319232
The density of \(N{H_4}OH\) solution is found to be 0.6 g/mL. It contains 35% by mass of \(N{H_4}OH\). The normality of the solution is:
1 10 N
2 4.8 N
3 0.6 N
4 6 N
Explanation:
\({\rm{Volume}}{\mkern 1mu} {\mkern 1mu} {\rm{of}}{\mkern 1mu} {\mkern 1mu} {\rm{solution = }}\frac{{{\rm{Mass}}}}{{{\rm{Density}}}}{\rm{ = }}\frac{{{\rm{100}}}}{{{\rm{0}}{\rm{.6}}}}{\rm{mL}}\) Number of equivalent of solute \({\rm{ = }}\frac{{{\rm{35}}}}{{{\rm{35}}}}{\rm{ = 1}}\) \({\rm{N = }}\frac{{{\rm{1 \times 1000}}}}{{{\rm{100/0}}{\rm{.6}}}}{\rm{ = 6}}\)
CHXII02:SOLUTIONS
319233
Concentrated nitric acid used in the laboratory work is \(68\% \) nitric acid by mass in aqueous solution. What should be the molarity of a sample of acid if the density of the solution is \({\rm{1}}{\rm{.504}}\,\,{\rm{gm}}{{\rm{L}}^{{\rm{ - 1}}}}\) ?
319234
Two bottles \({\rm{A}}\) and \({\rm{B}}\) contains \({\rm{1\,M}}\) and \({\rm{1\,m}}\) aqueous solution of sulphuric acid respectively.
1 \({\rm{A}}\) is more concentrated than \({\rm{B}}\)
2 \({\rm{B}}\) is more concentrated than \({\rm{A}}\)
3 concentration of \({\rm{A}}\) is equal to concentration of \({\rm{B}}\)
4 it is not possible to compare the concentration
Explanation:
\({\rm{1\,M}}\,{{\rm{H}}_{{\rm{2}}\,}}{\rm{S}}{{\rm{O}}_{\rm{4}}}\) means 1 mole \({{\rm{H}}_{{\rm{2}}\,}}{\rm{S}}{{\rm{O}}_{\rm{4}}}\) present in \(1000 \mathrm{cc}\) of solution whereas \(1 \mathrm{~m}\) means 1 mole \(\mathrm{H}_{2} \mathrm{SO}_{4}\) present in \(1000 \mathrm{~g}\) of water ( \(=1000 \mathrm{cc}\) of water). Total volume of \(1 \mathrm{~m}\) solution will be \({\rm{ > 1000}}\,{\rm{cc}}\) due to extra 1 mole \({{\rm{H}}_{{\rm{2}}\,}}{\rm{S}}{{\rm{O}}_{\rm{4}}}.\) Hence, number of moles in \(1\, \mathrm{M} \,\mathrm{H}_{2} \mathrm{SO}_{4}\) solution is more than the number of moles in \(1 \,\mathrm{~m} \,\mathrm{H}_{2} \mathrm{SO}_{4}\) solution. Thus, \(1 \,\mathrm{M}\) is more concentrated than \(1 \mathrm{~m}\). \(\therefore\) Option (1) is correct.
CHXII02:SOLUTIONS
319235
M = molarity of the solution m = molality of the solution d = density of the solution (in \({\rm{g}}{\rm{.m}}{{\rm{l}}^{{\rm{ - 1}}}}\) \({{\rm{M}}^{\rm{1}}}{\rm{ = }}\) gram molecular weight of solute Which of the following relations is correct?
319236
Molality (m) of 3 M aqueous solution of NaCl is (Given : Density of solution \({\mathrm{=1.25 \mathrm{~g} \mathrm{~mL}^{-1}}}\), atomic mass in \({\mathrm{\mathrm{g} \,\mathrm{mol}^{-1}: \mathrm{Na}-23, \mathrm{Cl}-35.5}}\) )
1 2.79 m
2 1.90 m
3 3.85 m
4 2.90 m
Explanation:
\({\mathrm{m=\dfrac{1000 \mathrm{M}}{1000 \mathrm{~d}-\left(\mathrm{MM}^{\prime}\right)}}}\) \(\begin{aligned}& =\dfrac{1000 \times 3}{(1000 \times 1.25)-(3 \times 58.5)} \\& =2.79 \mathrm{~m}\end{aligned}\) So, the correct option is (1).
