QBManager

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306850 The vapour density of a chloride of an element is 39.5 . The $E w$ of the elements is 3.82 . The atomic weight of the element is ____ .

1 7.64

2 3.95

3 1

4 10

Explanation:

$M w$ of metal chloride $({MCl}_{x}) = M + x \times 35.5$
$\begin{aligned} = 2 \times VD = 2 \times 39.5 \\ = 79.0 \end{aligned}$
$M w = E w \times x$
Valency of metal $= x$
Atomic weight of element $= E w \times x$
${MCl}_{x} = 3.82 \times x + x \times 35.5 = 79$
$x = \frac{79}{3.82 + 35.5} \approx 2$
Atomic weight $= E w \times x = 3.82 \times 2 = 7.64$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306851 Match Column I with Column II and choose the correct combination from the given options given.
supporting img

1

A - P, Q, R, B - Q, C - Q, D - P, Q, R

2

A - Q, B - P, Q, R, C - P, D - P Q, R

3

A - P, Q, R, B - R, C - P, Q, D - Q

4

A - Q, B - P, C - P, Q, R, D - Q

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306852 When a metal is burnt, its weight is increased by 24 per cent. The equivalent weight of the metal will be

1 2

2 24

3 33.3

4 76

Explanation:

If $100 g$ of metal is burnt, then it combines with oxygen to form metal oxide. As its weight increases by $24 %$ , then the mass of combined oxygen would be 24 g.
Thus, the equivalent weight of metal
= $\frac{w e i g h t o f m e t a l}{w e i g h t o f o x y g e n c o m b i n e d} \times 8$
$= \frac{100}{24} \times 8$
$= 33.33 g e q m o l^{- 1}$
Hence, the equivalent weight of metal will be 33.3.

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306853 The number of molecules in $100 mL$ of $0 .02 N H_{2} S O_{4}$ is

1

6.023 \times 10^{22}

2

6.023 \times 10^{21}

3

6.023 \times 10^{20}

4

6.023 \times 10^{18}

Explanation:

As, normality = molarity $\times$ valence factor for $H_{2} S O_{4}$ ,
valence factor for $H_{2} S O_{4} = 2$
$∴ N o r m a l i t y, N = M \times 2$
$o r M = \frac{N}{2} = \frac{0 .02}{2} = 0 .01 M$
$M o l e s o f H_{2} S O_{4}$
$= \frac{0.01 \times 100}{1000} = 1 \times 10^{- 3}$ mol
$[∵ Molarity = \frac{Number of moles of solute}{Volume in solution (in mL)} \times 1000]$
$∴$ Number of molecules in 100mL of $0 .02 N H_{2} S O_{4}$
$= 1 \times 10^{- 3} \times N_{A}$
$= 1 \times 10^{- 3} \times 6.023 \times 10^{23}$
$= 6.023 \times 10^{20} m o l e c u l e s$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306850 The vapour density of a chloride of an element is 39.5 . The $E w$ of the elements is 3.82 . The atomic weight of the element is ____ .

1 7.64

2 3.95

3 1

4 10

Explanation:

$M w$ of metal chloride $({MCl}_{x}) = M + x \times 35.5$
$\begin{aligned} = 2 \times VD = 2 \times 39.5 \\ = 79.0 \end{aligned}$
$M w = E w \times x$
Valency of metal $= x$
Atomic weight of element $= E w \times x$
${MCl}_{x} = 3.82 \times x + x \times 35.5 = 79$
$x = \frac{79}{3.82 + 35.5} \approx 2$
Atomic weight $= E w \times x = 3.82 \times 2 = 7.64$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306851 Match Column I with Column II and choose the correct combination from the given options given.
supporting img

1

A - P, Q, R, B - Q, C - Q, D - P, Q, R

2

A - Q, B - P, Q, R, C - P, D - P Q, R

3

A - P, Q, R, B - R, C - P, Q, D - Q

4

A - Q, B - P, C - P, Q, R, D - Q

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306852 When a metal is burnt, its weight is increased by 24 per cent. The equivalent weight of the metal will be

1 2

2 24

3 33.3

4 76

Explanation:

If $100 g$ of metal is burnt, then it combines with oxygen to form metal oxide. As its weight increases by $24 %$ , then the mass of combined oxygen would be 24 g.
Thus, the equivalent weight of metal
= $\frac{w e i g h t o f m e t a l}{w e i g h t o f o x y g e n c o m b i n e d} \times 8$
$= \frac{100}{24} \times 8$
$= 33.33 g e q m o l^{- 1}$
Hence, the equivalent weight of metal will be 33.3.

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306853 The number of molecules in $100 mL$ of $0 .02 N H_{2} S O_{4}$ is

1

6.023 \times 10^{22}

2

6.023 \times 10^{21}

3

6.023 \times 10^{20}

4

6.023 \times 10^{18}

Explanation:

As, normality = molarity $\times$ valence factor for $H_{2} S O_{4}$ ,
valence factor for $H_{2} S O_{4} = 2$
$∴ N o r m a l i t y, N = M \times 2$
$o r M = \frac{N}{2} = \frac{0 .02}{2} = 0 .01 M$
$M o l e s o f H_{2} S O_{4}$
$= \frac{0.01 \times 100}{1000} = 1 \times 10^{- 3}$ mol
$[∵ Molarity = \frac{Number of moles of solute}{Volume in solution (in mL)} \times 1000]$
$∴$ Number of molecules in 100mL of $0 .02 N H_{2} S O_{4}$
$= 1 \times 10^{- 3} \times N_{A}$
$= 1 \times 10^{- 3} \times 6.023 \times 10^{23}$
$= 6.023 \times 10^{20} m o l e c u l e s$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306850 The vapour density of a chloride of an element is 39.5 . The $E w$ of the elements is 3.82 . The atomic weight of the element is ____ .

