319220
The volume of water added to \({\text{500}}\,{\text{mL}}\,\,{\text{0}}{\text{.5M}}\,{\text{NaOH}}\) so that its strength becomes \({\text{10mg}}\,{\text{NaOH}}\) per mL is
1 \(100 \mathrm{~mL}\)
2 \(200 \mathrm{~mL}\)
3 \(250 \mathrm{~mL}\)
4 \(500 \mathrm{~mL}\)
Explanation:
1 mL has 10 mg NaOH 1000 mL will have \({\rm{ = 10 \times 1000mg = 10\;g}}\,\,{\rm{NaOH}}\) Normality \({\rm{ = }}\frac{{{\rm{ gm / litre }}}}{{{\rm{ eq}}{\rm{.wt}}{\rm{. }}}}{\rm{ = }}\frac{{{\rm{10}}}}{{{\rm{40}}}}{\rm{\;N = 0}}{\rm{.25\;N}}\) Now using normality equation, \({{\rm{N}}_{\rm{1}}}{{\rm{V}}_{\rm{1}}}{\rm{ = }}{{\rm{N}}_{\rm{2}}}{{\rm{V}}_{\rm{2}}}\) \({\rm{0}}{\rm{.5 \times 500 = 0}}{\rm{.25 \times }}{{\rm{V}}_{\rm{2}}}\) \({\rm{(}}{\rm{0}}{\rm{.5\,M}}\,\,{\rm{NaOH}} \equiv {\rm{0}}{\rm{.5\;N}}\,\,{\rm{NaOH)}}\) \({{\rm{V}}_{\rm{2}}}{\rm{ = }}\frac{{{\rm{0}}{\rm{.5 \times 500}}}}{{{\rm{0}}{\rm{.25}}}}{\rm{ = 1000\;mL}}\) Volume of water added = 1000 - 500 = 500 mL
CHXII02:SOLUTIONS
319221
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 500ml of 0.2N solution is
319222
1 kg of NaOH is added to 10 ml of 0.1N HCl, the resulting solution will
1 red litumus paper turns blue
2 turn phenolphthalein solution pink
3 turn methyl orange red
4 will have no effect on red or blue litmus paper
Explanation:
Number of moles of NaOH \({\rm{ = }}\frac{{{\rm{Given}}{\mkern 1mu} {\mkern 1mu} {\rm{mass}}}}{{{\rm{Molar}}{\mkern 1mu} {\mkern 1mu} {\rm{mass}}}}{\rm{ = }}\frac{{{\rm{1000}}}}{{{\rm{40}}}}{\rm{ = 25moles}}\) Molarity of HCl = Normality of HCl = 0.1 M Number of moles of HCl = \({\rm{M \times V}}\left( {\rm{L}} \right){\rm{ = 0}}{\rm{.1 \times 0}}{\rm{.01 = 0}}{\rm{.001}}{\mkern 1mu} {\mkern 1mu} {\rm{moles}}\) Due to a greater number of moles of NaOH, solution is basic in nature. Hence, turns red litmus paper to blue.
CHXII02:SOLUTIONS
319223
The normality of orthophosphoric acid having purity of \(70 \%\) by weight and specific gravity 1.54 is
1 \(11 \mathrm{~N}\)
2 \(22 \mathrm{~N}\)
3 \(33 \mathrm{~N}\)
4 \(44 \mathrm{~N}\)
Explanation:
\(70 \%\) by weight means \(70 \mathrm{~g}\) of orthophosphoric acid is present in \(100 \mathrm{~g}\) of acid. \(\mathrm{N}=\dfrac{\mathrm{w}}{\mathrm{Eq} \cdot \mathrm{wt}} \times \dfrac{1000}{\mathrm{~V}_{(\mathrm{cc})}}\) \(\Rightarrow \mathrm{w}=70 \mathrm{~g}\) Orthophosphoric acid is \(\mathrm{H}_{3} \mathrm{PO}_{4}\). It is tribasic acid. \(\therefore\) E.q. wt. \(=\dfrac{\text { molecular mass }}{\text { number of replaceable } \mathrm{H} \text { - atoms }}=\dfrac{98}{3}\) \(\mathrm{V}=\dfrac{\text { mass }}{\text { density }}=\dfrac{100}{1.54}\) Putting values in Eq. (1) \(\mathrm{N}=\dfrac{70 \times 3 \times 1000 \times 1.54}{98 \times 100}=33 \mathrm{~N}\) (OR) \(\mathrm{N}=\dfrac{\% \mathrm{w} / \mathrm{w} \times \mathrm{d} \times 10}{\mathrm{GEW}}=\dfrac{70 \times 1.54 \times 10}{98 / 3}=33 \mathrm{~N}\)
