306884
The weight of \({{\rm{H}}_{\rm{2}}}{{\rm{C}}_{\rm{2}}}{{\rm{O}}_{\rm{4}}}{\rm{.2}}{{\rm{H}}_{\rm{2}}}{\rm{O}}\) required to prepare 500 mL of 0.2N solution is
306886
Number of milli equivalents of solute in 0.5 litres of 0.2N solution is
1 10
2 1
3 100
4 1000
Explanation:
No of milli equivalents \({\rm{ = Normality \times volume}}{\mkern 1mu} {\mkern 1mu} {\rm{in}}{\mkern 1mu} {\mkern 1mu} {\rm{mL = 0}}{\rm{.2 \times 500 = 100}}\).
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306887
The number of milli equivalents of acid in 100 mL of 0.5 NHCl solution is
1 50
2 100
3 25
4 200
Explanation:
Number of milli equivalents = NV(mL) \({\rm{ = 0}}{\rm{.5 \times 100 = 50}}\)
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306888
6.90 gm of a metal carbonate were dissolved in 60 mL of 2N HCl. The excess acid was neutralised by 20 mL of 1N NaOH. What is the equivalent wt. of metal ?
1 40
2 20
3 19
4 39
Explanation:
Equivalents of HCl taken \({\rm{ = 60 \times 2 \times 1}}{{\rm{0}}^{{\rm{ - 3}}}}\) Equivalents of HCl present after the reaction \({\rm{ = 20 \times 1 \times 1}}{{\rm{0}}^{{\rm{ - 3}}}}\) \(\therefore \) Equivalents of HCl utilised \({\rm{ = 100 \times 1}}{{\rm{0}}^{{\rm{ - 3}}}}\) \(\therefore {\rm{100 \times 1}}{{\rm{0}}^{{\rm{ - 3}}}}\) equivalents of metal carbonate \({\rm{ = 6}}{\rm{.90}}{\mkern 1mu} {\mkern 1mu} {\rm{gm}}\) \(\therefore 1\) equivalents of metal carbonate \({\rm{ = }}\frac{{{\rm{6}}{\rm{.90}}}}{{{\rm{1}}{{\rm{0}}^{{\rm{ - 1}}}}}}{\rm{ = 69 gm}}\) \(\therefore \) Equivalent weight of metal = 69 – 30 = 39 [because equivalent weight of carbonate = 30]
306884
The weight of \({{\rm{H}}_{\rm{2}}}{{\rm{C}}_{\rm{2}}}{{\rm{O}}_{\rm{4}}}{\rm{.2}}{{\rm{H}}_{\rm{2}}}{\rm{O}}\) required to prepare 500 mL of 0.2N solution is
306886
Number of milli equivalents of solute in 0.5 litres of 0.2N solution is
1 10
2 1
3 100
4 1000
Explanation:
No of milli equivalents \({\rm{ = Normality \times volume}}{\mkern 1mu} {\mkern 1mu} {\rm{in}}{\mkern 1mu} {\mkern 1mu} {\rm{mL = 0}}{\rm{.2 \times 500 = 100}}\).
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306887
The number of milli equivalents of acid in 100 mL of 0.5 NHCl solution is
1 50
2 100
3 25
4 200
Explanation:
Number of milli equivalents = NV(mL) \({\rm{ = 0}}{\rm{.5 \times 100 = 50}}\)
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306888
6.90 gm of a metal carbonate were dissolved in 60 mL of 2N HCl. The excess acid was neutralised by 20 mL of 1N NaOH. What is the equivalent wt. of metal ?
