is due to
1 $\mathrm{Cl}^{-}$ion in the lattice
2 $\mathrm{Li}^{+}$ion in the lattice
3 electrons in the lattice
4 Both the ions in the lattice
Explanation:
Metal excess defect caused due to anion vacancies is responsible for the pink colour of $\mathrm{LiCl}$. When $\mathrm{LiCl}$ is heated, $\mathrm{Li}$ atoms gets deposited on the surface of the crystal. The $\mathrm{Cl}^{-}$ions diffuse into the surface and combine with $\mathrm{Li}$ atoms to give $\mathrm{LiCl}$. This is because of loss of electrons $\mathrm{Li}$ atoms to forms $\mathrm{Li}^{+}$. The released electrons diffuse excess into crystal and occupy anionic sites. As a result, there is an excess of $\mathrm{Li}$. The anionic sites occupied by unpaired electrons imparts a pink color to LiCl crystals.
228888
$20 \mathrm{~mL}$ of acetic acid reacts with $20 \mathrm{~mL}$ of ethyl alcohol to form ethyl acetate. The density of acid and alcohol are $1 \mathrm{~g} / \mathrm{mL}$ and $0.7 \mathrm{~g} / \mathrm{mL}$ respectively. The limiting reagent in this reaction is
1 acetic acid
2 ethyl alcohol
3 acetic acid and ethyl alcohol
4 ester.
Explanation:
Mass of $20 \mathrm{~mL}$ of acetic acid $=20 \times 1=20 \mathrm{~g}$ Mass of $20 \mathrm{~mL}$ of ethyl alcohol $=20 \times 0.7=14 \mathrm{~g}$ $60 \mathrm{~g}$ of acetic acid $=46 \mathrm{~g}$ of ethyl alcohol $20 \mathrm{~g}$ of acetic acid $=15.3 \mathrm{~g}$ of ethyl alcohol Therefore, ethyl alcohol is the limiting reagent.
COMEDK-2015
Some Basic Concepts of Chemistry
228891
If $\mathrm{NaCl}$ is doped with $10^{-3} \mathrm{~mol} \%$ of $\mathrm{SrCl}_2$, the number of cation vacancies will be
1 $6.023 \times 10^{18}$
2 $1 \times 10^{-3}$
3 $6 \times 10^{12}$
4 $6.023 \times 10^{23}$
Explanation:
1 cation of $\mathrm{Sr}^{2+}$ will create 1 cation vacancy in $\mathrm{NaCl}$. Therefore, the number of cation vacancy created in the lattice of $\mathrm{NaCl}$ is equal to the number of divalent $\mathrm{Sr}^{2+}$ ions added. Concentration of cation vacancy on being doped with $10^{-3} \mathrm{~mol} \%$ of $\mathrm{SrCl}_2$. $=10^{-3} \mathrm{~mol} \%=10^{-3} / 100=10^{-5} \mathrm{~mol}$ $=10^{-5} \times 6.023 \times 10^{23}=6.023 \times 10^{18}$ Therefore, Number of cation vacancies $=6.023 \times 10^{18}$
CG PET- 2015
Some Basic Concepts of Chemistry
228894
The number of moles of electrons required to deposit $36 \mathrm{~g}$ of $\mathrm{Al}$ from an aqueous solution of $\mathrm{Al}\left(\mathrm{NO}_3\right)_3$ is (At. wt. of $\mathrm{Al}=27$ )
1 4
2 2
3 3
4 1
Explanation:
Mass of $\mathrm{Al}$ in $1 \mathrm{~mol}$ of $\mathrm{Al}\left(\mathrm{NO}_3\right)_3$ is the $27 \mathrm{~g}$. For 1 mole of $\mathrm{Al}(27 \mathrm{~g}), 3$ mole of electrons are required to deposit $27 \mathrm{~g}$ of $\mathrm{Al}$. As 3 mole of electrons required to deposit $\rightarrow 27 \mathrm{~g}$ of $\mathrm{Al}$ So, for $36 \mathrm{~g} \rightarrow 36 \times \frac{3}{27}=\frac{36}{9}=4 \mathrm{moles}$
