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CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

307008 The amount of arsenic pentasulphide that can be obtained when 35.5 g arsenic acid is treated with excess $H_{2} S$ in the presence of conc. HCl (assuming 100% conversion) is:

1 0.25 mol

2 0.50 mol

3 0.333 mol

4 0.125 mol

Explanation:

\[\mathop {{\rm{2}}{{\rm{H}}_{\rm{3}}}{\rm{As}}{{\rm{O}}_{\rm{4}}}}\limits_{ $\begin{matrix} 2 m o l e s \\ 1 m o l e \end{matrix}$ } {\rm{ + 5}}{{\rm{H}}_{\rm{2}}}{\rm{S}}\mathop {{\rm{A}}{{\rm{s}}_{\rm{2}}}{{\rm{S}}_{\rm{5}}}}\limits_{ $\begin{matrix} 1 m o l e \\ 1 / 2 m o l e \end{matrix}$ } {\rm{ + 8}}{{\rm{H}}_{\rm{2}}}{\rm{O}}
$∴$ Number of moles of $H_{3} A s O_{4} = \frac{35.5}{142} = 0.25$ moles
$∴$ Number of moles of $A s_{2} S_{5} = \frac{0.25}{2}$
$= 0.125 m o l e s$

JEE - 2016

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

307009 Which of the following statements is correct about the reaction given below?
$4 F e (s) + 3 O_{2} (g) \to 2 F e_{2} O_{3} (g)$

1 Total mass of iron and oxygen in reactants = total mass of iron and oxygen on product therefore it follows law of conservation of mass.

2 Total mass of reactants = total massof product; therefore, law of multiple proportions is followed.

3 Amount of

F e_{2} O_{3}

can be increased by taking any one of the reactants (iron or oxygen) in excess.

4 Amount of

F e_{2} O_{3}

produced will decrease if the amount of any one of the reactants (iron or oxygen) is taken in excess.

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

307010 n g of a substance X reacts with m g of substance Y to form p g of substance R and q g of substance S.
This reaction can be represented as follow : $X + Y = R + S$
The relation which can be established in the amounts of the reactants and the products will be

1 n – m = p – q

2 n + m = p + q

3 n = m

4 p = q

Explanation:

$\underset{n g}{X} + \underset{m g}{Y} = \underset{p g}{R} + \underset{q g}{S}$
n + m = p + q by law of conservation of mass

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

307011 Liquid benzene burns in oxygen according to following equation
$2 C_{6} H_{6} + 15 O_{2} (g) \to 12 C O_{2} (g) + 6 H_{2} O (g)$
How many litres of oxygen are required for complete combustion of 39 g of liquid benzene?

1 11.2

2 22.4

3 42

4 84

Explanation:

From equation, $2 \times 78 g o f C_{6} H_{6}$ requires
$15 \times 22 .4 L o f O_{2}$
$39 g o f C_{6} H_{6} r e q u i r e s \frac{15 \times 22 .4 \times 39}{2 \times 78} = 84 L a t S T P .$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

307012 What volume of ${CO}_{2}$ will be liberated at $NTP$ , if $12 g$ of carbon is burnt in excess of oxygen?

1

11.2 L

2

22.4 L

3

2.24 L

4

1.12 L

Explanation:

$\underset{\begin{matrix} 1mol \\ 12g \end{matrix}}{C (g)} + O_{2} (g) \to \underset{\begin{matrix} 1mol \\ 22 . 4L \end{matrix}}{C O_{2} (g)}$
$∴$ If 12g of carbon is burnt in excess of oxygen, $22.4 L C O_{2}$ will be liberated.

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

307008 The amount of arsenic pentasulphide that can be obtained when 35.5 g arsenic acid is treated with excess $H_{2} S$ in the presence of conc. HCl (assuming 100% conversion) is:

1 0.25 mol

2 0.50 mol

3 0.333 mol

4 0.125 mol

Explanation:

\[\mathop {{\rm{2}}{{\rm{H}}_{\rm{3}}}{\rm{As}}{{\rm{O}}_{\rm{4}}}}\limits_{ $\begin{matrix} 2 m o l e s \\ 1 m o l e \end{matrix}$ } {\rm{ + 5}}{{\rm{H}}_{\rm{2}}}{\rm{S}}\mathop {{\rm{A}}{{\rm{s}}_{\rm{2}}}{{\rm{S}}_{\rm{5}}}}\limits_{ $\begin{matrix} 1 m o l e \\ 1 / 2 m o l e \end{matrix}$ } {\rm{ + 8}}{{\rm{H}}_{\rm{2}}}{\rm{O}}
$∴$ Number of moles of $H_{3} A s O_{4} = \frac{35.5}{142} = 0.25$ moles
$∴$ Number of moles of $A s_{2} S_{5} = \frac{0.25}{2}$
$= 0.125 m o l e s$

JEE - 2016

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

307009 Which of the following statements is correct about the reaction given below?
$4 F e (s) + 3 O_{2} (g) \to 2 F e_{2} O_{3} (g)$

1 Total mass of iron and oxygen in reactants = total mass of iron and oxygen on product therefore it follows law of conservation of mass.

2 Total mass of reactants = total massof product; therefore, law of multiple proportions is followed.

3 Amount of

F e_{2} O_{3}

can be increased by taking any one of the reactants (iron or oxygen) in excess.

4 Amount of

F e_{2} O_{3}

produced will decrease if the amount of any one of the reactants (iron or oxygen) is taken in excess.

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

307010 n g of a substance X reacts with m g of substance Y to form p g of substance R and q g of substance S.
This reaction can be represented as follow : $X + Y = R + S$
The relation which can be established in the amounts of the reactants and the products will be

1 n – m = p – q

2 n + m = p + q

3 n = m

4 p = q

Explanation:

$\underset{n g}{X} + \underset{m g}{Y} = \underset{p g}{R} + \underset{q g}{S}$
n + m = p + q by law of conservation of mass

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

307011 Liquid benzene burns in oxygen according to following equation
$2 C_{6} H_{6} + 15 O_{2} (g) \to 12 C O_{2} (g) + 6 H_{2} O (g)$
How many litres of oxygen are required for complete combustion of 39 g of liquid benzene?

1 11.2

2 22.4

3 42

4 84

Explanation:

From equation, $2 \times 78 g o f C_{6} H_{6}$ requires
$15 \times 22 .4 L o f O_{2}$
$39 g o f C_{6} H_{6} r e q u i r e s \frac{15 \times 22 .4 \times 39}{2 \times 78} = 84 L a t S T P .$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

307012 What volume of ${CO}_{2}$ will be liberated at $NTP$ , if $12 g$ of carbon is burnt in excess of oxygen?

1

11.2 L

2

22.4 L

3

2.24 L

4

1.12 L

Explanation:

$\underset{\begin{matrix} 1mol \\ 12g \end{matrix}}{C (g)} + O_{2} (g) \to \underset{\begin{matrix} 1mol \\ 22 . 4L \end{matrix}}{C O_{2} (g)}$
$∴$ If 12g of carbon is burnt in excess of oxygen, $22.4 L C O_{2}$ will be liberated.

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

307008 The amount of arsenic pentasulphide that can be obtained when 35.5 g arsenic acid is treated with excess $H_{2} S$ in the presence of conc. HCl (assuming 100% conversion) is:

1 0.25 mol

2 0.50 mol

3 0.333 mol

4 0.125 mol

Explanation:

\[\mathop {{\rm{2}}{{\rm{H}}_{\rm{3}}}{\rm{As}}{{\rm{O}}_{\rm{4}}}}\limits_{ $\begin{matrix} 2 m o l e s \\ 1 m o l e \end{matrix}$ } {\rm{ + 5}}{{\rm{H}}_{\rm{2}}}{\rm{S}}\mathop {{\rm{A}}{{\rm{s}}_{\rm{2}}}{{\rm{S}}_{\rm{5}}}}\limits_{ $\begin{matrix} 1 m o l e \\ 1 / 2 m o l e \end{matrix}$ } {\rm{ + 8}}{{\rm{H}}_{\rm{2}}}{\rm{O}}
$∴$ Number of moles of $H_{3} A s O_{4} = \frac{35.5}{142} = 0.25$ moles
$∴$ Number of moles of $A s_{2} S_{5} = \frac{0.25}{2}$
$= 0.125 m o l e s$

JEE - 2016

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

307009 Which of the following statements is correct about the reaction given below?
$4 F e (s) + 3 O_{2} (g) \to 2 F e_{2} O_{3} (g)$

1 Total mass of iron and oxygen in reactants = total mass of iron and oxygen on product therefore it follows law of conservation of mass.

2 Total mass of reactants = total massof product; therefore, law of multiple proportions is followed.

3 Amount of

F e_{2} O_{3}

can be increased by taking any one of the reactants (iron or oxygen) in excess.

