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CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306600 The vapour of gas A is three times that of gas B. If the molecular weight of A is M, the molecular weight of B is

1

3 M

2

\sqrt{3 M}

3

\frac{M}{3}

4

\frac{M}{\sqrt{3}}

Explanation:

$\frac{M_{A}}{M_{B}} = \frac{d_{A}}{d_{B}} . (d = v a p o u r d e n s i t y)$
$G i v e n d_{A} = 3 d_{B,} \frac{d A}{d B} \Rightarrow \frac{M_{A}}{M_{B}} = 3 o r$
$M_{B} = \frac{M_{A}}{3} = \frac{M}{3}$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306601 The sulphate of a metal M contains 9.87% of M. This sulphate is isomorphous with $Z n S O_{4} .7 H_{2} O$ . The atomic weight of M is

1 40.3

2 36.3

3 24.3

4 11.3

Explanation:

As the given sulphate is isomorphous with $Z n S O_{4} .7 H_{2} O$ , its formula would be $M S O_{4} .7 H_{2} O$ . If m is the atomic weight of M, molecular weight of $M S O_{4} .7 H_{2} O = m + 32 + 64 + 126 = m + 222 .$
Hence, $% o f M = \frac{m}{m + 222} \times 100 = 9 .87$ (Given)
or $100 m = 9 .87 m + 222 \times 9 .87$
or $90 .13 m = 222 \times 9 .87 o r m = 24 .3$ .

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306602 $5.6 L$ of a gas at S.T.P. weighs equal to $8 g$ . The vapour density of gas is

1 32

2 16

3 8

4 40

Explanation:

Mass of $22.4 L$ gas at S.T.P. $= \frac{8 \times 22.4}{5.6} =$ $32 g (1$ mole $)$
$∴$ Mol. mass $= 32 g / mole$
Vapour density is half of molar mass
$∴$ V.D $= \frac{32}{2} = 16$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306603 Molecular mass is defined as the

1 Mass of one atom compared with the mass of one molecule

2 Mass of one atom compared with the mass of one atom of hydrogen

3 Mass of one molecule of any substance compared with the mass of one atom of C-12.

4 None of the above

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306604 Vapour density of a metal chloride is 66. The molecular weight of the metal is

1 21

2 54

3 136

4 66

Explanation:

$M o l e c u l a r w e i g h t = 2 \times v a p o u r d e n s i t y$
$= 2 \times 66$
$= 136$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306600 The vapour of gas A is three times that of gas B. If the molecular weight of A is M, the molecular weight of B is

1

3 M

2

\sqrt{3 M}

3

\frac{M}{3}

4

\frac{M}{\sqrt{3}}

Explanation:

$\frac{M_{A}}{M_{B}} = \frac{d_{A}}{d_{B}} . (d = v a p o u r d e n s i t y)$
$G i v e n d_{A} = 3 d_{B,} \frac{d A}{d B} \Rightarrow \frac{M_{A}}{M_{B}} = 3 o r$
$M_{B} = \frac{M_{A}}{3} = \frac{M}{3}$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306601 The sulphate of a metal M contains 9.87% of M. This sulphate is isomorphous with $Z n S O_{4} .7 H_{2} O$ . The atomic weight of M is

1 40.3

2 36.3

3 24.3

4 11.3

Explanation:

As the given sulphate is isomorphous with $Z n S O_{4} .7 H_{2} O$ , its formula would be $M S O_{4} .7 H_{2} O$ . If m is the atomic weight of M, molecular weight of $M S O_{4} .7 H_{2} O = m + 32 + 64 + 126 = m + 222 .$
Hence, $% o f M = \frac{m}{m + 222} \times 100 = 9 .87$ (Given)
or $100 m = 9 .87 m + 222 \times 9 .87$
or $90 .13 m = 222 \times 9 .87 o r m = 24 .3$ .