AP EAMCET (Engg.) - 2012
Some Basic Concepts of Chemistry
228906
The weight of one molecule of a compound of molecular formula $\mathrm{C}_{60} \mathrm{H}_{122}$ is
1 $1.2 \times 10^{-20} \mathrm{~g}$
2 $5.025 \times 10^{-23} \mathrm{~g}$
3 $1.4 \times 10^{-21} \mathrm{~g}$
4 $6.023 \times 10^{-20} \mathrm{~g}$
Explanation:
The molecular weight of $\mathrm{C}_{60} \mathrm{H}_{122}$ is $842 \mathrm{~g} / \mathrm{mol}$ one mole of $\mathrm{C}_{60} \mathrm{H}_{122}$ weight $6.02 \times 10^{23} \mathrm{~g}$. Weight of one molecule of $\mathrm{C}_{60} \mathrm{H}_{122}$ is $\frac{842}{6.02 \times 10^{23}}=1.4 \times 10^{-21} \mathrm{~g}$
AIIMS-2002
Some Basic Concepts of Chemistry
228885
Pink colour of non-stoichiometric $\mathrm{LiCl}$ is due to
1 $\mathrm{Cl}^{-}$ion in the lattice
2 $\mathrm{Li}^{+}$ion in the lattice
3 electrons in the lattice
4 Both the ions in the lattice
Explanation:
Metal excess defect caused due to anion vacancies is responsible for the pink colour of $\mathrm{LiCl}$. When $\mathrm{LiCl}$ is heated, $\mathrm{Li}$ atoms gets deposited on the surface of the crystal. The $\mathrm{Cl}^{-}$ions diffuse into the surface and combine with $\mathrm{Li}$ atoms to give $\mathrm{LiCl}$. This is because of loss of electrons $\mathrm{Li}$ atoms to forms $\mathrm{Li}^{+}$. The released electrons diffuse excess into crystal and occupy anionic sites. As a result, there is an excess of $\mathrm{Li}$. The anionic sites occupied by unpaired electrons imparts a pink color to LiCl crystals.
228888
$20 \mathrm{~mL}$ of acetic acid reacts with $20 \mathrm{~mL}$ of ethyl alcohol to form ethyl acetate. The density of acid and alcohol are $1 \mathrm{~g} / \mathrm{mL}$ and $0.7 \mathrm{~g} / \mathrm{mL}$ respectively. The limiting reagent in this reaction is
1 acetic acid
2 ethyl alcohol
3 acetic acid and ethyl alcohol
4 ester.
Explanation:
Mass of $20 \mathrm{~mL}$ of acetic acid $=20 \times 1=20 \mathrm{~g}$ Mass of $20 \mathrm{~mL}$ of ethyl alcohol $=20 \times 0.7=14 \mathrm{~g}$ $60 \mathrm{~g}$ of acetic acid $=46 \mathrm{~g}$ of ethyl alcohol $20 \mathrm{~g}$ of acetic acid $=15.3 \mathrm{~g}$ of ethyl alcohol Therefore, ethyl alcohol is the limiting reagent.
COMEDK-2015
Some Basic Concepts of Chemistry
228891
If $\mathrm{NaCl}$ is doped with $10^{-3} \mathrm{~mol} \%$ of $\mathrm{SrCl}_2$, the number of cation vacancies will be
1 $6.023 \times 10^{18}$
2 $1 \times 10^{-3}$
3 $6 \times 10^{12}$
4 $6.023 \times 10^{23}$
Explanation:
1 cation of $\mathrm{Sr}^{2+}$ will create 1 cation vacancy in $\mathrm{NaCl}$. Therefore, the number of cation vacancy created in the lattice of $\mathrm{NaCl}$ is equal to the number of divalent $\mathrm{Sr}^{2+}$ ions added. Concentration of cation vacancy on being doped with $10^{-3} \mathrm{~mol} \%$ of $\mathrm{SrCl}_2$. $=10^{-3} \mathrm{~mol} \%=10^{-3} / 100=10^{-5} \mathrm{~mol}$ $=10^{-5} \times 6.023 \times 10^{23}=6.023 \times 10^{18}$ Therefore, Number of cation vacancies $=6.023 \times 10^{18}$