4 Amount of

F e_{2} O_{3}

produced will decrease if the amount of any one of the reactants (iron or oxygen) is taken in excess.

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

307010 n g of a substance X reacts with m g of substance Y to form p g of substance R and q g of substance S.
This reaction can be represented as follow : $X + Y = R + S$
The relation which can be established in the amounts of the reactants and the products will be

1 n – m = p – q

2 n + m = p + q

3 n = m

4 p = q

Explanation:

$\underset{n g}{X} + \underset{m g}{Y} = \underset{p g}{R} + \underset{q g}{S}$
n + m = p + q by law of conservation of mass

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

307011 Liquid benzene burns in oxygen according to following equation
$2 C_{6} H_{6} + 15 O_{2} (g) \to 12 C O_{2} (g) + 6 H_{2} O (g)$
How many litres of oxygen are required for complete combustion of 39 g of liquid benzene?

1 11.2

2 22.4

3 42

4 84

Explanation:

From equation, $2 \times 78 g o f C_{6} H_{6}$ requires
$15 \times 22 .4 L o f O_{2}$
$39 g o f C_{6} H_{6} r e q u i r e s \frac{15 \times 22 .4 \times 39}{2 \times 78} = 84 L a t S T P .$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

307012 What volume of ${CO}_{2}$ will be liberated at $NTP$ , if $12 g$ of carbon is burnt in excess of oxygen?

1

11.2 L

2

22.4 L

3

2.24 L

4

1.12 L

Explanation:

$\underset{\begin{matrix} 1mol \\ 12g \end{matrix}}{C (g)} + O_{2} (g) \to \underset{\begin{matrix} 1mol \\ 22 . 4L \end{matrix}}{C O_{2} (g)}$
$∴$ If 12g of carbon is burnt in excess of oxygen, $22.4 L C O_{2}$ will be liberated.

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

307008 The amount of arsenic pentasulphide that can be obtained when 35.5 g arsenic acid is treated with excess $H_{2} S$ in the presence of conc. HCl (assuming 100% conversion) is:

1 0.25 mol

2 0.50 mol

3 0.333 mol

4 0.125 mol

Explanation:

\[\mathop {{\rm{2}}{{\rm{H}}_{\rm{3}}}{\rm{As}}{{\rm{O}}_{\rm{4}}}}\limits_{ $\begin{matrix} 2 m o l e s \\ 1 m o l e \end{matrix}$ } {\rm{ + 5}}{{\rm{H}}_{\rm{2}}}{\rm{S}}\mathop {{\rm{A}}{{\rm{s}}_{\rm{2}}}{{\rm{S}}_{\rm{5}}}}\limits_{ $\begin{matrix} 1 m o l e \\ 1 / 2 m o l e \end{matrix}$ } {\rm{ + 8}}{{\rm{H}}_{\rm{2}}}{\rm{O}}
$∴$ Number of moles of $H_{3} A s O_{4} = \frac{35.5}{142} = 0.25$ moles
$∴$ Number of moles of $A s_{2} S_{5} = \frac{0.25}{2}$
$= 0.125 m o l e s$

JEE - 2016

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

307009 Which of the following statements is correct about the reaction given below?
$4 F e (s) + 3 O_{2} (g) \to 2 F e_{2} O_{3} (g)$

1 Total mass of iron and oxygen in reactants = total mass of iron and oxygen on product therefore it follows law of conservation of mass.

2 Total mass of reactants = total massof product; therefore, law of multiple proportions is followed.

3 Amount of

F e_{2} O_{3}

can be increased by taking any one of the reactants (iron or oxygen) in excess.

4 Amount of

F e_{2} O_{3}

produced will decrease if the amount of any one of the reactants (iron or oxygen) is taken in excess.

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

307010 n g of a substance X reacts with m g of substance Y to form p g of substance R and q g of substance S.
This reaction can be represented as follow : $X + Y = R + S$
The relation which can be established in the amounts of the reactants and the products will be

1 n – m = p – q

2 n + m = p + q

3 n = m

4 p = q

Explanation:

$\underset{n g}{X} + \underset{m g}{Y} = \underset{p g}{R} + \underset{q g}{S}$
n + m = p + q by law of conservation of mass

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

307011 Liquid benzene burns in oxygen according to following equation
$2 C_{6} H_{6} + 15 O_{2} (g) \to 12 C O_{2} (g) + 6 H_{2} O (g)$
How many litres of oxygen are required for complete combustion of 39 g of liquid benzene?

