NEET Test Series from KOTA - 10 Papers In MS WORD
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CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306778
A sample of potato starch was ground to give a starch like molecule. The product analysed 0.086% phosphorus. If each molecule is assumed to contain one atom of phosphorus, what is the average molecular mass of the material?
1 36000 amu
2 3600 amu
3 360 amu
4 36 amu
Explanation:
Given that, 0.086 g of phosphorus present in 100 g of starch One atom of phosphorus weighs 31 amu (Average atomic mass of Phosphorus) 31 amu of phosphorus present in \(\frac{{{\rm{100 \times 31}}}}{{{\rm{0}}{\rm{.086}}}}{\rm{ = 36000amu = }}\) Average atomic mass of starch
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306768
How many valence electrons are present in 0.53 g of \({\rm{N}}{{\rm{a}}_{\rm{2}}}{\rm{C}}{{\rm{O}}_{\rm{3}}}\) ?
In \({\rm{N}}{{\rm{a}}_{\rm{2}}}{\rm{C}}{{\rm{O}}_{\rm{3}}}\), sodium exists as \({\rm{N}}{{\rm{a}}^{\rm{ + }}}\) and carbonate exists as \({\rm{CO}}_{\rm{3}}^{{\rm{2 - }}}\). In \({\rm{N}}{{\rm{a}}^{\rm{ + }}}\), number of valence electrons are 8. In \({\rm{CO}}_{\rm{3}}^{{\rm{2 - }}}\), number of valence electrons are 24. Total number of valence electrons in \({\rm{N}}{{\rm{a}}_{\rm{2}}}{\rm{C}}{{\rm{O}}_{\rm{3}}}{\rm{ = 2}}\left( {\rm{8}} \right){\rm{ + 24 = 40}}\) Hence, 106 g of \({\rm{N}}{{\rm{a}}_{\rm{2}}}{\rm{C}}{{\rm{O}}_{\rm{3}}}\) contains \({\rm{40 \times 6}}{\rm{.023 \times 1}}{{\rm{0}}^{{\rm{23}}}}\) valence electrons. \( \Rightarrow {\rm{0}}{\rm{.53}}{\mkern 1mu} {\mkern 1mu} {\rm{g}}{\mkern 1mu} {\mkern 1mu} {\rm{of}}{\mkern 1mu} {\mkern 1mu} {\rm{N}}{{\rm{a}}_{\rm{2}}}{\rm{C}}{{\rm{O}}_{\rm{3}}}{\mkern 1mu} {\mkern 1mu} {\rm{contains}}\) \({\rm{0}}{\rm{.53 \times }}\frac{{{\rm{40}}}}{{{\rm{106}}}}{\rm{ \times 6}}{\rm{.023 \times 1}}{{\rm{0}}^{{\rm{23}}}}{\rm{ = 0}}{\rm{.2 \times 6}}{\rm{.023 \times 1}}{{\rm{0}}^{{\rm{23}}}}\) valence electrons. Note : Number of valence electrons in Na is 1 but in \({\rm{N}}{{\rm{a}}^{\rm{ + }}}\) is 8.
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306769
The weight of \({\mathrm{1 \times 10^{22}}}\) molecules of \({\mathrm{\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}}}\) is ____(approx) .
1 1
2 15
3 10
4 4
Explanation:
Weight of \({\mathrm{6.023 \times 10^{23}}}\) (Avogadro's number)\(={\mathrm{M w}}\) of \({\mathrm{\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}=249 \mathrm{~g}}}\) \(=1 \mathrm{~mol} \text { of } \mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\) Weight of \({\mathrm{1 \times 10^{22}}}\) molecules of \({\mathrm{\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}}}\) \(=\dfrac{249 \times 1 \times 10^{22}}{6.023 \times 10^{23}}=4.14 \mathrm{~g}\)
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306770
How many years it would take to spend Avogadro’s number of rupees at the rate of 1 million rupees per one second?
306778
A sample of potato starch was ground to give a starch like molecule. The product analysed 0.086% phosphorus. If each molecule is assumed to contain one atom of phosphorus, what is the average molecular mass of the material?
1 36000 amu
2 3600 amu
3 360 amu
4 36 amu
Explanation:
Given that, 0.086 g of phosphorus present in 100 g of starch One atom of phosphorus weighs 31 amu (Average atomic mass of Phosphorus) 31 amu of phosphorus present in \(\frac{{{\rm{100 \times 31}}}}{{{\rm{0}}{\rm{.086}}}}{\rm{ = 36000amu = }}\) Average atomic mass of starch
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306768
How many valence electrons are present in 0.53 g of \({\rm{N}}{{\rm{a}}_{\rm{2}}}{\rm{C}}{{\rm{O}}_{\rm{3}}}\) ?