1 40
2 20
3 19
4 39
Explanation:
Equivalents of HCl taken \({\rm{ = 60 \times 2 \times 1}}{{\rm{0}}^{{\rm{ - 3}}}}\) Equivalents of HCl present after the reaction \({\rm{ = 20 \times 1 \times 1}}{{\rm{0}}^{{\rm{ - 3}}}}\) \(\therefore \) Equivalents of HCl utilised \({\rm{ = 100 \times 1}}{{\rm{0}}^{{\rm{ - 3}}}}\) \(\therefore {\rm{100 \times 1}}{{\rm{0}}^{{\rm{ - 3}}}}\) equivalents of metal carbonate \({\rm{ = 6}}{\rm{.90}}{\mkern 1mu} {\mkern 1mu} {\rm{gm}}\) \(\therefore 1\) equivalents of metal carbonate \({\rm{ = }}\frac{{{\rm{6}}{\rm{.90}}}}{{{\rm{1}}{{\rm{0}}^{{\rm{ - 1}}}}}}{\rm{ = 69 gm}}\) \(\therefore \) Equivalent weight of metal = 69 – 30 = 39 [because equivalent weight of carbonate = 30]
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306884
The weight of \({{\rm{H}}_{\rm{2}}}{{\rm{C}}_{\rm{2}}}{{\rm{O}}_{\rm{4}}}{\rm{.2}}{{\rm{H}}_{\rm{2}}}{\rm{O}}\) required to prepare 500 mL of 0.2N solution is
306886
Number of milli equivalents of solute in 0.5 litres of 0.2N solution is
1 10
2 1
3 100
4 1000
Explanation:
No of milli equivalents \({\rm{ = Normality \times volume}}{\mkern 1mu} {\mkern 1mu} {\rm{in}}{\mkern 1mu} {\mkern 1mu} {\rm{mL = 0}}{\rm{.2 \times 500 = 100}}\).
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306887
The number of milli equivalents of acid in 100 mL of 0.5 NHCl solution is
1 50
2 100
3 25
4 200
Explanation:
Number of milli equivalents = NV(mL) \({\rm{ = 0}}{\rm{.5 \times 100 = 50}}\)
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306888
6.90 gm of a metal carbonate were dissolved in 60 mL of 2N HCl. The excess acid was neutralised by 20 mL of 1N NaOH. What is the equivalent wt. of metal ?
1 40
2 20
3 19
4 39
Explanation:
Equivalents of HCl taken \({\rm{ = 60 \times 2 \times 1}}{{\rm{0}}^{{\rm{ - 3}}}}\) Equivalents of HCl present after the reaction \({\rm{ = 20 \times 1 \times 1}}{{\rm{0}}^{{\rm{ - 3}}}}\) \(\therefore \) Equivalents of HCl utilised \({\rm{ = 100 \times 1}}{{\rm{0}}^{{\rm{ - 3}}}}\) \(\therefore {\rm{100 \times 1}}{{\rm{0}}^{{\rm{ - 3}}}}\) equivalents of metal carbonate \({\rm{ = 6}}{\rm{.90}}{\mkern 1mu} {\mkern 1mu} {\rm{gm}}\) \(\therefore 1\) equivalents of metal carbonate \({\rm{ = }}\frac{{{\rm{6}}{\rm{.90}}}}{{{\rm{1}}{{\rm{0}}^{{\rm{ - 1}}}}}}{\rm{ = 69 gm}}\) \(\therefore \) Equivalent weight of metal = 69 – 30 = 39 [because equivalent weight of carbonate = 30]
306884
The weight of \({{\rm{H}}_{\rm{2}}}{{\rm{C}}_{\rm{2}}}{{\rm{O}}_{\rm{4}}}{\rm{.2}}{{\rm{H}}_{\rm{2}}}{\rm{O}}\) required to prepare 500 mL of 0.2N solution is
306886
Number of milli equivalents of solute in 0.5 litres of 0.2N solution is
1 10
2 1
3 100
4 1000
Explanation:
No of milli equivalents \({\rm{ = Normality \times volume}}{\mkern 1mu} {\mkern 1mu} {\rm{in}}{\mkern 1mu} {\mkern 1mu} {\rm{mL = 0}}{\rm{.2 \times 500 = 100}}\).