319232
The density of \(N{H_4}OH\) solution is found to be 0.6 g/mL. It contains 35% by mass of \(N{H_4}OH\). The normality of the solution is:
1 10 N
2 4.8 N
3 0.6 N
4 6 N
Explanation:
\({\rm{Volume}}{\mkern 1mu} {\mkern 1mu} {\rm{of}}{\mkern 1mu} {\mkern 1mu} {\rm{solution = }}\frac{{{\rm{Mass}}}}{{{\rm{Density}}}}{\rm{ = }}\frac{{{\rm{100}}}}{{{\rm{0}}{\rm{.6}}}}{\rm{mL}}\) Number of equivalent of solute \({\rm{ = }}\frac{{{\rm{35}}}}{{{\rm{35}}}}{\rm{ = 1}}\) \({\rm{N = }}\frac{{{\rm{1 \times 1000}}}}{{{\rm{100/0}}{\rm{.6}}}}{\rm{ = 6}}\)
CHXII02:SOLUTIONS
319233
Concentrated nitric acid used in the laboratory work is \(68\% \) nitric acid by mass in aqueous solution. What should be the molarity of a sample of acid if the density of the solution is \({\rm{1}}{\rm{.504}}\,\,{\rm{gm}}{{\rm{L}}^{{\rm{ - 1}}}}\) ?
319234
Two bottles \({\rm{A}}\) and \({\rm{B}}\) contains \({\rm{1\,M}}\) and \({\rm{1\,m}}\) aqueous solution of sulphuric acid respectively.
1 \({\rm{A}}\) is more concentrated than \({\rm{B}}\)
2 \({\rm{B}}\) is more concentrated than \({\rm{A}}\)
3 concentration of \({\rm{A}}\) is equal to concentration of \({\rm{B}}\)
4 it is not possible to compare the concentration
Explanation:
\({\rm{1\,M}}\,{{\rm{H}}_{{\rm{2}}\,}}{\rm{S}}{{\rm{O}}_{\rm{4}}}\) means 1 mole \({{\rm{H}}_{{\rm{2}}\,}}{\rm{S}}{{\rm{O}}_{\rm{4}}}\) present in \(1000 \mathrm{cc}\) of solution whereas \(1 \mathrm{~m}\) means 1 mole \(\mathrm{H}_{2} \mathrm{SO}_{4}\) present in \(1000 \mathrm{~g}\) of water ( \(=1000 \mathrm{cc}\) of water). Total volume of \(1 \mathrm{~m}\) solution will be \({\rm{ > 1000}}\,{\rm{cc}}\) due to extra 1 mole \({{\rm{H}}_{{\rm{2}}\,}}{\rm{S}}{{\rm{O}}_{\rm{4}}}.\) Hence, number of moles in \(1\, \mathrm{M} \,\mathrm{H}_{2} \mathrm{SO}_{4}\) solution is more than the number of moles in \(1 \,\mathrm{~m} \,\mathrm{H}_{2} \mathrm{SO}_{4}\) solution. Thus, \(1 \,\mathrm{M}\) is more concentrated than \(1 \mathrm{~m}\). \(\therefore\) Option (1) is correct.
CHXII02:SOLUTIONS
319235
M = molarity of the solution m = molality of the solution d = density of the solution (in \({\rm{g}}{\rm{.m}}{{\rm{l}}^{{\rm{ - 1}}}}\) \({{\rm{M}}^{\rm{1}}}{\rm{ = }}\) gram molecular weight of solute Which of the following relations is correct?
319236
Molality (m) of 3 M aqueous solution of NaCl is (Given : Density of solution \({\mathrm{=1.25 \mathrm{~g} \mathrm{~mL}^{-1}}}\), atomic mass in \({\mathrm{\mathrm{g} \,\mathrm{mol}^{-1}: \mathrm{Na}-23, \mathrm{Cl}-35.5}}\) )
1 2.79 m
2 1.90 m
3 3.85 m
4 2.90 m
Explanation:
\({\mathrm{m=\dfrac{1000 \mathrm{M}}{1000 \mathrm{~d}-\left(\mathrm{MM}^{\prime}\right)}}}\) \(\begin{aligned}& =\dfrac{1000 \times 3}{(1000 \times 1.25)-(3 \times 58.5)} \\& =2.79 \mathrm{~m}\end{aligned}\) So, the correct option is (1).