In \({\rm{N}}{{\rm{a}}_{\rm{2}}}{\rm{C}}{{\rm{O}}_{\rm{3}}}\), sodium exists as \({\rm{N}}{{\rm{a}}^{\rm{ + }}}\) and carbonate exists as \({\rm{CO}}_{\rm{3}}^{{\rm{2 - }}}\). In \({\rm{N}}{{\rm{a}}^{\rm{ + }}}\), number of valence electrons are 8. In \({\rm{CO}}_{\rm{3}}^{{\rm{2 - }}}\), number of valence electrons are 24. Total number of valence electrons in \({\rm{N}}{{\rm{a}}_{\rm{2}}}{\rm{C}}{{\rm{O}}_{\rm{3}}}{\rm{ = 2}}\left( {\rm{8}} \right){\rm{ + 24 = 40}}\) Hence, 106 g of \({\rm{N}}{{\rm{a}}_{\rm{2}}}{\rm{C}}{{\rm{O}}_{\rm{3}}}\) contains \({\rm{40 \times 6}}{\rm{.023 \times 1}}{{\rm{0}}^{{\rm{23}}}}\) valence electrons. \( \Rightarrow {\rm{0}}{\rm{.53}}{\mkern 1mu} {\mkern 1mu} {\rm{g}}{\mkern 1mu} {\mkern 1mu} {\rm{of}}{\mkern 1mu} {\mkern 1mu} {\rm{N}}{{\rm{a}}_{\rm{2}}}{\rm{C}}{{\rm{O}}_{\rm{3}}}{\mkern 1mu} {\mkern 1mu} {\rm{contains}}\) \({\rm{0}}{\rm{.53 \times }}\frac{{{\rm{40}}}}{{{\rm{106}}}}{\rm{ \times 6}}{\rm{.023 \times 1}}{{\rm{0}}^{{\rm{23}}}}{\rm{ = 0}}{\rm{.2 \times 6}}{\rm{.023 \times 1}}{{\rm{0}}^{{\rm{23}}}}\) valence electrons. Note : Number of valence electrons in Na is 1 but in \({\rm{N}}{{\rm{a}}^{\rm{ + }}}\) is 8.
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306769
The weight of \({\mathrm{1 \times 10^{22}}}\) molecules of \({\mathrm{\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}}}\) is ____(approx) .
1 1
2 15
3 10
4 4
Explanation:
Weight of \({\mathrm{6.023 \times 10^{23}}}\) (Avogadro's number)\(={\mathrm{M w}}\) of \({\mathrm{\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}=249 \mathrm{~g}}}\) \(=1 \mathrm{~mol} \text { of } \mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\) Weight of \({\mathrm{1 \times 10^{22}}}\) molecules of \({\mathrm{\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}}}\) \(=\dfrac{249 \times 1 \times 10^{22}}{6.023 \times 10^{23}}=4.14 \mathrm{~g}\)
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306770
How many years it would take to spend Avogadro’s number of rupees at the rate of 1 million rupees per one second?
306778
A sample of potato starch was ground to give a starch like molecule. The product analysed 0.086% phosphorus. If each molecule is assumed to contain one atom of phosphorus, what is the average molecular mass of the material?
1 36000 amu
2 3600 amu
3 360 amu
4 36 amu
Explanation:
Given that, 0.086 g of phosphorus present in 100 g of starch One atom of phosphorus weighs 31 amu (Average atomic mass of Phosphorus) 31 amu of phosphorus present in \(\frac{{{\rm{100 \times 31}}}}{{{\rm{0}}{\rm{.086}}}}{\rm{ = 36000amu = }}\) Average atomic mass of starch
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306768
How many valence electrons are present in 0.53 g of \({\rm{N}}{{\rm{a}}_{\rm{2}}}{\rm{C}}{{\rm{O}}_{\rm{3}}}\) ?
In \({\rm{N}}{{\rm{a}}_{\rm{2}}}{\rm{C}}{{\rm{O}}_{\rm{3}}}\), sodium exists as \({\rm{N}}{{\rm{a}}^{\rm{ + }}}\) and carbonate exists as \({\rm{CO}}_{\rm{3}}^{{\rm{2 - }}}\). In \({\rm{N}}{{\rm{a}}^{\rm{ + }}}\), number of valence electrons are 8. In \({\rm{CO}}_{\rm{3}}^{{\rm{2 - }}}\), number of valence electrons are 24. Total number of valence electrons in \({\rm{N}}{{\rm{a}}_{\rm{2}}}{\rm{C}}{{\rm{O}}_{\rm{3}}}{\rm{ = 2}}\left( {\rm{8}} \right){\rm{ + 24 = 40}}\) Hence, 106 g of \({\rm{N}}{{\rm{a}}_{\rm{2}}}{\rm{C}}{{\rm{O}}_{\rm{3}}}\) contains \({\rm{40 \times 6}}{\rm{.023 \times 1}}{{\rm{0}}^{{\rm{23}}}}\) valence electrons. \( \Rightarrow {\rm{0}}{\rm{.53}}{\mkern 1mu} {\mkern 1mu} {\rm{g}}{\mkern 1mu} {\mkern 1mu} {\rm{of}}{\mkern 1mu} {\mkern 1mu} {\rm{N}}{{\rm{a}}_{\rm{2}}}{\rm{C}}{{\rm{O}}_{\rm{3}}}{\mkern 1mu} {\mkern 1mu} {\rm{contains}}\) \({\rm{0}}{\rm{.53 \times }}\frac{{{\rm{40}}}}{{{\rm{106}}}}{\rm{ \times 6}}{\rm{.023 \times 1}}{{\rm{0}}^{{\rm{23}}}}{\rm{ = 0}}{\rm{.2 \times 6}}{\rm{.023 \times 1}}{{\rm{0}}^{{\rm{23}}}}\) valence electrons. Note : Number of valence electrons in Na is 1 but in \({\rm{N}}{{\rm{a}}^{\rm{ + }}}\) is 8.