CG PET- 2015
Some Basic Concepts of Chemistry
228894
The number of moles of electrons required to deposit $36 \mathrm{~g}$ of $\mathrm{Al}$ from an aqueous solution of $\mathrm{Al}\left(\mathrm{NO}_3\right)_3$ is (At. wt. of $\mathrm{Al}=27$ )
1 4
2 2
3 3
4 1
Explanation:
Mass of $\mathrm{Al}$ in $1 \mathrm{~mol}$ of $\mathrm{Al}\left(\mathrm{NO}_3\right)_3$ is the $27 \mathrm{~g}$. For 1 mole of $\mathrm{Al}(27 \mathrm{~g}), 3$ mole of electrons are required to deposit $27 \mathrm{~g}$ of $\mathrm{Al}$. As 3 mole of electrons required to deposit $\rightarrow 27 \mathrm{~g}$ of $\mathrm{Al}$ So, for $36 \mathrm{~g} \rightarrow 36 \times \frac{3}{27}=\frac{36}{9}=4 \mathrm{moles}$
AP EAMCET (Engg.) - 2012
Some Basic Concepts of Chemistry
228906
The weight of one molecule of a compound of molecular formula $\mathrm{C}_{60} \mathrm{H}_{122}$ is
1 $1.2 \times 10^{-20} \mathrm{~g}$
2 $5.025 \times 10^{-23} \mathrm{~g}$
3 $1.4 \times 10^{-21} \mathrm{~g}$
4 $6.023 \times 10^{-20} \mathrm{~g}$
Explanation:
The molecular weight of $\mathrm{C}_{60} \mathrm{H}_{122}$ is $842 \mathrm{~g} / \mathrm{mol}$ one mole of $\mathrm{C}_{60} \mathrm{H}_{122}$ weight $6.02 \times 10^{23} \mathrm{~g}$. Weight of one molecule of $\mathrm{C}_{60} \mathrm{H}_{122}$ is $\frac{842}{6.02 \times 10^{23}}=1.4 \times 10^{-21} \mathrm{~g}$
AIIMS-2002
Some Basic Concepts of Chemistry
228885
Pink colour of non-stoichiometric $\mathrm{LiCl}$ is due to
1 $\mathrm{Cl}^{-}$ion in the lattice
2 $\mathrm{Li}^{+}$ion in the lattice
3 electrons in the lattice
4 Both the ions in the lattice
Explanation:
Metal excess defect caused due to anion vacancies is responsible for the pink colour of $\mathrm{LiCl}$. When $\mathrm{LiCl}$ is heated, $\mathrm{Li}$ atoms gets deposited on the surface of the crystal. The $\mathrm{Cl}^{-}$ions diffuse into the surface and combine with $\mathrm{Li}$ atoms to give $\mathrm{LiCl}$. This is because of loss of electrons $\mathrm{Li}$ atoms to forms $\mathrm{Li}^{+}$. The released electrons diffuse excess into crystal and occupy anionic sites. As a result, there is an excess of $\mathrm{Li}$. The anionic sites occupied by unpaired electrons imparts a pink color to LiCl crystals.
228888
$20 \mathrm{~mL}$ of acetic acid reacts with $20 \mathrm{~mL}$ of ethyl alcohol to form ethyl acetate. The density of acid and alcohol are $1 \mathrm{~g} / \mathrm{mL}$ and $0.7 \mathrm{~g} / \mathrm{mL}$ respectively. The limiting reagent in this reaction is
1 acetic acid
2 ethyl alcohol
3 acetic acid and ethyl alcohol
4 ester.
Explanation:
Mass of $20 \mathrm{~mL}$ of acetic acid $=20 \times 1=20 \mathrm{~g}$ Mass of $20 \mathrm{~mL}$ of ethyl alcohol $=20 \times 0.7=14 \mathrm{~g}$ $60 \mathrm{~g}$ of acetic acid $=46 \mathrm{~g}$ of ethyl alcohol $20 \mathrm{~g}$ of acetic acid $=15.3 \mathrm{~g}$ of ethyl alcohol Therefore, ethyl alcohol is the limiting reagent.
COMEDK-2015
Some Basic Concepts of Chemistry
228891
If $\mathrm{NaCl}$ is doped with $10^{-3} \mathrm{~mol} \%$ of $\mathrm{SrCl}_2$, the number of cation vacancies will be
1 $6.023 \times 10^{18}$
2 $1 \times 10^{-3}$
3 $6 \times 10^{12}$
4 $6.023 \times 10^{23}$
Explanation:
1 cation of $\mathrm{Sr}^{2+}$ will create 1 cation vacancy in $\mathrm{NaCl}$. Therefore, the number of cation vacancy created in the lattice of $\mathrm{NaCl}$ is equal to the number of divalent $\mathrm{Sr}^{2+}$ ions added. Concentration of cation vacancy on being doped with $10^{-3} \mathrm{~mol} \%$ of $\mathrm{SrCl}_2$. $=10^{-3} \mathrm{~mol} \%=10^{-3} / 100=10^{-5} \mathrm{~mol}$ $=10^{-5} \times 6.023 \times 10^{23}=6.023 \times 10^{18}$ Therefore, Number of cation vacancies $=6.023 \times 10^{18}$
CG PET- 2015
Some Basic Concepts of Chemistry
228894
The number of moles of electrons required to deposit $36 \mathrm{~g}$ of $\mathrm{Al}$ from an aqueous solution of $\mathrm{Al}\left(\mathrm{NO}_3\right)_3$ is (At. wt. of $\mathrm{Al}=27$ )
1 4
2 2
3 3
4 1
Explanation:
Mass of $\mathrm{Al}$ in $1 \mathrm{~mol}$ of $\mathrm{Al}\left(\mathrm{NO}_3\right)_3$ is the $27 \mathrm{~g}$. For 1 mole of $\mathrm{Al}(27 \mathrm{~g}), 3$ mole of electrons are required to deposit $27 \mathrm{~g}$ of $\mathrm{Al}$. As 3 mole of electrons required to deposit $\rightarrow 27 \mathrm{~g}$ of $\mathrm{Al}$ So, for $36 \mathrm{~g} \rightarrow 36 \times \frac{3}{27}=\frac{36}{9}=4 \mathrm{moles}$
AP EAMCET (Engg.) - 2012
Some Basic Concepts of Chemistry
228906
The weight of one molecule of a compound of molecular formula $\mathrm{C}_{60} \mathrm{H}_{122}$ is
1 $1.2 \times 10^{-20} \mathrm{~g}$
2 $5.025 \times 10^{-23} \mathrm{~g}$
3 $1.4 \times 10^{-21} \mathrm{~g}$
4 $6.023 \times 10^{-20} \mathrm{~g}$
Explanation:
The molecular weight of $\mathrm{C}_{60} \mathrm{H}_{122}$ is $842 \mathrm{~g} / \mathrm{mol}$ one mole of $\mathrm{C}_{60} \mathrm{H}_{122}$ weight $6.02 \times 10^{23} \mathrm{~g}$. Weight of one molecule of $\mathrm{C}_{60} \mathrm{H}_{122}$ is $\frac{842}{6.02 \times 10^{23}}=1.4 \times 10^{-21} \mathrm{~g}$
AIIMS-2002
Some Basic Concepts of Chemistry
228885
Pink colour of non-stoichiometric $\mathrm{LiCl}$ is due to
1 $\mathrm{Cl}^{-}$ion in the lattice
2 $\mathrm{Li}^{+}$ion in the lattice
3 electrons in the lattice
4 Both the ions in the lattice
Explanation:
Metal excess defect caused due to anion vacancies is responsible for the pink colour of $\mathrm{LiCl}$. When $\mathrm{LiCl}$ is heated, $\mathrm{Li}$ atoms gets deposited on the surface of the crystal. The $\mathrm{Cl}^{-}$ions diffuse into the surface and combine with $\mathrm{Li}$ atoms to give $\mathrm{LiCl}$. This is because of loss of electrons $\mathrm{Li}$ atoms to forms $\mathrm{Li}^{+}$. The released electrons diffuse excess into crystal and occupy anionic sites. As a result, there is an excess of $\mathrm{Li}$. The anionic sites occupied by unpaired electrons imparts a pink color to LiCl crystals.
228888
$20 \mathrm{~mL}$ of acetic acid reacts with $20 \mathrm{~mL}$ of ethyl alcohol to form ethyl acetate. The density of acid and alcohol are $1 \mathrm{~g} / \mathrm{mL}$ and $0.7 \mathrm{~g} / \mathrm{mL}$ respectively. The limiting reagent in this reaction is
1 acetic acid
2 ethyl alcohol
3 acetic acid and ethyl alcohol
4 ester.
Explanation:
Mass of $20 \mathrm{~mL}$ of acetic acid $=20 \times 1=20 \mathrm{~g}$ Mass of $20 \mathrm{~mL}$ of ethyl alcohol $=20 \times 0.7=14 \mathrm{~g}$ $60 \mathrm{~g}$ of acetic acid $=46 \mathrm{~g}$ of ethyl alcohol $20 \mathrm{~g}$ of acetic acid $=15.3 \mathrm{~g}$ of ethyl alcohol Therefore, ethyl alcohol is the limiting reagent.