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306602 $5.6 L$ of a gas at S.T.P. weighs equal to $8 g$ . The vapour density of gas is

1 32

2 16

3 8

4 40

Explanation:

Mass of $22.4 L$ gas at S.T.P. $= \frac{8 \times 22.4}{5.6} =$ $32 g (1$ mole $)$
$∴$ Mol. mass $= 32 g / mole$
Vapour density is half of molar mass
$∴$ V.D $= \frac{32}{2} = 16$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306603 Molecular mass is defined as the

1 Mass of one atom compared with the mass of one molecule

2 Mass of one atom compared with the mass of one atom of hydrogen

3 Mass of one molecule of any substance compared with the mass of one atom of C-12.

4 None of the above

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306604 Vapour density of a metal chloride is 66. The molecular weight of the metal is

1 21

2 54

3 136

4 66

Explanation:

$M o l e c u l a r w e i g h t = 2 \times v a p o u r d e n s i t y$
$= 2 \times 66$
$= 136$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306600 The vapour of gas A is three times that of gas B. If the molecular weight of A is M, the molecular weight of B is

1

3 M

2

\sqrt{3 M}

3

\frac{M}{3}

4

\frac{M}{\sqrt{3}}

Explanation:

$\frac{M_{A}}{M_{B}} = \frac{d_{A}}{d_{B}} . (d = v a p o u r d e n s i t y)$
$G i v e n d_{A} = 3 d_{B,} \frac{d A}{d B} \Rightarrow \frac{M_{A}}{M_{B}} = 3 o r$
$M_{B} = \frac{M_{A}}{3} = \frac{M}{3}$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306601 The sulphate of a metal M contains 9.87% of M. This sulphate is isomorphous with $Z n S O_{4} .7 H_{2} O$ . The atomic weight of M is

1 40.3

2 36.3

3 24.3

4 11.3

Explanation:

As the given sulphate is isomorphous with $Z n S O_{4} .7 H_{2} O$ , its formula would be $M S O_{4} .7 H_{2} O$ . If m is the atomic weight of M, molecular weight of $M S O_{4} .7 H_{2} O = m + 32 + 64 + 126 = m + 222 .$
Hence, $% o f M = \frac{m}{m + 222} \times 100 = 9 .87$ (Given)
or $100 m = 9 .87 m + 222 \times 9 .87$
or $90 .13 m = 222 \times 9 .87 o r m = 24 .3$ .

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306602 $5.6 L$ of a gas at S.T.P. weighs equal to $8 g$ . The vapour density of gas is

1 32

2 16

3 8

4 40

Explanation:

Mass of $22.4 L$ gas at S.T.P. $= \frac{8 \times 22.4}{5.6} =$ $32 g (1$ mole $)$
$∴$ Mol. mass $= 32 g / mole$
Vapour density is half of molar mass
$∴$ V.D $= \frac{32}{2} = 16$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306603 Molecular mass is defined as the

1 Mass of one atom compared with the mass of one molecule

2 Mass of one atom compared with the mass of one atom of hydrogen

3 Mass of one molecule of any substance compared with the mass of one atom of C-12.

4 None of the above

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306604 Vapour density of a metal chloride is 66. The molecular weight of the metal is

1 21

2 54

3 136

4 66

Explanation:

$M o l e c u l a r w e i g h t = 2 \times v a p o u r d e n s i t y$
$= 2 \times 66$
$= 136$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306600 The vapour of gas A is three times that of gas B. If the molecular weight of A is M, the molecular weight of B is

1

3 M

2

\sqrt{3 M}

3

\frac{M}{3}

4

\frac{M}{\sqrt{3}}

Explanation:

$\frac{M_{A}}{M_{B}} = \frac{d_{A}}{d_{B}} . (d = v a p o u r d e n s i t y)$
$G i v e n d_{A} = 3 d_{B,} \frac{d A}{d B} \Rightarrow \frac{M_{A}}{M_{B}} = 3 o r$
$M_{B} = \frac{M_{A}}{3} = \frac{M}{3}$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306601 The sulphate of a metal M contains 9.87% of M. This sulphate is isomorphous with $Z n S O_{4} .7 H_{2} O$ . The atomic weight of M is