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306769
The weight of \({\mathrm{1 \times 10^{22}}}\) molecules of \({\mathrm{\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}}}\) is ____(approx) .
1 1
2 15
3 10
4 4
Explanation:
Weight of \({\mathrm{6.023 \times 10^{23}}}\) (Avogadro's number)\(={\mathrm{M w}}\) of \({\mathrm{\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}=249 \mathrm{~g}}}\) \(=1 \mathrm{~mol} \text { of } \mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\) Weight of \({\mathrm{1 \times 10^{22}}}\) molecules of \({\mathrm{\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}}}\) \(=\dfrac{249 \times 1 \times 10^{22}}{6.023 \times 10^{23}}=4.14 \mathrm{~g}\)
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306770
How many years it would take to spend Avogadro’s number of rupees at the rate of 1 million rupees per one second?
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306778
A sample of potato starch was ground to give a starch like molecule. The product analysed 0.086% phosphorus. If each molecule is assumed to contain one atom of phosphorus, what is the average molecular mass of the material?
1 36000 amu
2 3600 amu
3 360 amu
4 36 amu
Explanation:
Given that, 0.086 g of phosphorus present in 100 g of starch One atom of phosphorus weighs 31 amu (Average atomic mass of Phosphorus) 31 amu of phosphorus present in \(\frac{{{\rm{100 \times 31}}}}{{{\rm{0}}{\rm{.086}}}}{\rm{ = 36000amu = }}\) Average atomic mass of starch
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306768
How many valence electrons are present in 0.53 g of \({\rm{N}}{{\rm{a}}_{\rm{2}}}{\rm{C}}{{\rm{O}}_{\rm{3}}}\) ?
In \({\rm{N}}{{\rm{a}}_{\rm{2}}}{\rm{C}}{{\rm{O}}_{\rm{3}}}\), sodium exists as \({\rm{N}}{{\rm{a}}^{\rm{ + }}}\) and carbonate exists as \({\rm{CO}}_{\rm{3}}^{{\rm{2 - }}}\). In \({\rm{N}}{{\rm{a}}^{\rm{ + }}}\), number of valence electrons are 8. In \({\rm{CO}}_{\rm{3}}^{{\rm{2 - }}}\), number of valence electrons are 24. Total number of valence electrons in \({\rm{N}}{{\rm{a}}_{\rm{2}}}{\rm{C}}{{\rm{O}}_{\rm{3}}}{\rm{ = 2}}\left( {\rm{8}} \right){\rm{ + 24 = 40}}\) Hence, 106 g of \({\rm{N}}{{\rm{a}}_{\rm{2}}}{\rm{C}}{{\rm{O}}_{\rm{3}}}\) contains \({\rm{40 \times 6}}{\rm{.023 \times 1}}{{\rm{0}}^{{\rm{23}}}}\) valence electrons. \( \Rightarrow {\rm{0}}{\rm{.53}}{\mkern 1mu} {\mkern 1mu} {\rm{g}}{\mkern 1mu} {\mkern 1mu} {\rm{of}}{\mkern 1mu} {\mkern 1mu} {\rm{N}}{{\rm{a}}_{\rm{2}}}{\rm{C}}{{\rm{O}}_{\rm{3}}}{\mkern 1mu} {\mkern 1mu} {\rm{contains}}\) \({\rm{0}}{\rm{.53 \times }}\frac{{{\rm{40}}}}{{{\rm{106}}}}{\rm{ \times 6}}{\rm{.023 \times 1}}{{\rm{0}}^{{\rm{23}}}}{\rm{ = 0}}{\rm{.2 \times 6}}{\rm{.023 \times 1}}{{\rm{0}}^{{\rm{23}}}}\) valence electrons. Note : Number of valence electrons in Na is 1 but in \({\rm{N}}{{\rm{a}}^{\rm{ + }}}\) is 8.
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306769
The weight of \({\mathrm{1 \times 10^{22}}}\) molecules of \({\mathrm{\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}}}\) is ____(approx) .
1 1
2 15
3 10
4 4
Explanation:
Weight of \({\mathrm{6.023 \times 10^{23}}}\) (Avogadro's number)\(={\mathrm{M w}}\) of \({\mathrm{\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}=249 \mathrm{~g}}}\) \(=1 \mathrm{~mol} \text { of } \mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\) Weight of \({\mathrm{1 \times 10^{22}}}\) molecules of \({\mathrm{\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}}}\) \(=\dfrac{249 \times 1 \times 10^{22}}{6.023 \times 10^{23}}=4.14 \mathrm{~g}\)
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306770
How many years it would take to spend Avogadro’s number of rupees at the rate of 1 million rupees per one second?