306671
Copper sulphate pentahydrate contains \({\mathrm{25.45 \%}}\) of copper and \({\mathrm{36.08 \%}}\) of water. If the law of constant proportions is true, then the weight of copper required to produce 20 g of copper sulphate pentahydrate will be ____ g.
1 5
2 25
3 20
4 0.5
Explanation:
100 g of copper sulphate pentahydrate is obtained from 25.45 g of copper. \({\mathrm{\therefore \quad 1 \mathrm{~g}}}\) of copper sulphate pentahydrate will be obtained from \({\mathrm{=\dfrac{25.45}{100} \mathrm{~g}}}\) of copper. \({\mathrm{\therefore \quad 20 \mathrm{~g}}}\) of copper sulphate pentahydrate will be obtained from \({\mathrm{=\dfrac{25.45}{100} \times 20=5.09 \mathrm{~g}}}\) of copper. \( \approx 5\,{\rm{g}}\,\) of copper.
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306672
Irrespective of the source, pure sample of water always yields 88.89% mass of oxygen and 11.11% mass of hydrogen. This is explained by the law of
1 Conservation of mass
2 Constant composition
3 Multiple proportions
4 Constant volume
Explanation:
\({{\rm{H}}_{\rm{2}}}{\rm{O}}\) contains H and O in a fixed ratio by mass. It illustrates law of constant composition.
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306673
Zinc sulphate contains 22.65% of zinc and 43.9% of water of crystallisation. If the law of constant proportions is true, then the weight of zinc required to produce 20 g of the crystals will be
1 45.3 g
2 4.53 g
3 0.453 g
4 453 g
Explanation:
To prepare 20 g of the crystals, zinc required \({\rm{ = }}\frac{{{\rm{22}}{\rm{.65}}}}{{{\rm{100}}}}{\rm{ \times 20 = 4}}{\rm{.53g}}\).
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306674
A sample of \(\mathrm{CaCO}_{3}\) has \(\mathrm{Ca}=40 \%, \mathrm{C}=12 \%\) and \(\mathrm{O}=48 \%\). If the law of constant proportions is true, then the mass of Ca in 5 g of \(\mathrm{CaCO}_{3}\) from another source will be
1 2.0 g
2 0.2 g
3 0.02 g
4 20.0 g
Explanation:
Wt. of \(\mathrm{Ca}=\dfrac{5 \times 40}{100}=2\) gram Calcium.
306671
Copper sulphate pentahydrate contains \({\mathrm{25.45 \%}}\) of copper and \({\mathrm{36.08 \%}}\) of water. If the law of constant proportions is true, then the weight of copper required to produce 20 g of copper sulphate pentahydrate will be ____ g.
1 5
2 25
3 20
4 0.5
Explanation:
100 g of copper sulphate pentahydrate is obtained from 25.45 g of copper. \({\mathrm{\therefore \quad 1 \mathrm{~g}}}\) of copper sulphate pentahydrate will be obtained from \({\mathrm{=\dfrac{25.45}{100} \mathrm{~g}}}\) of copper. \({\mathrm{\therefore \quad 20 \mathrm{~g}}}\) of copper sulphate pentahydrate will be obtained from \({\mathrm{=\dfrac{25.45}{100} \times 20=5.09 \mathrm{~g}}}\) of copper. \( \approx 5\,{\rm{g}}\,\) of copper.
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306672
Irrespective of the source, pure sample of water always yields 88.89% mass of oxygen and 11.11% mass of hydrogen. This is explained by the law of
1 Conservation of mass
2 Constant composition
3 Multiple proportions
4 Constant volume
Explanation:
\({{\rm{H}}_{\rm{2}}}{\rm{O}}\) contains H and O in a fixed ratio by mass. It illustrates law of constant composition.
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306673
Zinc sulphate contains 22.65% of zinc and 43.9% of water of crystallisation. If the law of constant proportions is true, then the weight of zinc required to produce 20 g of the crystals will be
1 45.3 g
2 4.53 g
3 0.453 g
4 453 g
Explanation:
To prepare 20 g of the crystals, zinc required \({\rm{ = }}\frac{{{\rm{22}}{\rm{.65}}}}{{{\rm{100}}}}{\rm{ \times 20 = 4}}{\rm{.53g}}\).
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306674
A sample of \(\mathrm{CaCO}_{3}\) has \(\mathrm{Ca}=40 \%, \mathrm{C}=12 \%\) and \(\mathrm{O}=48 \%\). If the law of constant proportions is true, then the mass of Ca in 5 g of \(\mathrm{CaCO}_{3}\) from another source will be
1 2.0 g
2 0.2 g
3 0.02 g
4 20.0 g
Explanation:
Wt. of \(\mathrm{Ca}=\dfrac{5 \times 40}{100}=2\) gram Calcium.
306671
Copper sulphate pentahydrate contains \({\mathrm{25.45 \%}}\) of copper and \({\mathrm{36.08 \%}}\) of water. If the law of constant proportions is true, then the weight of copper required to produce 20 g of copper sulphate pentahydrate will be ____ g.
1 5
2 25
3 20
4 0.5
Explanation:
100 g of copper sulphate pentahydrate is obtained from 25.45 g of copper. \({\mathrm{\therefore \quad 1 \mathrm{~g}}}\) of copper sulphate pentahydrate will be obtained from \({\mathrm{=\dfrac{25.45}{100} \mathrm{~g}}}\) of copper. \({\mathrm{\therefore \quad 20 \mathrm{~g}}}\) of copper sulphate pentahydrate will be obtained from \({\mathrm{=\dfrac{25.45}{100} \times 20=5.09 \mathrm{~g}}}\) of copper. \( \approx 5\,{\rm{g}}\,\) of copper.
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306672
Irrespective of the source, pure sample of water always yields 88.89% mass of oxygen and 11.11% mass of hydrogen. This is explained by the law of
1 Conservation of mass
2 Constant composition
3 Multiple proportions
4 Constant volume
Explanation:
\({{\rm{H}}_{\rm{2}}}{\rm{O}}\) contains H and O in a fixed ratio by mass. It illustrates law of constant composition.
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306673
Zinc sulphate contains 22.65% of zinc and 43.9% of water of crystallisation. If the law of constant proportions is true, then the weight of zinc required to produce 20 g of the crystals will be
1 45.3 g
2 4.53 g
3 0.453 g
4 453 g
Explanation:
To prepare 20 g of the crystals, zinc required \({\rm{ = }}\frac{{{\rm{22}}{\rm{.65}}}}{{{\rm{100}}}}{\rm{ \times 20 = 4}}{\rm{.53g}}\).
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306674
A sample of \(\mathrm{CaCO}_{3}\) has \(\mathrm{Ca}=40 \%, \mathrm{C}=12 \%\) and \(\mathrm{O}=48 \%\). If the law of constant proportions is true, then the mass of Ca in 5 g of \(\mathrm{CaCO}_{3}\) from another source will be
1 2.0 g
2 0.2 g
3 0.02 g
4 20.0 g
Explanation:
Wt. of \(\mathrm{Ca}=\dfrac{5 \times 40}{100}=2\) gram Calcium.
306671
Copper sulphate pentahydrate contains \({\mathrm{25.45 \%}}\) of copper and \({\mathrm{36.08 \%}}\) of water. If the law of constant proportions is true, then the weight of copper required to produce 20 g of copper sulphate pentahydrate will be ____ g.
1 5
2 25
3 20
4 0.5
Explanation:
100 g of copper sulphate pentahydrate is obtained from 25.45 g of copper. \({\mathrm{\therefore \quad 1 \mathrm{~g}}}\) of copper sulphate pentahydrate will be obtained from \({\mathrm{=\dfrac{25.45}{100} \mathrm{~g}}}\) of copper. \({\mathrm{\therefore \quad 20 \mathrm{~g}}}\) of copper sulphate pentahydrate will be obtained from \({\mathrm{=\dfrac{25.45}{100} \times 20=5.09 \mathrm{~g}}}\) of copper. \( \approx 5\,{\rm{g}}\,\) of copper.
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306672
Irrespective of the source, pure sample of water always yields 88.89% mass of oxygen and 11.11% mass of hydrogen. This is explained by the law of
1 Conservation of mass
2 Constant composition
3 Multiple proportions
4 Constant volume
Explanation:
\({{\rm{H}}_{\rm{2}}}{\rm{O}}\) contains H and O in a fixed ratio by mass. It illustrates law of constant composition.
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306673
Zinc sulphate contains 22.65% of zinc and 43.9% of water of crystallisation. If the law of constant proportions is true, then the weight of zinc required to produce 20 g of the crystals will be
1 45.3 g
2 4.53 g
3 0.453 g
4 453 g
Explanation:
To prepare 20 g of the crystals, zinc required \({\rm{ = }}\frac{{{\rm{22}}{\rm{.65}}}}{{{\rm{100}}}}{\rm{ \times 20 = 4}}{\rm{.53g}}\).
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306674
A sample of \(\mathrm{CaCO}_{3}\) has \(\mathrm{Ca}=40 \%, \mathrm{C}=12 \%\) and \(\mathrm{O}=48 \%\). If the law of constant proportions is true, then the mass of Ca in 5 g of \(\mathrm{CaCO}_{3}\) from another source will be
1 2.0 g
2 0.2 g
3 0.02 g
4 20.0 g
Explanation:
Wt. of \(\mathrm{Ca}=\dfrac{5 \times 40}{100}=2\) gram Calcium.
