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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 $\frac{100 \times 31}{0 .086} = 36000 a m u =$ Average atomic mass of starch

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306768 How many valence electrons are present in 0.53 g of $N a_{2} C O_{3}$ ?

1

40 \times 6 .023 \times 1 0^{23}

2

0 .2 \times 6 .023 \times 1 0^{23}

3

0 .4 \times 6 .023 \times 1 0^{23}

4

2 \times 6 .023 \times 1 0^{23}

Explanation:

In $N a_{2} C O_{3}$ , sodium exists as $N a^{+}$ and carbonate exists as ${C O}_{3}^{2 -}$ . In $N a^{+}$ , number of valence electrons are 8. In ${C O}_{3}^{2 -}$ , number of valence electrons are 24.
Total number of valence electrons in $N a_{2} C O_{3} = 2 (8) + 24 = 40$
Hence, 106 g of $N a_{2} C O_{3}$ contains $40 \times 6 .023 \times 1 0^{23}$ valence electrons.
$\Rightarrow 0 .53 g o f N a_{2} C O_{3} c o n t a i n s$
$0 .53 \times \frac{40}{106} \times 6 .023 \times 1 0^{23} = 0 .2 \times 6 .023 \times 1 0^{23}$ valence electrons.
Note : Number of valence electrons in Na is 1 but in $N a^{+}$ is 8.

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306769 The weight of $1 \times 10^{22}$ molecules of ${CuSO}_{4} \cdot 5 H_{2} O$ is ____(approx) .

1 1

2 15

3 10

4 4

Explanation:

Weight of $6.023 \times 10^{23}$ (Avogadro's number) $= M w$ of ${CuSO}_{4} \cdot 5 H_{2} O = 249 g$
$= 1 mol of {CuSO}_{4} \cdot 5 H_{2} O$
Weight of $1 \times 10^{22}$ molecules of ${CuSO}_{4} \cdot 5 H_{2} O$
$= \frac{249 \times 1 \times 10^{22}}{6.023 \times 10^{23}} = 4.14 g$

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?

1

19 .098 \times 1 0^{19} y e a r s

2

19 .098 y e a r s

3

19 .098 \times 1 0^{9} y e a r s

4 None of these

Explanation:

$∵$ $1 0^{6}$ rupees are spent in 1sec
$∴ 6 .022 \times 1 0^{23}$ rupees are spent in
$= \frac{1 \times 6 .022 \times 1 0^{23}}{1 0^{6}} s e c$
$= \frac{1 \times 60 .22 \times 1 0^{23}}{1 0^{6} \times 60 \times 60 \times 24 \times 365} y e a r s$
$19 .098 \times 1 0^{9} y e a r s$

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 $\frac{100 \times 31}{0 .086} = 36000 a m u =$ Average atomic mass of starch

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306768 How many valence electrons are present in 0.53 g of $N a_{2} C O_{3}$ ?

1

40 \times 6 .023 \times 1 0^{23}

2

0 .2 \times 6 .023 \times 1 0^{23}

3

0 .4 \times 6 .023 \times 1 0^{23}

4

2 \times 6 .023 \times 1 0^{23}

Explanation:

In $N a_{2} C O_{3}$ , sodium exists as $N a^{+}$ and carbonate exists as ${C O}_{3}^{2 -}$ . In $N a^{+}$ , number of valence electrons are 8. In ${C O}_{3}^{2 -}$ , number of valence electrons are 24.
Total number of valence electrons in $N a_{2} C O_{3} = 2 (8) + 24 = 40$
Hence, 106 g of $N a_{2} C O_{3}$ contains $40 \times 6 .023 \times 1 0^{23}$ valence electrons.
$\Rightarrow 0 .53 g o f N a_{2} C O_{3} c o n t a i n s$
$0 .53 \times \frac{40}{106} \times 6 .023 \times 1 0^{23} = 0 .2 \times 6 .023 \times 1 0^{23}$ valence electrons.
Note : Number of valence electrons in Na is 1 but in $N a^{+}$ is 8.

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306769 The weight of $1 \times 10^{22}$ molecules of ${CuSO}_{4} \cdot 5 H_{2} O$ is ____(approx) .

1 1

2 15

3 10

4 4

Explanation:

Weight of $6.023 \times 10^{23}$ (Avogadro's number) $= M w$ of ${CuSO}_{4} \cdot 5 H_{2} O = 249 g$
$= 1 mol of {CuSO}_{4} \cdot 5 H_{2} O$
Weight of $1 \times 10^{22}$ molecules of ${CuSO}_{4} \cdot 5 H_{2} O$
$= \frac{249 \times 1 \times 10^{22}}{6.023 \times 10^{23}} = 4.14 g$

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?

1

19 .098 \times 1 0^{19} y e a r s

2

19 .098 y e a r s

3

19 .098 \times 1 0^{9} y e a r s

4 None of these

Explanation:

$∵$ $1 0^{6}$ rupees are spent in 1sec
$∴ 6 .022 \times 1 0^{23}$ rupees are spent in
$= \frac{1 \times 6 .022 \times 1 0^{23}}{1 0^{6}} s e c$
$= \frac{1 \times 60 .22 \times 1 0^{23}}{1 0^{6} \times 60 \times 60 \times 24 \times 365} y e a r s$
$19 .098 \times 1 0^{9} y e a r s$

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 $\frac{100 \times 31}{0 .086} = 36000 a m u =$ Average atomic mass of starch

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306768 How many valence electrons are present in 0.53 g of $N a_{2} C O_{3}$ ?