COMEDK-2015
Some Basic Concepts of Chemistry
228891
If $\mathrm{NaCl}$ is doped with $10^{-3} \mathrm{~mol} \%$ of $\mathrm{SrCl}_2$, the number of cation vacancies will be
1 $6.023 \times 10^{18}$
2 $1 \times 10^{-3}$
3 $6 \times 10^{12}$
4 $6.023 \times 10^{23}$
Explanation:
1 cation of $\mathrm{Sr}^{2+}$ will create 1 cation vacancy in $\mathrm{NaCl}$. Therefore, the number of cation vacancy created in the lattice of $\mathrm{NaCl}$ is equal to the number of divalent $\mathrm{Sr}^{2+}$ ions added. Concentration of cation vacancy on being doped with $10^{-3} \mathrm{~mol} \%$ of $\mathrm{SrCl}_2$. $=10^{-3} \mathrm{~mol} \%=10^{-3} / 100=10^{-5} \mathrm{~mol}$ $=10^{-5} \times 6.023 \times 10^{23}=6.023 \times 10^{18}$ Therefore, Number of cation vacancies $=6.023 \times 10^{18}$
CG PET- 2015
Some Basic Concepts of Chemistry
228894
The number of moles of electrons required to deposit $36 \mathrm{~g}$ of $\mathrm{Al}$ from an aqueous solution of $\mathrm{Al}\left(\mathrm{NO}_3\right)_3$ is (At. wt. of $\mathrm{Al}=27$ )
1 4
2 2
3 3
4 1
Explanation:
Mass of $\mathrm{Al}$ in $1 \mathrm{~mol}$ of $\mathrm{Al}\left(\mathrm{NO}_3\right)_3$ is the $27 \mathrm{~g}$. For 1 mole of $\mathrm{Al}(27 \mathrm{~g}), 3$ mole of electrons are required to deposit $27 \mathrm{~g}$ of $\mathrm{Al}$. As 3 mole of electrons required to deposit $\rightarrow 27 \mathrm{~g}$ of $\mathrm{Al}$ So, for $36 \mathrm{~g} \rightarrow 36 \times \frac{3}{27}=\frac{36}{9}=4 \mathrm{moles}$
AP EAMCET (Engg.) - 2012
Some Basic Concepts of Chemistry
228906
The weight of one molecule of a compound of molecular formula $\mathrm{C}_{60} \mathrm{H}_{122}$ is
1 $1.2 \times 10^{-20} \mathrm{~g}$
2 $5.025 \times 10^{-23} \mathrm{~g}$
3 $1.4 \times 10^{-21} \mathrm{~g}$
4 $6.023 \times 10^{-20} \mathrm{~g}$
Explanation:
The molecular weight of $\mathrm{C}_{60} \mathrm{H}_{122}$ is $842 \mathrm{~g} / \mathrm{mol}$ one mole of $\mathrm{C}_{60} \mathrm{H}_{122}$ weight $6.02 \times 10^{23} \mathrm{~g}$. Weight of one molecule of $\mathrm{C}_{60} \mathrm{H}_{122}$ is $\frac{842}{6.02 \times 10^{23}}=1.4 \times 10^{-21} \mathrm{~g}$
AIIMS-2002
Some Basic Concepts of Chemistry
228885
Pink colour of non-stoichiometric $\mathrm{LiCl}$ is due to
1 $\mathrm{Cl}^{-}$ion in the lattice
2 $\mathrm{Li}^{+}$ion in the lattice
3 electrons in the lattice
4 Both the ions in the lattice
Explanation:
Metal excess defect caused due to anion vacancies is responsible for the pink colour of $\mathrm{LiCl}$. When $\mathrm{LiCl}$ is heated, $\mathrm{Li}$ atoms gets deposited on the surface of the crystal. The $\mathrm{Cl}^{-}$ions diffuse into the surface and combine with $\mathrm{Li}$ atoms to give $\mathrm{LiCl}$. This is because of loss of electrons $\mathrm{Li}$ atoms to forms $\mathrm{Li}^{+}$. The released electrons diffuse excess into crystal and occupy anionic sites. As a result, there is an excess of $\mathrm{Li}$. The anionic sites occupied by unpaired electrons imparts a pink color to LiCl crystals.