1 7.64

2 3.95

3 1

4 10

Explanation:

$M w$ of metal chloride $({MCl}_{x}) = M + x \times 35.5$
$\begin{aligned} = 2 \times VD = 2 \times 39.5 \\ = 79.0 \end{aligned}$
$M w = E w \times x$
Valency of metal $= x$
Atomic weight of element $= E w \times x$
${MCl}_{x} = 3.82 \times x + x \times 35.5 = 79$
$x = \frac{79}{3.82 + 35.5} \approx 2$
Atomic weight $= E w \times x = 3.82 \times 2 = 7.64$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306851 Match Column I with Column II and choose the correct combination from the given options given.
supporting img

1

A - P, Q, R, B - Q, C - Q, D - P, Q, R

2

A - Q, B - P, Q, R, C - P, D - P Q, R

3

A - P, Q, R, B - R, C - P, Q, D - Q

4

A - Q, B - P, C - P, Q, R, D - Q

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306852 When a metal is burnt, its weight is increased by 24 per cent. The equivalent weight of the metal will be

1 2

2 24

3 33.3

4 76

Explanation:

If $100 g$ of metal is burnt, then it combines with oxygen to form metal oxide. As its weight increases by $24 %$ , then the mass of combined oxygen would be 24 g.
Thus, the equivalent weight of metal
= $\frac{w e i g h t o f m e t a l}{w e i g h t o f o x y g e n c o m b i n e d} \times 8$
$= \frac{100}{24} \times 8$
$= 33.33 g e q m o l^{- 1}$
Hence, the equivalent weight of metal will be 33.3.

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306853 The number of molecules in $100 mL$ of $0 .02 N H_{2} S O_{4}$ is

1

6.023 \times 10^{22}

2

6.023 \times 10^{21}

3

6.023 \times 10^{20}

4

6.023 \times 10^{18}

Explanation:

As, normality = molarity $\times$ valence factor for $H_{2} S O_{4}$ ,
valence factor for $H_{2} S O_{4} = 2$
$∴ N o r m a l i t y, N = M \times 2$
$o r M = \frac{N}{2} = \frac{0 .02}{2} = 0 .01 M$
$M o l e s o f H_{2} S O_{4}$
$= \frac{0.01 \times 100}{1000} = 1 \times 10^{- 3}$ mol
$[∵ Molarity = \frac{Number of moles of solute}{Volume in solution (in mL)} \times 1000]$
$∴$ Number of molecules in 100mL of $0 .02 N H_{2} S O_{4}$
$= 1 \times 10^{- 3} \times N_{A}$
$= 1 \times 10^{- 3} \times 6.023 \times 10^{23}$
$= 6.023 \times 10^{20} m o l e c u l e s$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306850 The vapour density of a chloride of an element is 39.5 . The $E w$ of the elements is 3.82 . The atomic weight of the element is ____ .

1 7.64

2 3.95

3 1

4 10

Explanation:

$M w$ of metal chloride $({MCl}_{x}) = M + x \times 35.5$
$\begin{aligned} = 2 \times VD = 2 \times 39.5 \\ = 79.0 \end{aligned}$
$M w = E w \times x$
Valency of metal $= x$
Atomic weight of element $= E w \times x$
${MCl}_{x} = 3.82 \times x + x \times 35.5 = 79$
$x = \frac{79}{3.82 + 35.5} \approx 2$
Atomic weight $= E w \times x = 3.82 \times 2 = 7.64$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306851 Match Column I with Column II and choose the correct combination from the given options given.
supporting img

1

A - P, Q, R, B - Q, C - Q, D - P, Q, R

2

A - Q, B - P, Q, R, C - P, D - P Q, R

3

A - P, Q, R, B - R, C - P, Q, D - Q

4

A - Q, B - P, C - P, Q, R, D - Q

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306852 When a metal is burnt, its weight is increased by 24 per cent. The equivalent weight of the metal will be

1 2

2 24

3 33.3

4 76

Explanation:

If $100 g$ of metal is burnt, then it combines with oxygen to form metal oxide. As its weight increases by $24 %$ , then the mass of combined oxygen would be 24 g.
Thus, the equivalent weight of metal
= $\frac{w e i g h t o f m e t a l}{w e i g h t o f o x y g e n c o m b i n e d} \times 8$
$= \frac{100}{24} \times 8$
$= 33.33 g e q m o l^{- 1}$
Hence, the equivalent weight of metal will be 33.3.

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306853 The number of molecules in $100 mL$ of $0 .02 N H_{2} S O_{4}$ is

1

6.023 \times 10^{22}

2

6.023 \times 10^{21}

3

6.023 \times 10^{20}

4

6.023 \times 10^{18}

Explanation:

As, normality = molarity $\times$ valence factor for $H_{2} S O_{4}$ ,
valence factor for $H_{2} S O_{4} = 2$
$∴ N o r m a l i t y, N = M \times 2$
$o r M = \frac{N}{2} = \frac{0 .02}{2} = 0 .01 M$
$M o l e s o f H_{2} S O_{4}$
$= \frac{0.01 \times 100}{1000} = 1 \times 10^{- 3}$ mol
$[∵ Molarity = \frac{Number of moles of solute}{Volume in solution (in mL)} \times 1000]$
$∴$ Number of molecules in 100mL of $0 .02 N H_{2} S O_{4}$
$= 1 \times 10^{- 3} \times N_{A}$
$= 1 \times 10^{- 3} \times 6.023 \times 10^{23}$
$= 6.023 \times 10^{20} m o l e c u l e s$

Question Bank Series

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation: