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CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306785 The ratio of masses of oxygen and nitrogen in a particular gaseous mixture is 1:4. The ratio of number of their molecules is

1

3 : 16

2

1 : 4

3

7 : 32

4

1 : 8

Explanation:

Ratio of masses of $O_{2}$ and $N_{2} = 1 : 4$
Ratio of moles of $O_{2}$ and $N_{2} = \frac{1}{32} : \frac{4}{28} = \frac{1}{32} : \frac{1}{7} = 7 : 32$
$∴$ Ratio of molecules of $O_{2}$ and $N_{2} = 7 : 32$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306786 Mass of 0.1 mole of methane is

1 1.6 g

2 0.1 g

3 1 g

4 16 g

Explanation:

$N o . o f m o l e s = \frac{m a s s o f s u b s t a n c e}{m o l e c u l a r m a s s o f s u b s t a n c e}$
$0 .1 = \frac{w}{M_{C H_{4}}}$
$0 .1 = \frac{w}{16}$ $(∵ M_{C H_{4}} = 16)$
$\Rightarrow w = 1 .6 g$

KCET - 2008

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306787 If $10^{21}$ molecules are removed from 140 mg of $N_{2}$ , then the number of moles of $N_{2}$ left is $3.34 \times 10^{X}$ moles. The value of ' X ' is ____.

1

-

1

2

-

2

3

-

3

4

-

4

Explanation:

140 mg of $N_{2} = 140 \times 10^{- 3} = 0.14 g$
28 g of $N_{2} = 6.022 \times 10^{23}$ molecules
$∴ 0.14 g$ of $N_{2} = \frac{6.022 \times 10^{23}}{28} \times 0.14$ $\begin{aligned} = 0.03011 \times 10^{23} \\ = 3.011 \times 10^{21} molecules \end{aligned}$
Now, $10^{21}$ molecules are removed,
So, remaining molecules $= 3.011 \times 10^{21} - 10^{21}$ $\begin{aligned} = 10^{21} (3.011 - 1) \\ = 2.011 \times 10^{21} molecules \end{aligned}$
Now, $6.022 \times 10^{23}$ molecules $= 1$ mole
$∴ 2.011 \times 10^{21}$ molecules $= \frac{2.011 \times 10^{21}}{6.022 \times 10^{23}}$
$\begin{aligned} = 0.334 \times 10^{- 2} \\ = 3.34 \times 10^{- 3} moles \end{aligned}$
$∴$ Value of ' X ' is -3 .

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306797 Number of atoms in the following samples of substances is the largest in:

1 4.0 g of hydrogen

2 7.10 g of chlorine

3 127.0 g of iodine

4 48.0 g of magnesium

Explanation:

(1) 4g of hydrogen = 4 mole of hydrogen atoms $= 4 \times 6.023 \times 10^{23} a t o m s$
(2) 71.0 gm of chlorine $= \frac{71.0}{71.0}$
= 1 moleof chlorine
$= 6.023 \times 10^{23} a t o m s$
(3) 127 gm of iodine =1 mole of I atoms $= 6.023 \times 10^{23} a t o m s$
(4) 48.0 gm of magnesium $= \frac{48.0}{24.0}$
2 moles of Mg $= 2 \times 6.023 \times 10^{23} a t o m s$
$∴$ 4.0 gm $H_{2}$ has the largest number of atoms.

JEE - 2013

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306785 The ratio of masses of oxygen and nitrogen in a particular gaseous mixture is 1:4. The ratio of number of their molecules is

1

3 : 16

2

1 : 4

3

7 : 32

4

1 : 8

Explanation:

Ratio of masses of $O_{2}$ and $N_{2} = 1 : 4$
Ratio of moles of $O_{2}$ and $N_{2} = \frac{1}{32} : \frac{4}{28} = \frac{1}{32} : \frac{1}{7} = 7 : 32$
$∴$ Ratio of molecules of $O_{2}$ and $N_{2} = 7 : 32$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306786 Mass of 0.1 mole of methane is

1 1.6 g

2 0.1 g

3 1 g

4 16 g

Explanation:

$N o . o f m o l e s = \frac{m a s s o f s u b s t a n c e}{m o l e c u l a r m a s s o f s u b s t a n c e}$
$0 .1 = \frac{w}{M_{C H_{4}}}$
$0 .1 = \frac{w}{16}$ $(∵ M_{C H_{4}} = 16)$
$\Rightarrow w = 1 .6 g$

KCET - 2008

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306787 If $10^{21}$ molecules are removed from 140 mg of $N_{2}$ , then the number of moles of $N_{2}$ left is $3.34 \times 10^{X}$ moles. The value of ' X ' is ____.

1

-

1

2

-

2

3

-

3

4

-

4

Explanation:

140 mg of $N_{2} = 140 \times 10^{- 3} = 0.14 g$
28 g of $N_{2} = 6.022 \times 10^{23}$ molecules
$∴ 0.14 g$ of $N_{2} = \frac{6.022 \times 10^{23}}{28} \times 0.14$ $\begin{aligned} = 0.03011 \times 10^{23} \\ = 3.011 \times 10^{21} molecules \end{aligned}$
Now, $10^{21}$ molecules are removed,
So, remaining molecules $= 3.011 \times 10^{21} - 10^{21}$ $\begin{aligned} = 10^{21} (3.011 - 1) \\ = 2.011 \times 10^{21} molecules \end{aligned}$
Now, $6.022 \times 10^{23}$ molecules $= 1$ mole
$∴ 2.011 \times 10^{21}$ molecules $= \frac{2.011 \times 10^{21}}{6.022 \times 10^{23}}$
$\begin{aligned} = 0.334 \times 10^{- 2} \\ = 3.34 \times 10^{- 3} moles \end{aligned}$
$∴$ Value of ' X ' is -3 .

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306797 Number of atoms in the following samples of substances is the largest in:

1 4.0 g of hydrogen

2 7.10 g of chlorine

3 127.0 g of iodine

4 48.0 g of magnesium

Explanation:

(1) 4g of hydrogen = 4 mole of hydrogen atoms $= 4 \times 6.023 \times 10^{23} a t o m s$
(2) 71.0 gm of chlorine $= \frac{71.0}{71.0}$
= 1 moleof chlorine
$= 6.023 \times 10^{23} a t o m s$
(3) 127 gm of iodine =1 mole of I atoms $= 6.023 \times 10^{23} a t o m s$
(4) 48.0 gm of magnesium $= \frac{48.0}{24.0}$
2 moles of Mg $= 2 \times 6.023 \times 10^{23} a t o m s$
$∴$ 4.0 gm $H_{2}$ has the largest number of atoms.

JEE - 2013

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306785 The ratio of masses of oxygen and nitrogen in a particular gaseous mixture is 1:4. The ratio of number of their molecules is

1

3 : 16

2

1 : 4

3

7 : 32

4

1 : 8

Explanation:

Ratio of masses of $O_{2}$ and $N_{2} = 1 : 4$
Ratio of moles of $O_{2}$ and $N_{2} = \frac{1}{32} : \frac{4}{28} = \frac{1}{32} : \frac{1}{7} = 7 : 32$
$∴$ Ratio of molecules of $O_{2}$ and $N_{2} = 7 : 32$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306786 Mass of 0.1 mole of methane is

1 1.6 g

2 0.1 g

3 1 g

4 16 g

Explanation:

$N o . o f m o l e s = \frac{m a s s o f s u b s t a n c e}{m o l e c u l a r m a s s o f s u b s t a n c e}$
$0 .1 = \frac{w}{M_{C H_{4}}}$
$0 .1 = \frac{w}{16}$ $(∵ M_{C H_{4}} = 16)$
$\Rightarrow w = 1 .6 g$

KCET - 2008

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306787 If $10^{21}$ molecules are removed from 140 mg of $N_{2}$ , then the number of moles of $N_{2}$ left is $3.34 \times 10^{X}$ moles. The value of ' X ' is ____.

1

-

1

2

-

2

3

-

3

4

-

4

Explanation:

140 mg of $N_{2} = 140 \times 10^{- 3} = 0.14 g$
28 g of $N_{2} = 6.022 \times 10^{23}$ molecules
$∴ 0.14 g$ of $N_{2} = \frac{6.022 \times 10^{23}}{28} \times 0.14$ $\begin{aligned} = 0.03011 \times 10^{23} \\ = 3.011 \times 10^{21} molecules \end{aligned}$
Now, $10^{21}$ molecules are removed,
So, remaining molecules $= 3.011 \times 10^{21} - 10^{21}$ $\begin{aligned} = 10^{21} (3.011 - 1) \\ = 2.011 \times 10^{21} molecules \end{aligned}$
Now, $6.022 \times 10^{23}$ molecules $= 1$ mole
$∴ 2.011 \times 10^{21}$ molecules $= \frac{2.011 \times 10^{21}}{6.022 \times 10^{23}}$
$\begin{aligned} = 0.334 \times 10^{- 2} \\ = 3.34 \times 10^{- 3} moles \end{aligned}$
$∴$ Value of ' X ' is -3 .

