320887
The gold number of gelatin is 0.01 . Calculate the amount of gelatin to be added to 1000 mL of a colloidal sol of gold to prevent its coagulation, before adding 1 mL of \({\rm{10}}\% \,\,{\rm{NaCl}}\) solution.
1 1
2 0.5
3 2
4 4
Explanation:
Gold number of gelatin \({\mathrm{=0.01}}\) or 0.01 mg gelatin required to be added to 10 mL of gold sol to completely prevent coagulation of 1 mL of \({\mathrm{10 \%}}\) NaCl solution. Therefore gelatin added to 1000 mL of gold sol to prevent coagulation \({\mathrm{=\dfrac{0.01 \times 1000}{10}=1 \mathrm{mg}}}\).
CHXII05:SURFACE CHEMISTRY
320888
Highest flocculating power for the coagulation of negatively charged sol is -
1 \(\mathrm{Na}^{+}\)
2 \(\mathrm{Be}^{2+}\)
3 \(\mathrm{PO}_{4}^{3-}\)
4 \(\mathrm{SO}_{4}^{2-}\)
Explanation:
Conceptual Questions
JEE - 2021
CHXII05:SURFACE CHEMISTRY
320889
Which of the following is most effective in causing the coagulation of ferric hydroxide sol?
\(\mathrm{Fe}(\mathrm{OH})_{3}\) is positive sol. \({{\rm{K}}_{\rm{4}}}\left[ {{\rm{Fe(CN}}{{\rm{)}}_{\rm{6}}}} \right]\) will provide \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}\) for coagulation having highest magnitude of -ve charge among given options.
CHXII05:SURFACE CHEMISTRY
320890
The gold number of three substances \({\text{ A, B}}\) and \(\mathrm{C}\) are \(0.05,0.8\) and 0.3 . The substance with maximum protective power is
1 A
2 B
3 \(\mathrm{C}\)
4 All of these
Explanation:
Protective power \(\alpha \dfrac{1}{\text { Gold number }}\) = \(\frac{{\text{1}}}{{{\text{0}}{\text{.05}}}}{\text{ = 20}}\)
CHXII05:SURFACE CHEMISTRY
320891
When excess of electrolyte is added to a colloid it
320887
The gold number of gelatin is 0.01 . Calculate the amount of gelatin to be added to 1000 mL of a colloidal sol of gold to prevent its coagulation, before adding 1 mL of \({\rm{10}}\% \,\,{\rm{NaCl}}\) solution.
1 1
2 0.5
3 2
4 4
Explanation:
Gold number of gelatin \({\mathrm{=0.01}}\) or 0.01 mg gelatin required to be added to 10 mL of gold sol to completely prevent coagulation of 1 mL of \({\mathrm{10 \%}}\) NaCl solution. Therefore gelatin added to 1000 mL of gold sol to prevent coagulation \({\mathrm{=\dfrac{0.01 \times 1000}{10}=1 \mathrm{mg}}}\).
CHXII05:SURFACE CHEMISTRY
320888
Highest flocculating power for the coagulation of negatively charged sol is -
1 \(\mathrm{Na}^{+}\)
2 \(\mathrm{Be}^{2+}\)
3 \(\mathrm{PO}_{4}^{3-}\)
4 \(\mathrm{SO}_{4}^{2-}\)
Explanation:
Conceptual Questions
JEE - 2021
CHXII05:SURFACE CHEMISTRY
320889
Which of the following is most effective in causing the coagulation of ferric hydroxide sol?
\(\mathrm{Fe}(\mathrm{OH})_{3}\) is positive sol. \({{\rm{K}}_{\rm{4}}}\left[ {{\rm{Fe(CN}}{{\rm{)}}_{\rm{6}}}} \right]\) will provide \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}\) for coagulation having highest magnitude of -ve charge among given options.
CHXII05:SURFACE CHEMISTRY
320890
The gold number of three substances \({\text{ A, B}}\) and \(\mathrm{C}\) are \(0.05,0.8\) and 0.3 . The substance with maximum protective power is
1 A
2 B
3 \(\mathrm{C}\)
4 All of these
Explanation:
Protective power \(\alpha \dfrac{1}{\text { Gold number }}\) = \(\frac{{\text{1}}}{{{\text{0}}{\text{.05}}}}{\text{ = 20}}\)
CHXII05:SURFACE CHEMISTRY
320891
When excess of electrolyte is added to a colloid it
320887
The gold number of gelatin is 0.01 . Calculate the amount of gelatin to be added to 1000 mL of a colloidal sol of gold to prevent its coagulation, before adding 1 mL of \({\rm{10}}\% \,\,{\rm{NaCl}}\) solution.
1 1
2 0.5
3 2
4 4
Explanation:
Gold number of gelatin \({\mathrm{=0.01}}\) or 0.01 mg gelatin required to be added to 10 mL of gold sol to completely prevent coagulation of 1 mL of \({\mathrm{10 \%}}\) NaCl solution. Therefore gelatin added to 1000 mL of gold sol to prevent coagulation \({\mathrm{=\dfrac{0.01 \times 1000}{10}=1 \mathrm{mg}}}\).
CHXII05:SURFACE CHEMISTRY
320888
Highest flocculating power for the coagulation of negatively charged sol is -
1 \(\mathrm{Na}^{+}\)
2 \(\mathrm{Be}^{2+}\)
3 \(\mathrm{PO}_{4}^{3-}\)
4 \(\mathrm{SO}_{4}^{2-}\)
Explanation:
Conceptual Questions
JEE - 2021
CHXII05:SURFACE CHEMISTRY
320889
Which of the following is most effective in causing the coagulation of ferric hydroxide sol?