319220
The volume of water added to \({\text{500}}\,{\text{mL}}\,\,{\text{0}}{\text{.5M}}\,{\text{NaOH}}\) so that its strength becomes \({\text{10mg}}\,{\text{NaOH}}\) per mL is
1 \(100 \mathrm{~mL}\)
2 \(200 \mathrm{~mL}\)
3 \(250 \mathrm{~mL}\)
4 \(500 \mathrm{~mL}\)
Explanation:
1 mL has 10 mg NaOH 1000 mL will have \({\rm{ = 10 \times 1000mg = 10\;g}}\,\,{\rm{NaOH}}\) Normality \({\rm{ = }}\frac{{{\rm{ gm / litre }}}}{{{\rm{ eq}}{\rm{.wt}}{\rm{. }}}}{\rm{ = }}\frac{{{\rm{10}}}}{{{\rm{40}}}}{\rm{\;N = 0}}{\rm{.25\;N}}\) Now using normality equation, \({{\rm{N}}_{\rm{1}}}{{\rm{V}}_{\rm{1}}}{\rm{ = }}{{\rm{N}}_{\rm{2}}}{{\rm{V}}_{\rm{2}}}\) \({\rm{0}}{\rm{.5 \times 500 = 0}}{\rm{.25 \times }}{{\rm{V}}_{\rm{2}}}\) \({\rm{(}}{\rm{0}}{\rm{.5\,M}}\,\,{\rm{NaOH}} \equiv {\rm{0}}{\rm{.5\;N}}\,\,{\rm{NaOH)}}\) \({{\rm{V}}_{\rm{2}}}{\rm{ = }}\frac{{{\rm{0}}{\rm{.5 \times 500}}}}{{{\rm{0}}{\rm{.25}}}}{\rm{ = 1000\;mL}}\) Volume of water added = 1000 - 500 = 500 mL
CHXII02:SOLUTIONS
319221
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 500ml of 0.2N solution is
319222
1 kg of NaOH is added to 10 ml of 0.1N HCl, the resulting solution will
1 red litumus paper turns blue
2 turn phenolphthalein solution pink
3 turn methyl orange red
4 will have no effect on red or blue litmus paper
Explanation:
Number of moles of NaOH \({\rm{ = }}\frac{{{\rm{Given}}{\mkern 1mu} {\mkern 1mu} {\rm{mass}}}}{{{\rm{Molar}}{\mkern 1mu} {\mkern 1mu} {\rm{mass}}}}{\rm{ = }}\frac{{{\rm{1000}}}}{{{\rm{40}}}}{\rm{ = 25moles}}\) Molarity of HCl = Normality of HCl = 0.1 M Number of moles of HCl = \({\rm{M \times V}}\left( {\rm{L}} \right){\rm{ = 0}}{\rm{.1 \times 0}}{\rm{.01 = 0}}{\rm{.001}}{\mkern 1mu} {\mkern 1mu} {\rm{moles}}\) Due to a greater number of moles of NaOH, solution is basic in nature. Hence, turns red litmus paper to blue.
CHXII02:SOLUTIONS
319223
The normality of orthophosphoric acid having purity of \(70 \%\) by weight and specific gravity 1.54 is
1 \(11 \mathrm{~N}\)
2 \(22 \mathrm{~N}\)
3 \(33 \mathrm{~N}\)
4 \(44 \mathrm{~N}\)
Explanation:
\(70 \%\) by weight means \(70 \mathrm{~g}\) of orthophosphoric acid is present in \(100 \mathrm{~g}\) of acid. \(\mathrm{N}=\dfrac{\mathrm{w}}{\mathrm{Eq} \cdot \mathrm{wt}} \times \dfrac{1000}{\mathrm{~V}_{(\mathrm{cc})}}\) \(\Rightarrow \mathrm{w}=70 \mathrm{~g}\) Orthophosphoric acid is \(\mathrm{H}_{3} \mathrm{PO}_{4}\). It is tribasic acid. \(\therefore\) E.q. wt. \(=\dfrac{\text { molecular mass }}{\text { number of replaceable } \mathrm{H} \text { - atoms }}=\dfrac{98}{3}\) \(\mathrm{V}=\dfrac{\text { mass }}{\text { density }}=\dfrac{100}{1.54}\) Putting values in Eq. (1) \(\mathrm{N}=\dfrac{70 \times 3 \times 1000 \times 1.54}{98 \times 100}=33 \mathrm{~N}\) (OR) \(\mathrm{N}=\dfrac{\% \mathrm{w} / \mathrm{w} \times \mathrm{d} \times 10}{\mathrm{GEW}}=\dfrac{70 \times 1.54 \times 10}{98 / 3}=33 \mathrm{~N}\)