1 11.2

2 22.4

3 42

4 84

Explanation:

From equation, $2 \times 78 g o f C_{6} H_{6}$ requires
$15 \times 22 .4 L o f O_{2}$
$39 g o f C_{6} H_{6} r e q u i r e s \frac{15 \times 22 .4 \times 39}{2 \times 78} = 84 L a t S T P .$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

307012 What volume of ${CO}_{2}$ will be liberated at $NTP$ , if $12 g$ of carbon is burnt in excess of oxygen?

1

11.2 L

2

22.4 L

3

2.24 L

4

1.12 L

Explanation:

$\underset{\begin{matrix} 1mol \\ 12g \end{matrix}}{C (g)} + O_{2} (g) \to \underset{\begin{matrix} 1mol \\ 22 . 4L \end{matrix}}{C O_{2} (g)}$
$∴$ If 12g of carbon is burnt in excess of oxygen, $22.4 L C O_{2}$ will be liberated.

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

307008 The amount of arsenic pentasulphide that can be obtained when 35.5 g arsenic acid is treated with excess $H_{2} S$ in the presence of conc. HCl (assuming 100% conversion) is:

1 0.25 mol

2 0.50 mol

3 0.333 mol

4 0.125 mol

Explanation:

\[\mathop {{\rm{2}}{{\rm{H}}_{\rm{3}}}{\rm{As}}{{\rm{O}}_{\rm{4}}}}\limits_{ $\begin{matrix} 2 m o l e s \\ 1 m o l e \end{matrix}$ } {\rm{ + 5}}{{\rm{H}}_{\rm{2}}}{\rm{S}}\mathop {{\rm{A}}{{\rm{s}}_{\rm{2}}}{{\rm{S}}_{\rm{5}}}}\limits_{ $\begin{matrix} 1 m o l e \\ 1 / 2 m o l e \end{matrix}$ } {\rm{ + 8}}{{\rm{H}}_{\rm{2}}}{\rm{O}}
$∴$ Number of moles of $H_{3} A s O_{4} = \frac{35.5}{142} = 0.25$ moles
$∴$ Number of moles of $A s_{2} S_{5} = \frac{0.25}{2}$
$= 0.125 m o l e s$

JEE - 2016

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

307009 Which of the following statements is correct about the reaction given below?
$4 F e (s) + 3 O_{2} (g) \to 2 F e_{2} O_{3} (g)$

1 Total mass of iron and oxygen in reactants = total mass of iron and oxygen on product therefore it follows law of conservation of mass.

2 Total mass of reactants = total massof product; therefore, law of multiple proportions is followed.

3 Amount of

F e_{2} O_{3}

can be increased by taking any one of the reactants (iron or oxygen) in excess.

4 Amount of

F e_{2} O_{3}

produced will decrease if the amount of any one of the reactants (iron or oxygen) is taken in excess.

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

307010 n g of a substance X reacts with m g of substance Y to form p g of substance R and q g of substance S.
This reaction can be represented as follow : $X + Y = R + S$
The relation which can be established in the amounts of the reactants and the products will be

1 n – m = p – q

2 n + m = p + q

3 n = m

4 p = q

Explanation:

$\underset{n g}{X} + \underset{m g}{Y} = \underset{p g}{R} + \underset{q g}{S}$
n + m = p + q by law of conservation of mass

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

307011 Liquid benzene burns in oxygen according to following equation
$2 C_{6} H_{6} + 15 O_{2} (g) \to 12 C O_{2} (g) + 6 H_{2} O (g)$
How many litres of oxygen are required for complete combustion of 39 g of liquid benzene?

1 11.2

2 22.4

3 42

4 84

Explanation:

From equation, $2 \times 78 g o f C_{6} H_{6}$ requires
$15 \times 22 .4 L o f O_{2}$
$39 g o f C_{6} H_{6} r e q u i r e s \frac{15 \times 22 .4 \times 39}{2 \times 78} = 84 L a t S T P .$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

307012 What volume of ${CO}_{2}$ will be liberated at $NTP$ , if $12 g$ of carbon is burnt in excess of oxygen?

1

11.2 L

2

22.4 L

3

2.24 L

4

1.12 L

Explanation:

$\underset{\begin{matrix} 1mol \\ 12g \end{matrix}}{C (g)} + O_{2} (g) \to \underset{\begin{matrix} 1mol \\ 22 . 4L \end{matrix}}{C O_{2} (g)}$
$∴$ If 12g of carbon is burnt in excess of oxygen, $22.4 L C O_{2}$ will be liberated.

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