QBManager

Some Basic Concepts of Chemistry

228339 Production of iron in blast furnace follows the following equation:
${Fe}_{3} O_{4} (s) + 4 CO (g) \to 3 Fe (1) + 4 {CO}_{2} (g)$
When $4.640 kg$ of ${Fe}_{3} O_{4}$ and $2.520 kg$ of $CO$ are allowed to react then the amount of iron (in g) produced is:
[Given: Molar Atomic mass $(g {mol}^{- 1}) : Fe =$ 56
Molar Atomic mass $(g {mol}^{- 1}) : O = 16$
Molar Atomic mass $(g {mol}^{- 1}) : C = 12]$

1 1400

2 2200

3 3360

4 4200

Explanation:

Given, The reaction is -
$\begin{aligned} {Fe}_{3} O_{4} (s) + 4 CO (g) ⟶ 3 Fe (l) + 4 {CO}_{2} (g) \\ {Fe}_{3} O_{4} = 4.640 kg . \\ CO = 2.520 kg . \end{aligned}$
$\begin{array}{r} Molar mass of {Fe}_{3} O_{4} = 3 \times 56 + 4 \times 16 \\ = 232 g \end{array}$
1 mole of ${Fe}_{3} O_{4}$ is produced 3 mole of $Fe$
$= 3 \times 56 = 168 g Fe$
$∵ 232 g {FeO}_{4}$ produce $168 g Fe$ .$
$\begin{aligned} ∵ 4.640 kg Fe O_{4} will produce & = \frac{168}{232} \times 4.640 kg \\ = 3.36 kg \\ = 3360 g \end{aligned}$
So, the correct option is (c)

Shift-I

Some Basic Concepts of Chemistry

228340 The maximum number of molecules is present in

1

15 L

of

H_{2}

gas at

STP

2

5 L

of

N_{2}

gas at

STP

3

0.5 g

of

H_{2}

gas

4

10 g

of

H_{2}

gas

Explanation:

1 mole of gas at STP occupies a volume of $22.4 L$ Option (a)- $15 L$ of gas corresponds to
$\frac{15 L}{22.4 L} = 0.67 moles$
Option (b)- $5 L$ of $N_{2}$ gas at STP:
$5 L$ of gas corresponds to $\frac{5 L}{22.4 L} = 0.22$ moles
Option (c)- $0.5 g$ of $H_{2}$ gas (molecular weight $2 g / mole$ ) corresponds to $\frac{0.5 g}{2 g / mol} = 0.25$ moles
Option (d)- $10 g$ of $O_{2}$ gas (molecular weight $3 g / mol$ ) corresponds to $\frac{10 g}{32 g / mol} = 0.3125 moles$
Higher is the number of moles of a gas, higher will be its number of molecules.

JCECE - 2008 UPTU/UPSEE-2007 NEET-2004

Some Basic Concepts of Chemistry

228341 $10 gm$ of hydrogen and $64 g$ of oxygen were kept in a steel vessel and exploded. Amount of water produced in this reaction will be

1

2 mol

2

4 mol

3

8 mol

4

10 mol

Explanation:

$2 H_{2} + O_{2} \to 2 H_{2} O$
$\begin{aligned} 10 g of hydrogen (molar mass 2 g / mol) = \frac{10 g}{2 g / mol} \\ = 5 & mol \end{aligned}$
$64 g$ of oxygen $($ molar mass $32 g / mol) = \frac{64}{32 g / mol}$
$= 2 mol$
2 moles of oxygen will react with $2 \times 2 = 4$ moles of hydrogen to form 4 moles of water.

UP CPMT-2012 NEET-2009

Some Basic Concepts of Chemistry

228342 The mass of one mole of electron is

1

9.1 \times 10^{- 28} g

2

0.55 mg

3

9.1 \times 10^{- 24} g

4

610^{- 12} g

Explanation:

As we know that, Mass of one electron $= 9.1 \times 10^{- 31} kg$ $\begin{aligned} = 9.1 \times 10^{- 28} g \end{aligned}$ 1 mole $= 6.023 \times 10^{23}$ electron So, 1 mole of electron has mass of
$\begin{aligned} = 9.1 \times 10^{- 28} \times 6.023 \times 10^{23} \\ = 5.48 \times 10^{- 4} \\ = 0.55 mg \end{aligned}$

UP CPMT-2010 UPTU/UPSEE-2006

Some Basic Concepts of Chemistry

228339 Production of iron in blast furnace follows the following equation:
${Fe}_{3} O_{4} (s) + 4 CO (g) \to 3 Fe (1) + 4 {CO}_{2} (g)$
When $4.640 kg$ of ${Fe}_{3} O_{4}$ and $2.520 kg$ of $CO$ are allowed to react then the amount of iron (in g) produced is:
[Given: Molar Atomic mass $(g {mol}^{- 1}) : Fe =$ 56
Molar Atomic mass $(g {mol}^{- 1}) : O = 16$
Molar Atomic mass $(g {mol}^{- 1}) : C = 12]$

1 1400

2 2200

3 3360

4 4200

Explanation:

Given, The reaction is -
$\begin{aligned} {Fe}_{3} O_{4} (s) + 4 CO (g) ⟶ 3 Fe (l) + 4 {CO}_{2} (g) \\ {Fe}_{3} O_{4} = 4.640 kg . \\ CO = 2.520 kg . \end{aligned}$
$\begin{array}{r} Molar mass of {Fe}_{3} O_{4} = 3 \times 56 + 4 \times 16 \\ = 232 g \end{array}$
1 mole of ${Fe}_{3} O_{4}$ is produced 3 mole of $Fe$
$= 3 \times 56 = 168 g Fe$
$∵ 232 g {FeO}_{4}$ produce $168 g Fe$ .$
$\begin{aligned} ∵ 4.640 kg Fe O_{4} will produce & = \frac{168}{232} \times 4.640 kg \\ = 3.36 kg \\ = 3360 g \end{aligned}$
So, the correct option is (c)

Shift-I

Some Basic Concepts of Chemistry

228340 The maximum number of molecules is present in

1

15 L

of

H_{2}

gas at

STP

2

5 L

of

N_{2}

gas at

STP

3

0.5 g

of

H_{2}

gas

4

10 g

of

H_{2}

gas

Explanation:

1 mole of gas at STP occupies a volume of $22.4 L$ Option (a)- $15 L$ of gas corresponds to
$\frac{15 L}{22.4 L} = 0.67 moles$
Option (b)- $5 L$ of $N_{2}$ gas at STP:
$5 L$ of gas corresponds to $\frac{5 L}{22.4 L} = 0.22$ moles
Option (c)- $0.5 g$ of $H_{2}$ gas (molecular weight $2 g / mole$ ) corresponds to $\frac{0.5 g}{2 g / mol} = 0.25$ moles
Option (d)- $10 g$ of $O_{2}$ gas (molecular weight $3 g / mol$ ) corresponds to $\frac{10 g}{32 g / mol} = 0.3125 moles$
Higher is the number of moles of a gas, higher will be its number of molecules.

JCECE - 2008 UPTU/UPSEE-2007 NEET-2004

Some Basic Concepts of Chemistry

228341 $10 gm$ of hydrogen and $64 g$ of oxygen were kept in a steel vessel and exploded. Amount of water produced in this reaction will be

1

2 mol

2

4 mol

3

8 mol

4

10 mol

Explanation:

$2 H_{2} + O_{2} \to 2 H_{2} O$
$\begin{aligned} 10 g of hydrogen (molar mass 2 g / mol) = \frac{10 g}{2 g / mol} \\ = 5 & mol \end{aligned}$
$64 g$ of oxygen $($ molar mass $32 g / mol) = \frac{64}{32 g / mol}$
$= 2 mol$
2 moles of oxygen will react with $2 \times 2 = 4$ moles of hydrogen to form 4 moles of water.

UP CPMT-2012 NEET-2009

Some Basic Concepts of Chemistry

228342 The mass of one mole of electron is

1

9.1 \times 10^{- 28} g

2

0.55 mg

3

9.1 \times 10^{- 24} g

4

610^{- 12} g

Explanation:

As we know that, Mass of one electron $= 9.1 \times 10^{- 31} kg$ $\begin{aligned} = 9.1 \times 10^{- 28} g \end{aligned}$ 1 mole $= 6.023 \times 10^{23}$ electron So, 1 mole of electron has mass of
$\begin{aligned} = 9.1 \times 10^{- 28} \times 6.023 \times 10^{23} \\ = 5.48 \times 10^{- 4} \\ = 0.55 mg \end{aligned}$

UP CPMT-2010 UPTU/UPSEE-2006

Some Basic Concepts of Chemistry

228339 Production of iron in blast furnace follows the following equation:
${Fe}_{3} O_{4} (s) + 4 CO (g) \to 3 Fe (1) + 4 {CO}_{2} (g)$
When $4.640 kg$ of ${Fe}_{3} O_{4}$ and $2.520 kg$ of $CO$ are allowed to react then the amount of iron (in g) produced is:
[Given: Molar Atomic mass $(g {mol}^{- 1}) : Fe =$ 56
Molar Atomic mass $(g {mol}^{- 1}) : O = 16$
Molar Atomic mass $(g {mol}^{- 1}) : C = 12]$

1 1400

2 2200

3 3360

4 4200

Explanation:

Given, The reaction is -
$\begin{aligned} {Fe}_{3} O_{4} (s) + 4 CO (g) ⟶ 3 Fe (l) + 4 {CO}_{2} (g) \\ {Fe}_{3} O_{4} = 4.640 kg . \\ CO = 2.520 kg . \end{aligned}$
$\begin{array}{r} Molar mass of {Fe}_{3} O_{4} = 3 \times 56 + 4 \times 16 \\ = 232 g \end{array}$
1 mole of ${Fe}_{3} O_{4}$ is produced 3 mole of $Fe$
$= 3 \times 56 = 168 g Fe$
$∵ 232 g {FeO}_{4}$ produce $168 g Fe$ .$
$\begin{aligned} ∵ 4.640 kg Fe O_{4} will produce & = \frac{168}{232} \times 4.640 kg \\ = 3.36 kg \\ = 3360 g \end{aligned}$
So, the correct option is (c)

Shift-I

Some Basic Concepts of Chemistry

228340 The maximum number of molecules is present in

1

15 L

of

H_{2}

gas at

STP

2

5 L

of

N_{2}

gas at

STP

3

0.5 g

of

H_{2}

gas

4

10 g

of

H_{2}

gas

Explanation:

1 mole of gas at STP occupies a volume of $22.4 L$ Option (a)- $15 L$ of gas corresponds to
$\frac{15 L}{22.4 L} = 0.67 moles$
Option (b)- $5 L$ of $N_{2}$ gas at STP:
$5 L$ of gas corresponds to $\frac{5 L}{22.4 L} = 0.22$ moles
Option (c)- $0.5 g$ of $H_{2}$ gas (molecular weight $2 g / mole$ ) corresponds to $\frac{0.5 g}{2 g / mol} = 0.25$ moles
Option (d)- $10 g$ of $O_{2}$ gas (molecular weight $3 g / mol$ ) corresponds to $\frac{10 g}{32 g / mol} = 0.3125 moles$
Higher is the number of moles of a gas, higher will be its number of molecules.

JCECE - 2008 UPTU/UPSEE-2007 NEET-2004

Some Basic Concepts of Chemistry

228341 $10 gm$ of hydrogen and $64 g$ of oxygen were kept in a steel vessel and exploded. Amount of water produced in this reaction will be

1

2 mol

2

4 mol

3

8 mol

4

10 mol

Explanation:

$2 H_{2} + O_{2} \to 2 H_{2} O$
$\begin{aligned} 10 g of hydrogen (molar mass 2 g / mol) = \frac{10 g}{2 g / mol} \\ = 5 & mol \end{aligned}$
$64 g$ of oxygen $($ molar mass $32 g / mol) = \frac{64}{32 g / mol}$
$= 2 mol$
2 moles of oxygen will react with $2 \times 2 = 4$ moles of hydrogen to form 4 moles of water.

UP CPMT-2012 NEET-2009

Some Basic Concepts of Chemistry

228342 The mass of one mole of electron is

1

9.1 \times 10^{- 28} g

2

0.55 mg

3

9.1 \times 10^{- 24} g

4

610^{- 12} g

Explanation:

As we know that, Mass of one electron $= 9.1 \times 10^{- 31} kg$ $\begin{aligned} = 9.1 \times 10^{- 28} g \end{aligned}$ 1 mole $= 6.023 \times 10^{23}$ electron So, 1 mole of electron has mass of
$\begin{aligned} = 9.1 \times 10^{- 28} \times 6.023 \times 10^{23} \\ = 5.48 \times 10^{- 4} \\ = 0.55 mg \end{aligned}$

UP CPMT-2010 UPTU/UPSEE-2006

Some Basic Concepts of Chemistry

228339 Production of iron in blast furnace follows the following equation:
${Fe}_{3} O_{4} (s) + 4 CO (g) \to 3 Fe (1) + 4 {CO}_{2} (g)$
When $4.640 kg$ of ${Fe}_{3} O_{4}$ and $2.520 kg$ of $CO$ are allowed to react then the amount of iron (in g) produced is:
[Given: Molar Atomic mass $(g {mol}^{- 1}) : Fe =$ 56
Molar Atomic mass $(g {mol}^{- 1}) : O = 16$
Molar Atomic mass $(g {mol}^{- 1}) : C = 12]$

1 1400

2 2200

3 3360

4 4200

Explanation:

Given, The reaction is -
$\begin{aligned} {Fe}_{3} O_{4} (s) + 4 CO (g) ⟶ 3 Fe (l) + 4 {CO}_{2} (g) \\ {Fe}_{3} O_{4} = 4.640 kg . \\ CO = 2.520 kg . \end{aligned}$
$\begin{array}{r} Molar mass of {Fe}_{3} O_{4} = 3 \times 56 + 4 \times 16 \\ = 232 g \end{array}$
1 mole of ${Fe}_{3} O_{4}$ is produced 3 mole of $Fe$
$= 3 \times 56 = 168 g Fe$
$∵ 232 g {FeO}_{4}$ produce $168 g Fe$ .$
$\begin{aligned} ∵ 4.640 kg Fe O_{4} will produce & = \frac{168}{232} \times 4.640 kg \\ = 3.36 kg \\ = 3360 g \end{aligned}$
So, the correct option is (c)

Shift-I

Some Basic Concepts of Chemistry

228340 The maximum number of molecules is present in

1

15 L

of

H_{2}

gas at

STP

2

5 L

of

N_{2}

gas at

STP

3

0.5 g

of

H_{2}

gas

4

10 g

of

H_{2}

gas

Explanation:

1 mole of gas at STP occupies a volume of $22.4 L$ Option (a)- $15 L$ of gas corresponds to
$\frac{15 L}{22.4 L} = 0.67 moles$
Option (b)- $5 L$ of $N_{2}$ gas at STP:
$5 L$ of gas corresponds to $\frac{5 L}{22.4 L} = 0.22$ moles
Option (c)- $0.5 g$ of $H_{2}$ gas (molecular weight $2 g / mole$ ) corresponds to $\frac{0.5 g}{2 g / mol} = 0.25$ moles
Option (d)- $10 g$ of $O_{2}$ gas (molecular weight $3 g / mol$ ) corresponds to $\frac{10 g}{32 g / mol} = 0.3125 moles$
Higher is the number of moles of a gas, higher will be its number of molecules.

JCECE - 2008 UPTU/UPSEE-2007 NEET-2004

Some Basic Concepts of Chemistry

228341 $10 gm$ of hydrogen and $64 g$ of oxygen were kept in a steel vessel and exploded. Amount of water produced in this reaction will be

1

2 mol

2

4 mol

3

8 mol

4

10 mol

Explanation:

$2 H_{2} + O_{2} \to 2 H_{2} O$
$\begin{aligned} 10 g of hydrogen (molar mass 2 g / mol) = \frac{10 g}{2 g / mol} \\ = 5 & mol \end{aligned}$
$64 g$ of oxygen $($ molar mass $32 g / mol) = \frac{64}{32 g / mol}$
$= 2 mol$
2 moles of oxygen will react with $2 \times 2 = 4$ moles of hydrogen to form 4 moles of water.

UP CPMT-2012 NEET-2009

Some Basic Concepts of Chemistry

228342 The mass of one mole of electron is

1

9.1 \times 10^{- 28} g

2

0.55 mg

3

9.1 \times 10^{- 24} g

4

610^{- 12} g

Explanation:

As we know that, Mass of one electron $= 9.1 \times 10^{- 31} kg$ $\begin{aligned} = 9.1 \times 10^{- 28} g \end{aligned}$ 1 mole $= 6.023 \times 10^{23}$ electron So, 1 mole of electron has mass of
$\begin{aligned} = 9.1 \times 10^{- 28} \times 6.023 \times 10^{23} \\ = 5.48 \times 10^{- 4} \\ = 0.55 mg \end{aligned}$

UP CPMT-2010 UPTU/UPSEE-2006

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