1 40.3

2 36.3

3 24.3

4 11.3

Explanation:

As the given sulphate is isomorphous with $Z n S O_{4} .7 H_{2} O$ , its formula would be $M S O_{4} .7 H_{2} O$ . If m is the atomic weight of M, molecular weight of $M S O_{4} .7 H_{2} O = m + 32 + 64 + 126 = m + 222 .$
Hence, $% o f M = \frac{m}{m + 222} \times 100 = 9 .87$ (Given)
or $100 m = 9 .87 m + 222 \times 9 .87$
or $90 .13 m = 222 \times 9 .87 o r m = 24 .3$ .

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306602 $5.6 L$ of a gas at S.T.P. weighs equal to $8 g$ . The vapour density of gas is

1 32

2 16

3 8

4 40

Explanation:

Mass of $22.4 L$ gas at S.T.P. $= \frac{8 \times 22.4}{5.6} =$ $32 g (1$ mole $)$
$∴$ Mol. mass $= 32 g / mole$
Vapour density is half of molar mass
$∴$ V.D $= \frac{32}{2} = 16$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306603 Molecular mass is defined as the

1 Mass of one atom compared with the mass of one molecule

2 Mass of one atom compared with the mass of one atom of hydrogen

3 Mass of one molecule of any substance compared with the mass of one atom of C-12.

4 None of the above

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306604 Vapour density of a metal chloride is 66. The molecular weight of the metal is

1 21

2 54

3 136

4 66

Explanation:

$M o l e c u l a r w e i g h t = 2 \times v a p o u r d e n s i t y$
$= 2 \times 66$
$= 136$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306600 The vapour of gas A is three times that of gas B. If the molecular weight of A is M, the molecular weight of B is

1

3 M

2

\sqrt{3 M}

3

\frac{M}{3}

4

\frac{M}{\sqrt{3}}

Explanation:

$\frac{M_{A}}{M_{B}} = \frac{d_{A}}{d_{B}} . (d = v a p o u r d e n s i t y)$
$G i v e n d_{A} = 3 d_{B,} \frac{d A}{d B} \Rightarrow \frac{M_{A}}{M_{B}} = 3 o r$
$M_{B} = \frac{M_{A}}{3} = \frac{M}{3}$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306601 The sulphate of a metal M contains 9.87% of M. This sulphate is isomorphous with $Z n S O_{4} .7 H_{2} O$ . The atomic weight of M is

1 40.3

2 36.3

3 24.3

4 11.3

Explanation:

As the given sulphate is isomorphous with $Z n S O_{4} .7 H_{2} O$ , its formula would be $M S O_{4} .7 H_{2} O$ . If m is the atomic weight of M, molecular weight of $M S O_{4} .7 H_{2} O = m + 32 + 64 + 126 = m + 222 .$
Hence, $% o f M = \frac{m}{m + 222} \times 100 = 9 .87$ (Given)
or $100 m = 9 .87 m + 222 \times 9 .87$
or $90 .13 m = 222 \times 9 .87 o r m = 24 .3$ .

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306602 $5.6 L$ of a gas at S.T.P. weighs equal to $8 g$ . The vapour density of gas is

1 32

2 16

3 8

4 40

Explanation:

Mass of $22.4 L$ gas at S.T.P. $= \frac{8 \times 22.4}{5.6} =$ $32 g (1$ mole $)$
$∴$ Mol. mass $= 32 g / mole$
Vapour density is half of molar mass
$∴$ V.D $= \frac{32}{2} = 16$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306603 Molecular mass is defined as the

1 Mass of one atom compared with the mass of one molecule

2 Mass of one atom compared with the mass of one atom of hydrogen

3 Mass of one molecule of any substance compared with the mass of one atom of C-12.

4 None of the above

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306604 Vapour density of a metal chloride is 66. The molecular weight of the metal is

1 21

2 54

3 136

4 66

Explanation:

$M o l e c u l a r w e i g h t = 2 \times v a p o u r d e n s i t y$
$= 2 \times 66$
$= 136$