319232
The density of \(N{H_4}OH\) solution is found to be 0.6 g/mL. It contains 35% by mass of \(N{H_4}OH\). The normality of the solution is:
1 10 N
2 4.8 N
3 0.6 N
4 6 N
Explanation:
\({\rm{Volume}}{\mkern 1mu} {\mkern 1mu} {\rm{of}}{\mkern 1mu} {\mkern 1mu} {\rm{solution = }}\frac{{{\rm{Mass}}}}{{{\rm{Density}}}}{\rm{ = }}\frac{{{\rm{100}}}}{{{\rm{0}}{\rm{.6}}}}{\rm{mL}}\) Number of equivalent of solute \({\rm{ = }}\frac{{{\rm{35}}}}{{{\rm{35}}}}{\rm{ = 1}}\) \({\rm{N = }}\frac{{{\rm{1 \times 1000}}}}{{{\rm{100/0}}{\rm{.6}}}}{\rm{ = 6}}\)
CHXII02:SOLUTIONS
319233
Concentrated nitric acid used in the laboratory work is \(68\% \) nitric acid by mass in aqueous solution. What should be the molarity of a sample of acid if the density of the solution is \({\rm{1}}{\rm{.504}}\,\,{\rm{gm}}{{\rm{L}}^{{\rm{ - 1}}}}\) ?
319234
Two bottles \({\rm{A}}\) and \({\rm{B}}\) contains \({\rm{1\,M}}\) and \({\rm{1\,m}}\) aqueous solution of sulphuric acid respectively.
1 \({\rm{A}}\) is more concentrated than \({\rm{B}}\)
2 \({\rm{B}}\) is more concentrated than \({\rm{A}}\)
3 concentration of \({\rm{A}}\) is equal to concentration of \({\rm{B}}\)
4 it is not possible to compare the concentration
Explanation:
\({\rm{1\,M}}\,{{\rm{H}}_{{\rm{2}}\,}}{\rm{S}}{{\rm{O}}_{\rm{4}}}\) means 1 mole \({{\rm{H}}_{{\rm{2}}\,}}{\rm{S}}{{\rm{O}}_{\rm{4}}}\) present in \(1000 \mathrm{cc}\) of solution whereas \(1 \mathrm{~m}\) means 1 mole \(\mathrm{H}_{2} \mathrm{SO}_{4}\) present in \(1000 \mathrm{~g}\) of water ( \(=1000 \mathrm{cc}\) of water). Total volume of \(1 \mathrm{~m}\) solution will be \({\rm{ > 1000}}\,{\rm{cc}}\) due to extra 1 mole \({{\rm{H}}_{{\rm{2}}\,}}{\rm{S}}{{\rm{O}}_{\rm{4}}}.\) Hence, number of moles in \(1\, \mathrm{M} \,\mathrm{H}_{2} \mathrm{SO}_{4}\) solution is more than the number of moles in \(1 \,\mathrm{~m} \,\mathrm{H}_{2} \mathrm{SO}_{4}\) solution. Thus, \(1 \,\mathrm{M}\) is more concentrated than \(1 \mathrm{~m}\). \(\therefore\) Option (1) is correct.
CHXII02:SOLUTIONS
319235
M = molarity of the solution m = molality of the solution d = density of the solution (in \({\rm{g}}{\rm{.m}}{{\rm{l}}^{{\rm{ - 1}}}}\) \({{\rm{M}}^{\rm{1}}}{\rm{ = }}\) gram molecular weight of solute Which of the following relations is correct?
319236
Molality (m) of 3 M aqueous solution of NaCl is (Given : Density of solution \({\mathrm{=1.25 \mathrm{~g} \mathrm{~mL}^{-1}}}\), atomic mass in \({\mathrm{\mathrm{g} \,\mathrm{mol}^{-1}: \mathrm{Na}-23, \mathrm{Cl}-35.5}}\) )
1 2.79 m
2 1.90 m
3 3.85 m
4 2.90 m
Explanation:
\({\mathrm{m=\dfrac{1000 \mathrm{M}}{1000 \mathrm{~d}-\left(\mathrm{MM}^{\prime}\right)}}}\) \(\begin{aligned}& =\dfrac{1000 \times 3}{(1000 \times 1.25)-(3 \times 58.5)} \\& =2.79 \mathrm{~m}\end{aligned}\) So, the correct option is (1).