1

40 \times 6 .023 \times 1 0^{23}

2

0 .2 \times 6 .023 \times 1 0^{23}

3

0 .4 \times 6 .023 \times 1 0^{23}

4

2 \times 6 .023 \times 1 0^{23}

Explanation:

In $N a_{2} C O_{3}$ , sodium exists as $N a^{+}$ and carbonate exists as ${C O}_{3}^{2 -}$ . In $N a^{+}$ , number of valence electrons are 8. In ${C O}_{3}^{2 -}$ , number of valence electrons are 24.
Total number of valence electrons in $N a_{2} C O_{3} = 2 (8) + 24 = 40$
Hence, 106 g of $N a_{2} C O_{3}$ contains $40 \times 6 .023 \times 1 0^{23}$ valence electrons.
$\Rightarrow 0 .53 g o f N a_{2} C O_{3} c o n t a i n s$
$0 .53 \times \frac{40}{106} \times 6 .023 \times 1 0^{23} = 0 .2 \times 6 .023 \times 1 0^{23}$ valence electrons.
Note : Number of valence electrons in Na is 1 but in $N a^{+}$ is 8.

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306769 The weight of $1 \times 10^{22}$ molecules of ${CuSO}_{4} \cdot 5 H_{2} O$ is ____(approx) .

1 1

2 15

3 10

4 4

Explanation:

Weight of $6.023 \times 10^{23}$ (Avogadro's number) $= M w$ of ${CuSO}_{4} \cdot 5 H_{2} O = 249 g$
$= 1 mol of {CuSO}_{4} \cdot 5 H_{2} O$
Weight of $1 \times 10^{22}$ molecules of ${CuSO}_{4} \cdot 5 H_{2} O$
$= \frac{249 \times 1 \times 10^{22}}{6.023 \times 10^{23}} = 4.14 g$

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?

1

19 .098 \times 1 0^{19} y e a r s

2

19 .098 y e a r s

3

19 .098 \times 1 0^{9} y e a r s

4 None of these

Explanation:

$∵$ $1 0^{6}$ rupees are spent in 1sec
$∴ 6 .022 \times 1 0^{23}$ rupees are spent in
$= \frac{1 \times 6 .022 \times 1 0^{23}}{1 0^{6}} s e c$
$= \frac{1 \times 60 .22 \times 1 0^{23}}{1 0^{6} \times 60 \times 60 \times 24 \times 365} y e a r s$
$19 .098 \times 1 0^{9} y e a r s$

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 $\frac{100 \times 31}{0 .086} = 36000 a m u =$ Average atomic mass of starch

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306768 How many valence electrons are present in 0.53 g of $N a_{2} C O_{3}$ ?

1

40 \times 6 .023 \times 1 0^{23}

2

0 .2 \times 6 .023 \times 1 0^{23}

3

0 .4 \times 6 .023 \times 1 0^{23}

4

2 \times 6 .023 \times 1 0^{23}

Explanation:

In $N a_{2} C O_{3}$ , sodium exists as $N a^{+}$ and carbonate exists as ${C O}_{3}^{2 -}$ . In $N a^{+}$ , number of valence electrons are 8. In ${C O}_{3}^{2 -}$ , number of valence electrons are 24.
Total number of valence electrons in $N a_{2} C O_{3} = 2 (8) + 24 = 40$
Hence, 106 g of $N a_{2} C O_{3}$ contains $40 \times 6 .023 \times 1 0^{23}$ valence electrons.
$\Rightarrow 0 .53 g o f N a_{2} C O_{3} c o n t a i n s$
$0 .53 \times \frac{40}{106} \times 6 .023 \times 1 0^{23} = 0 .2 \times 6 .023 \times 1 0^{23}$ valence electrons.
Note : Number of valence electrons in Na is 1 but in $N a^{+}$ is 8.

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306769 The weight of $1 \times 10^{22}$ molecules of ${CuSO}_{4} \cdot 5 H_{2} O$ is ____(approx) .

1 1

2 15

3 10

4 4

Explanation:

Weight of $6.023 \times 10^{23}$ (Avogadro's number) $= M w$ of ${CuSO}_{4} \cdot 5 H_{2} O = 249 g$
$= 1 mol of {CuSO}_{4} \cdot 5 H_{2} O$
Weight of $1 \times 10^{22}$ molecules of ${CuSO}_{4} \cdot 5 H_{2} O$
$= \frac{249 \times 1 \times 10^{22}}{6.023 \times 10^{23}} = 4.14 g$

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?

1

19 .098 \times 1 0^{19} y e a r s

2

19 .098 y e a r s

3

19 .098 \times 1 0^{9} y e a r s

4 None of these

Explanation:

$∵$ $1 0^{6}$ rupees are spent in 1sec
$∴ 6 .022 \times 1 0^{23}$ rupees are spent in
$= \frac{1 \times 6 .022 \times 1 0^{23}}{1 0^{6}} s e c$
$= \frac{1 \times 60 .22 \times 1 0^{23}}{1 0^{6} \times 60 \times 60 \times 24 \times 365} y e a r s$
$19 .098 \times 1 0^{9} y e a r s$

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