319220
The volume of water added to \({\text{500}}\,{\text{mL}}\,\,{\text{0}}{\text{.5M}}\,{\text{NaOH}}\) so that its strength becomes \({\text{10mg}}\,{\text{NaOH}}\) per mL is
1 \(100 \mathrm{~mL}\)
2 \(200 \mathrm{~mL}\)
3 \(250 \mathrm{~mL}\)
4 \(500 \mathrm{~mL}\)
Explanation:
1 mL has 10 mg NaOH 1000 mL will have \({\rm{ = 10 \times 1000mg = 10\;g}}\,\,{\rm{NaOH}}\) Normality \({\rm{ = }}\frac{{{\rm{ gm / litre }}}}{{{\rm{ eq}}{\rm{.wt}}{\rm{. }}}}{\rm{ = }}\frac{{{\rm{10}}}}{{{\rm{40}}}}{\rm{\;N = 0}}{\rm{.25\;N}}\) Now using normality equation, \({{\rm{N}}_{\rm{1}}}{{\rm{V}}_{\rm{1}}}{\rm{ = }}{{\rm{N}}_{\rm{2}}}{{\rm{V}}_{\rm{2}}}\) \({\rm{0}}{\rm{.5 \times 500 = 0}}{\rm{.25 \times }}{{\rm{V}}_{\rm{2}}}\) \({\rm{(}}{\rm{0}}{\rm{.5\,M}}\,\,{\rm{NaOH}} \equiv {\rm{0}}{\rm{.5\;N}}\,\,{\rm{NaOH)}}\) \({{\rm{V}}_{\rm{2}}}{\rm{ = }}\frac{{{\rm{0}}{\rm{.5 \times 500}}}}{{{\rm{0}}{\rm{.25}}}}{\rm{ = 1000\;mL}}\) Volume of water added = 1000 - 500 = 500 mL
CHXII02:SOLUTIONS
319221
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 500ml of 0.2N solution is
319222
1 kg of NaOH is added to 10 ml of 0.1N HCl, the resulting solution will
1 red litumus paper turns blue
2 turn phenolphthalein solution pink
3 turn methyl orange red
4 will have no effect on red or blue litmus paper
Explanation:
Number of moles of NaOH \({\rm{ = }}\frac{{{\rm{Given}}{\mkern 1mu} {\mkern 1mu} {\rm{mass}}}}{{{\rm{Molar}}{\mkern 1mu} {\mkern 1mu} {\rm{mass}}}}{\rm{ = }}\frac{{{\rm{1000}}}}{{{\rm{40}}}}{\rm{ = 25moles}}\) Molarity of HCl = Normality of HCl = 0.1 M Number of moles of HCl = \({\rm{M \times V}}\left( {\rm{L}} \right){\rm{ = 0}}{\rm{.1 \times 0}}{\rm{.01 = 0}}{\rm{.001}}{\mkern 1mu} {\mkern 1mu} {\rm{moles}}\) Due to a greater number of moles of NaOH, solution is basic in nature. Hence, turns red litmus paper to blue.
CHXII02:SOLUTIONS
319223
The normality of orthophosphoric acid having purity of \(70 \%\) by weight and specific gravity 1.54 is
1 \(11 \mathrm{~N}\)
2 \(22 \mathrm{~N}\)
3 \(33 \mathrm{~N}\)
4 \(44 \mathrm{~N}\)
Explanation:
\(70 \%\) by weight means \(70 \mathrm{~g}\) of orthophosphoric acid is present in \(100 \mathrm{~g}\) of acid. \(\mathrm{N}=\dfrac{\mathrm{w}}{\mathrm{Eq} \cdot \mathrm{wt}} \times \dfrac{1000}{\mathrm{~V}_{(\mathrm{cc})}}\) \(\Rightarrow \mathrm{w}=70 \mathrm{~g}\) Orthophosphoric acid is \(\mathrm{H}_{3} \mathrm{PO}_{4}\). It is tribasic acid. \(\therefore\) E.q. wt. \(=\dfrac{\text { molecular mass }}{\text { number of replaceable } \mathrm{H} \text { - atoms }}=\dfrac{98}{3}\) \(\mathrm{V}=\dfrac{\text { mass }}{\text { density }}=\dfrac{100}{1.54}\) Putting values in Eq. (1) \(\mathrm{N}=\dfrac{70 \times 3 \times 1000 \times 1.54}{98 \times 100}=33 \mathrm{~N}\) (OR) \(\mathrm{N}=\dfrac{\% \mathrm{w} / \mathrm{w} \times \mathrm{d} \times 10}{\mathrm{GEW}}=\dfrac{70 \times 1.54 \times 10}{98 / 3}=33 \mathrm{~N}\)