\(\mathrm{Fe}(\mathrm{OH})_{3}\) is positive sol. \({{\rm{K}}_{\rm{4}}}\left[ {{\rm{Fe(CN}}{{\rm{)}}_{\rm{6}}}} \right]\) will provide \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}\) for coagulation having highest magnitude of -ve charge among given options.
CHXII05:SURFACE CHEMISTRY
320890
The gold number of three substances \({\text{ A, B}}\) and \(\mathrm{C}\) are \(0.05,0.8\) and 0.3 . The substance with maximum protective power is
1 A
2 B
3 \(\mathrm{C}\)
4 All of these
Explanation:
Protective power \(\alpha \dfrac{1}{\text { Gold number }}\) = \(\frac{{\text{1}}}{{{\text{0}}{\text{.05}}}}{\text{ = 20}}\)
CHXII05:SURFACE CHEMISTRY
320891
When excess of electrolyte is added to a colloid it
NEET Test Series from KOTA - 10 Papers In MS WORD
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CHXII05:SURFACE CHEMISTRY
320887
The gold number of gelatin is 0.01 . Calculate the amount of gelatin to be added to 1000 mL of a colloidal sol of gold to prevent its coagulation, before adding 1 mL of \({\rm{10}}\% \,\,{\rm{NaCl}}\) solution.
1 1
2 0.5
3 2
4 4
Explanation:
Gold number of gelatin \({\mathrm{=0.01}}\) or 0.01 mg gelatin required to be added to 10 mL of gold sol to completely prevent coagulation of 1 mL of \({\mathrm{10 \%}}\) NaCl solution. Therefore gelatin added to 1000 mL of gold sol to prevent coagulation \({\mathrm{=\dfrac{0.01 \times 1000}{10}=1 \mathrm{mg}}}\).
CHXII05:SURFACE CHEMISTRY
320888
Highest flocculating power for the coagulation of negatively charged sol is -
1 \(\mathrm{Na}^{+}\)
2 \(\mathrm{Be}^{2+}\)
3 \(\mathrm{PO}_{4}^{3-}\)
4 \(\mathrm{SO}_{4}^{2-}\)
Explanation:
Conceptual Questions
JEE - 2021
CHXII05:SURFACE CHEMISTRY
320889
Which of the following is most effective in causing the coagulation of ferric hydroxide sol?
\(\mathrm{Fe}(\mathrm{OH})_{3}\) is positive sol. \({{\rm{K}}_{\rm{4}}}\left[ {{\rm{Fe(CN}}{{\rm{)}}_{\rm{6}}}} \right]\) will provide \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}\) for coagulation having highest magnitude of -ve charge among given options.
CHXII05:SURFACE CHEMISTRY
320890
The gold number of three substances \({\text{ A, B}}\) and \(\mathrm{C}\) are \(0.05,0.8\) and 0.3 . The substance with maximum protective power is
1 A
2 B
3 \(\mathrm{C}\)
4 All of these
Explanation:
Protective power \(\alpha \dfrac{1}{\text { Gold number }}\) = \(\frac{{\text{1}}}{{{\text{0}}{\text{.05}}}}{\text{ = 20}}\)
CHXII05:SURFACE CHEMISTRY
320891
When excess of electrolyte is added to a colloid it
320887
The gold number of gelatin is 0.01 . Calculate the amount of gelatin to be added to 1000 mL of a colloidal sol of gold to prevent its coagulation, before adding 1 mL of \({\rm{10}}\% \,\,{\rm{NaCl}}\) solution.
1 1
2 0.5
3 2
4 4
Explanation:
Gold number of gelatin \({\mathrm{=0.01}}\) or 0.01 mg gelatin required to be added to 10 mL of gold sol to completely prevent coagulation of 1 mL of \({\mathrm{10 \%}}\) NaCl solution. Therefore gelatin added to 1000 mL of gold sol to prevent coagulation \({\mathrm{=\dfrac{0.01 \times 1000}{10}=1 \mathrm{mg}}}\).
CHXII05:SURFACE CHEMISTRY
320888
Highest flocculating power for the coagulation of negatively charged sol is -
1 \(\mathrm{Na}^{+}\)
2 \(\mathrm{Be}^{2+}\)
3 \(\mathrm{PO}_{4}^{3-}\)
4 \(\mathrm{SO}_{4}^{2-}\)
Explanation:
Conceptual Questions
JEE - 2021
CHXII05:SURFACE CHEMISTRY
320889
Which of the following is most effective in causing the coagulation of ferric hydroxide sol?
\(\mathrm{Fe}(\mathrm{OH})_{3}\) is positive sol. \({{\rm{K}}_{\rm{4}}}\left[ {{\rm{Fe(CN}}{{\rm{)}}_{\rm{6}}}} \right]\) will provide \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}\) for coagulation having highest magnitude of -ve charge among given options.
CHXII05:SURFACE CHEMISTRY
320890
The gold number of three substances \({\text{ A, B}}\) and \(\mathrm{C}\) are \(0.05,0.8\) and 0.3 . The substance with maximum protective power is
1 A
2 B
3 \(\mathrm{C}\)
4 All of these
Explanation:
Protective power \(\alpha \dfrac{1}{\text { Gold number }}\) = \(\frac{{\text{1}}}{{{\text{0}}{\text{.05}}}}{\text{ = 20}}\)
CHXII05:SURFACE CHEMISTRY
320891
When excess of electrolyte is added to a colloid it