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CHXII02:SOLUTIONS
319232
The density of \(N{H_4}OH\) solution is found to be 0.6 g/mL. It contains 35% by mass of \(N{H_4}OH\). The normality of the solution is:
1 10 N
2 4.8 N
3 0.6 N
4 6 N
Explanation:
\({\rm{Volume}}{\mkern 1mu} {\mkern 1mu} {\rm{of}}{\mkern 1mu} {\mkern 1mu} {\rm{solution = }}\frac{{{\rm{Mass}}}}{{{\rm{Density}}}}{\rm{ = }}\frac{{{\rm{100}}}}{{{\rm{0}}{\rm{.6}}}}{\rm{mL}}\) Number of equivalent of solute \({\rm{ = }}\frac{{{\rm{35}}}}{{{\rm{35}}}}{\rm{ = 1}}\) \({\rm{N = }}\frac{{{\rm{1 \times 1000}}}}{{{\rm{100/0}}{\rm{.6}}}}{\rm{ = 6}}\)
CHXII02:SOLUTIONS
319233
Concentrated nitric acid used in the laboratory work is \(68\% \) nitric acid by mass in aqueous solution. What should be the molarity of a sample of acid if the density of the solution is \({\rm{1}}{\rm{.504}}\,\,{\rm{gm}}{{\rm{L}}^{{\rm{ - 1}}}}\) ?
319234
Two bottles \({\rm{A}}\) and \({\rm{B}}\) contains \({\rm{1\,M}}\) and \({\rm{1\,m}}\) aqueous solution of sulphuric acid respectively.
1 \({\rm{A}}\) is more concentrated than \({\rm{B}}\)
2 \({\rm{B}}\) is more concentrated than \({\rm{A}}\)
3 concentration of \({\rm{A}}\) is equal to concentration of \({\rm{B}}\)
4 it is not possible to compare the concentration
Explanation:
\({\rm{1\,M}}\,{{\rm{H}}_{{\rm{2}}\,}}{\rm{S}}{{\rm{O}}_{\rm{4}}}\) means 1 mole \({{\rm{H}}_{{\rm{2}}\,}}{\rm{S}}{{\rm{O}}_{\rm{4}}}\) present in \(1000 \mathrm{cc}\) of solution whereas \(1 \mathrm{~m}\) means 1 mole \(\mathrm{H}_{2} \mathrm{SO}_{4}\) present in \(1000 \mathrm{~g}\) of water ( \(=1000 \mathrm{cc}\) of water). Total volume of \(1 \mathrm{~m}\) solution will be \({\rm{ > 1000}}\,{\rm{cc}}\) due to extra 1 mole \({{\rm{H}}_{{\rm{2}}\,}}{\rm{S}}{{\rm{O}}_{\rm{4}}}.\) Hence, number of moles in \(1\, \mathrm{M} \,\mathrm{H}_{2} \mathrm{SO}_{4}\) solution is more than the number of moles in \(1 \,\mathrm{~m} \,\mathrm{H}_{2} \mathrm{SO}_{4}\) solution. Thus, \(1 \,\mathrm{M}\) is more concentrated than \(1 \mathrm{~m}\). \(\therefore\) Option (1) is correct.
CHXII02:SOLUTIONS
319235
M = molarity of the solution m = molality of the solution d = density of the solution (in \({\rm{g}}{\rm{.m}}{{\rm{l}}^{{\rm{ - 1}}}}\) \({{\rm{M}}^{\rm{1}}}{\rm{ = }}\) gram molecular weight of solute Which of the following relations is correct?
319236
Molality (m) of 3 M aqueous solution of NaCl is (Given : Density of solution \({\mathrm{=1.25 \mathrm{~g} \mathrm{~mL}^{-1}}}\), atomic mass in \({\mathrm{\mathrm{g} \,\mathrm{mol}^{-1}: \mathrm{Na}-23, \mathrm{Cl}-35.5}}\) )
1 2.79 m
2 1.90 m
3 3.85 m
4 2.90 m
Explanation:
\({\mathrm{m=\dfrac{1000 \mathrm{M}}{1000 \mathrm{~d}-\left(\mathrm{MM}^{\prime}\right)}}}\) \(\begin{aligned}& =\dfrac{1000 \times 3}{(1000 \times 1.25)-(3 \times 58.5)} \\& =2.79 \mathrm{~m}\end{aligned}\) So, the correct option is (1).