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306887
The number of milli equivalents of acid in 100 mL of 0.5 NHCl solution is
1 50
2 100
3 25
4 200
Explanation:
Number of milli equivalents = NV(mL) \({\rm{ = 0}}{\rm{.5 \times 100 = 50}}\)
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306888
6.90 gm of a metal carbonate were dissolved in 60 mL of 2N HCl. The excess acid was neutralised by 20 mL of 1N NaOH. What is the equivalent wt. of metal ?
1 40
2 20
3 19
4 39
Explanation:
Equivalents of HCl taken \({\rm{ = 60 \times 2 \times 1}}{{\rm{0}}^{{\rm{ - 3}}}}\) Equivalents of HCl present after the reaction \({\rm{ = 20 \times 1 \times 1}}{{\rm{0}}^{{\rm{ - 3}}}}\) \(\therefore \) Equivalents of HCl utilised \({\rm{ = 100 \times 1}}{{\rm{0}}^{{\rm{ - 3}}}}\) \(\therefore {\rm{100 \times 1}}{{\rm{0}}^{{\rm{ - 3}}}}\) equivalents of metal carbonate \({\rm{ = 6}}{\rm{.90}}{\mkern 1mu} {\mkern 1mu} {\rm{gm}}\) \(\therefore 1\) equivalents of metal carbonate \({\rm{ = }}\frac{{{\rm{6}}{\rm{.90}}}}{{{\rm{1}}{{\rm{0}}^{{\rm{ - 1}}}}}}{\rm{ = 69 gm}}\) \(\therefore \) Equivalent weight of metal = 69 – 30 = 39 [because equivalent weight of carbonate = 30]
306884
The weight of \({{\rm{H}}_{\rm{2}}}{{\rm{C}}_{\rm{2}}}{{\rm{O}}_{\rm{4}}}{\rm{.2}}{{\rm{H}}_{\rm{2}}}{\rm{O}}\) required to prepare 500 mL of 0.2N solution is
306886
Number of milli equivalents of solute in 0.5 litres of 0.2N solution is
1 10
2 1
3 100
4 1000
Explanation:
No of milli equivalents \({\rm{ = Normality \times volume}}{\mkern 1mu} {\mkern 1mu} {\rm{in}}{\mkern 1mu} {\mkern 1mu} {\rm{mL = 0}}{\rm{.2 \times 500 = 100}}\).
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306887
The number of milli equivalents of acid in 100 mL of 0.5 NHCl solution is
1 50
2 100
3 25
4 200
Explanation:
Number of milli equivalents = NV(mL) \({\rm{ = 0}}{\rm{.5 \times 100 = 50}}\)
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306888
6.90 gm of a metal carbonate were dissolved in 60 mL of 2N HCl. The excess acid was neutralised by 20 mL of 1N NaOH. What is the equivalent wt. of metal ?
1 40
2 20
3 19
4 39
Explanation:
Equivalents of HCl taken \({\rm{ = 60 \times 2 \times 1}}{{\rm{0}}^{{\rm{ - 3}}}}\) Equivalents of HCl present after the reaction \({\rm{ = 20 \times 1 \times 1}}{{\rm{0}}^{{\rm{ - 3}}}}\) \(\therefore \) Equivalents of HCl utilised \({\rm{ = 100 \times 1}}{{\rm{0}}^{{\rm{ - 3}}}}\) \(\therefore {\rm{100 \times 1}}{{\rm{0}}^{{\rm{ - 3}}}}\) equivalents of metal carbonate \({\rm{ = 6}}{\rm{.90}}{\mkern 1mu} {\mkern 1mu} {\rm{gm}}\) \(\therefore 1\) equivalents of metal carbonate \({\rm{ = }}\frac{{{\rm{6}}{\rm{.90}}}}{{{\rm{1}}{{\rm{0}}^{{\rm{ - 1}}}}}}{\rm{ = 69 gm}}\) \(\therefore \) Equivalent weight of metal = 69 – 30 = 39 [because equivalent weight of carbonate = 30]