319220
The volume of water added to \({\text{500}}\,{\text{mL}}\,\,{\text{0}}{\text{.5M}}\,{\text{NaOH}}\) so that its strength becomes \({\text{10mg}}\,{\text{NaOH}}\) per mL is
1 \(100 \mathrm{~mL}\)
2 \(200 \mathrm{~mL}\)
3 \(250 \mathrm{~mL}\)
4 \(500 \mathrm{~mL}\)
Explanation:
1 mL has 10 mg NaOH 1000 mL will have \({\rm{ = 10 \times 1000mg = 10\;g}}\,\,{\rm{NaOH}}\) Normality \({\rm{ = }}\frac{{{\rm{ gm / litre }}}}{{{\rm{ eq}}{\rm{.wt}}{\rm{. }}}}{\rm{ = }}\frac{{{\rm{10}}}}{{{\rm{40}}}}{\rm{\;N = 0}}{\rm{.25\;N}}\) Now using normality equation, \({{\rm{N}}_{\rm{1}}}{{\rm{V}}_{\rm{1}}}{\rm{ = }}{{\rm{N}}_{\rm{2}}}{{\rm{V}}_{\rm{2}}}\) \({\rm{0}}{\rm{.5 \times 500 = 0}}{\rm{.25 \times }}{{\rm{V}}_{\rm{2}}}\) \({\rm{(}}{\rm{0}}{\rm{.5\,M}}\,\,{\rm{NaOH}} \equiv {\rm{0}}{\rm{.5\;N}}\,\,{\rm{NaOH)}}\) \({{\rm{V}}_{\rm{2}}}{\rm{ = }}\frac{{{\rm{0}}{\rm{.5 \times 500}}}}{{{\rm{0}}{\rm{.25}}}}{\rm{ = 1000\;mL}}\) Volume of water added = 1000 - 500 = 500 mL
CHXII02:SOLUTIONS
319221
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 500ml of 0.2N solution is
319222
1 kg of NaOH is added to 10 ml of 0.1N HCl, the resulting solution will
1 red litumus paper turns blue
2 turn phenolphthalein solution pink
3 turn methyl orange red
4 will have no effect on red or blue litmus paper
Explanation:
Number of moles of NaOH \({\rm{ = }}\frac{{{\rm{Given}}{\mkern 1mu} {\mkern 1mu} {\rm{mass}}}}{{{\rm{Molar}}{\mkern 1mu} {\mkern 1mu} {\rm{mass}}}}{\rm{ = }}\frac{{{\rm{1000}}}}{{{\rm{40}}}}{\rm{ = 25moles}}\) Molarity of HCl = Normality of HCl = 0.1 M Number of moles of HCl = \({\rm{M \times V}}\left( {\rm{L}} \right){\rm{ = 0}}{\rm{.1 \times 0}}{\rm{.01 = 0}}{\rm{.001}}{\mkern 1mu} {\mkern 1mu} {\rm{moles}}\) Due to a greater number of moles of NaOH, solution is basic in nature. Hence, turns red litmus paper to blue.
CHXII02:SOLUTIONS
319223
The normality of orthophosphoric acid having purity of \(70 \%\) by weight and specific gravity 1.54 is
1 \(11 \mathrm{~N}\)
2 \(22 \mathrm{~N}\)
3 \(33 \mathrm{~N}\)
4 \(44 \mathrm{~N}\)
Explanation:
\(70 \%\) by weight means \(70 \mathrm{~g}\) of orthophosphoric acid is present in \(100 \mathrm{~g}\) of acid. \(\mathrm{N}=\dfrac{\mathrm{w}}{\mathrm{Eq} \cdot \mathrm{wt}} \times \dfrac{1000}{\mathrm{~V}_{(\mathrm{cc})}}\) \(\Rightarrow \mathrm{w}=70 \mathrm{~g}\) Orthophosphoric acid is \(\mathrm{H}_{3} \mathrm{PO}_{4}\). It is tribasic acid. \(\therefore\) E.q. wt. \(=\dfrac{\text { molecular mass }}{\text { number of replaceable } \mathrm{H} \text { - atoms }}=\dfrac{98}{3}\) \(\mathrm{V}=\dfrac{\text { mass }}{\text { density }}=\dfrac{100}{1.54}\) Putting values in Eq. (1) \(\mathrm{N}=\dfrac{70 \times 3 \times 1000 \times 1.54}{98 \times 100}=33 \mathrm{~N}\) (OR) \(\mathrm{N}=\dfrac{\% \mathrm{w} / \mathrm{w} \times \mathrm{d} \times 10}{\mathrm{GEW}}=\dfrac{70 \times 1.54 \times 10}{98 / 3}=33 \mathrm{~N}\)