319232
The density of \(N{H_4}OH\) solution is found to be 0.6 g/mL. It contains 35% by mass of \(N{H_4}OH\). The normality of the solution is:
1 10 N
2 4.8 N
3 0.6 N
4 6 N
Explanation:
\({\rm{Volume}}{\mkern 1mu} {\mkern 1mu} {\rm{of}}{\mkern 1mu} {\mkern 1mu} {\rm{solution = }}\frac{{{\rm{Mass}}}}{{{\rm{Density}}}}{\rm{ = }}\frac{{{\rm{100}}}}{{{\rm{0}}{\rm{.6}}}}{\rm{mL}}\) Number of equivalent of solute \({\rm{ = }}\frac{{{\rm{35}}}}{{{\rm{35}}}}{\rm{ = 1}}\) \({\rm{N = }}\frac{{{\rm{1 \times 1000}}}}{{{\rm{100/0}}{\rm{.6}}}}{\rm{ = 6}}\)
CHXII02:SOLUTIONS
319233
Concentrated nitric acid used in the laboratory work is \(68\% \) nitric acid by mass in aqueous solution. What should be the molarity of a sample of acid if the density of the solution is \({\rm{1}}{\rm{.504}}\,\,{\rm{gm}}{{\rm{L}}^{{\rm{ - 1}}}}\) ?
319234
Two bottles \({\rm{A}}\) and \({\rm{B}}\) contains \({\rm{1\,M}}\) and \({\rm{1\,m}}\) aqueous solution of sulphuric acid respectively.
1 \({\rm{A}}\) is more concentrated than \({\rm{B}}\)
2 \({\rm{B}}\) is more concentrated than \({\rm{A}}\)
3 concentration of \({\rm{A}}\) is equal to concentration of \({\rm{B}}\)
4 it is not possible to compare the concentration
Explanation:
\({\rm{1\,M}}\,{{\rm{H}}_{{\rm{2}}\,}}{\rm{S}}{{\rm{O}}_{\rm{4}}}\) means 1 mole \({{\rm{H}}_{{\rm{2}}\,}}{\rm{S}}{{\rm{O}}_{\rm{4}}}\) present in \(1000 \mathrm{cc}\) of solution whereas \(1 \mathrm{~m}\) means 1 mole \(\mathrm{H}_{2} \mathrm{SO}_{4}\) present in \(1000 \mathrm{~g}\) of water ( \(=1000 \mathrm{cc}\) of water). Total volume of \(1 \mathrm{~m}\) solution will be \({\rm{ > 1000}}\,{\rm{cc}}\) due to extra 1 mole \({{\rm{H}}_{{\rm{2}}\,}}{\rm{S}}{{\rm{O}}_{\rm{4}}}.\) Hence, number of moles in \(1\, \mathrm{M} \,\mathrm{H}_{2} \mathrm{SO}_{4}\) solution is more than the number of moles in \(1 \,\mathrm{~m} \,\mathrm{H}_{2} \mathrm{SO}_{4}\) solution. Thus, \(1 \,\mathrm{M}\) is more concentrated than \(1 \mathrm{~m}\). \(\therefore\) Option (1) is correct.
CHXII02:SOLUTIONS
319235
M = molarity of the solution m = molality of the solution d = density of the solution (in \({\rm{g}}{\rm{.m}}{{\rm{l}}^{{\rm{ - 1}}}}\) \({{\rm{M}}^{\rm{1}}}{\rm{ = }}\) gram molecular weight of solute Which of the following relations is correct?
319236
Molality (m) of 3 M aqueous solution of NaCl is (Given : Density of solution \({\mathrm{=1.25 \mathrm{~g} \mathrm{~mL}^{-1}}}\), atomic mass in \({\mathrm{\mathrm{g} \,\mathrm{mol}^{-1}: \mathrm{Na}-23, \mathrm{Cl}-35.5}}\) )
1 2.79 m
2 1.90 m
3 3.85 m
4 2.90 m
Explanation:
\({\mathrm{m=\dfrac{1000 \mathrm{M}}{1000 \mathrm{~d}-\left(\mathrm{MM}^{\prime}\right)}}}\) \(\begin{aligned}& =\dfrac{1000 \times 3}{(1000 \times 1.25)-(3 \times 58.5)} \\& =2.79 \mathrm{~m}\end{aligned}\) So, the correct option is (1).