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306797 Number of atoms in the following samples of substances is the largest in:

1 4.0 g of hydrogen

2 7.10 g of chlorine

3 127.0 g of iodine

4 48.0 g of magnesium

Explanation:

(1) 4g of hydrogen = 4 mole of hydrogen atoms $= 4 \times 6.023 \times 10^{23} a t o m s$
(2) 71.0 gm of chlorine $= \frac{71.0}{71.0}$
= 1 moleof chlorine
$= 6.023 \times 10^{23} a t o m s$
(3) 127 gm of iodine =1 mole of I atoms $= 6.023 \times 10^{23} a t o m s$
(4) 48.0 gm of magnesium $= \frac{48.0}{24.0}$
2 moles of Mg $= 2 \times 6.023 \times 10^{23} a t o m s$
$∴$ 4.0 gm $H_{2}$ has the largest number of atoms.

JEE - 2013

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306785 The ratio of masses of oxygen and nitrogen in a particular gaseous mixture is 1:4. The ratio of number of their molecules is

1

3 : 16

2

1 : 4

3

7 : 32

4

1 : 8

Explanation:

Ratio of masses of $O_{2}$ and $N_{2} = 1 : 4$
Ratio of moles of $O_{2}$ and $N_{2} = \frac{1}{32} : \frac{4}{28} = \frac{1}{32} : \frac{1}{7} = 7 : 32$
$∴$ Ratio of molecules of $O_{2}$ and $N_{2} = 7 : 32$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306786 Mass of 0.1 mole of methane is

1 1.6 g

2 0.1 g

3 1 g

4 16 g

Explanation:

$N o . o f m o l e s = \frac{m a s s o f s u b s t a n c e}{m o l e c u l a r m a s s o f s u b s t a n c e}$
$0 .1 = \frac{w}{M_{C H_{4}}}$
$0 .1 = \frac{w}{16}$ $(∵ M_{C H_{4}} = 16)$
$\Rightarrow w = 1 .6 g$

KCET - 2008

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306787 If $10^{21}$ molecules are removed from 140 mg of $N_{2}$ , then the number of moles of $N_{2}$ left is $3.34 \times 10^{X}$ moles. The value of ' X ' is ____.

1

-

1

2

-

2

3

-

3

4

-

4

Explanation:

140 mg of $N_{2} = 140 \times 10^{- 3} = 0.14 g$
28 g of $N_{2} = 6.022 \times 10^{23}$ molecules
$∴ 0.14 g$ of $N_{2} = \frac{6.022 \times 10^{23}}{28} \times 0.14$ $\begin{aligned} = 0.03011 \times 10^{23} \\ = 3.011 \times 10^{21} molecules \end{aligned}$
Now, $10^{21}$ molecules are removed,
So, remaining molecules $= 3.011 \times 10^{21} - 10^{21}$ $\begin{aligned} = 10^{21} (3.011 - 1) \\ = 2.011 \times 10^{21} molecules \end{aligned}$
Now, $6.022 \times 10^{23}$ molecules $= 1$ mole
$∴ 2.011 \times 10^{21}$ molecules $= \frac{2.011 \times 10^{21}}{6.022 \times 10^{23}}$
$\begin{aligned} = 0.334 \times 10^{- 2} \\ = 3.34 \times 10^{- 3} moles \end{aligned}$
$∴$ Value of ' X ' is -3 .

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306797 Number of atoms in the following samples of substances is the largest in:

1 4.0 g of hydrogen

2 7.10 g of chlorine

3 127.0 g of iodine

4 48.0 g of magnesium

Explanation:

(1) 4g of hydrogen = 4 mole of hydrogen atoms $= 4 \times 6.023 \times 10^{23} a t o m s$
(2) 71.0 gm of chlorine $= \frac{71.0}{71.0}$
= 1 moleof chlorine
$= 6.023 \times 10^{23} a t o m s$
(3) 127 gm of iodine =1 mole of I atoms $= 6.023 \times 10^{23} a t o m s$
(4) 48.0 gm of magnesium $= \frac{48.0}{24.0}$
2 moles of Mg $= 2 \times 6.023 \times 10^{23} a t o m s$
$∴$ 4.0 gm $H_{2}$ has the largest number of atoms.

JEE - 2013

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