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CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306974 When 22.4 L of $H_{2} (g)$ is mixed with 11.2 L of $C l_{2} (g)$ , each at STP, the moles of $H C l (g)$ formed is equal to

1 1 mole of HCl (g)

2 2 moles of HCl (g)

3 0.5 mole of HCl (g)

4 1.5 mole of HCl (g)

Explanation:

The given problem is related to the concept of stoichiometry of chemical equations. Thus, we have to convert the given volume into their moles and then identify the limiting reagent [possessing minimum number of moles and gets completely used up in the reaction]. The limiting reagent gives the moles of product formed in the reaction.
$H_{2} (g) + C l_{2} (g) \to 2 H C l (g)$
$I n i t i a l v o l . 22 .4 L 11 .2 L 2 m o l$
$∵ 22 .4L$ volume at STP is occupied by $C l_{2} = 1 m o l e$
$∴ 1102 L$ volume will be occupied by, $C l_{2} = \frac{1 \times 11 .2}{22 .4} m o l e = 0 .5 m o l$
Thus, $H_{2} (g) + C l_{2} (g) \to 2 H C l_{2} (g)$
$1 m o l 0 .5 m o l$
Since, $C l_{2}$ possesses minimum number of moles, thus it is the limiting reagent
As per equation,
$1 m o l C l_{2} \equiv 2 m o l H C l$
$∴ 0 .5 m o l C l_{2} = 2 \times 0 .5 m o l H C l = 1 .0 m o l H C l$
Hence 1.0 mole of HCl (g) is produced by 0.5 mole of $C l_{2} [11 .2 L]$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306975 What mass of 95% pure $C a C O_{3}$ will be required to neutralise 50 mL of 0.5 M HCl solution according to the following reaction?
$C a C O_{3 (s)} + 2 H C l_{(a q)} \to C a C L_{2 (g)} + H_{2} O_{(l)}$
[Calculate upto second place of decimal point]

1 1.32 g

2 3.65 g

3 9.50 g

4 1.25 g

Explanation:

$C a C O_{3 (s)} + 2 H C l_{(a q)} \to C a C L_{2 (g)} + C O_{2 (g)} + H_{2} O_{(ι)}$
$N o . o f m o l e s o f C a C O_{3} (p u r e) =$
$\frac{1}{2} \times m o l e o f H C l$
$[M o l e = m o l a r i t y \times v o l u m e (i n l t r .)]$
$= \frac{1}{2} \times 0 .5 \times \frac{50}{1000} = 0 .0125$
$W e i g h t o f C a C O_{3} (p u r e) = m o l e \times m o l . w t$
$= 0 .0125 \times 100 = 1 .25 g$
$% P u r i t y = \frac{w t . o f p u r e s u b s t a n c e}{w t . o f i m p u r e s a m p l e} \times 100$
$95 = \frac{125}{w t . o f i m p u r e s a m p l e} \times 100$
$95 = \frac{125}{w t . o f i m p u r e s a m p l e} \times 100$

NEET - 2022

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306976 The volume of ${Cl}_{2}$ at STP obtained on reacting $4.35 g {MnO}_{2}$ with conc. $HCl$ (At. wt. of $Mn = 55$ )

1 4.48 litre

2 2.24 litres

3 1.12 litre

4 0.56 litre

Explanation:

$\underset{87 g}{M n O_{2}} + 4 H C l \to M n C l_{2} + 2 H_{2} O + \underset{\begin{matrix} 22 .4 l i t r e s \\ a t S T P \end{matrix}}{C l_{2}}$
$1 g m M n O_{2} g i v e s = \frac{22 .4}{87} l i t r e s o f C l_{2}$
$4 .35 g M n O_{2} g i v e s = \frac{22 .4 \times 4 .35}{87}$
$= 1 .12 l i t . o f C l_{2} .$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306977 A sample of a hydrate of barium chloride weighing 61 g was heated until all the water of hydration is removed. The dried sample weighed 52 g. The formula of the hydrated salt is:
(atomic mass : Ba = 137 amu , Cl = 35.5 amu)

1

B a C l_{2} .4 H_{2} O

2

B a C l_{2} .3 H_{2} O

3

B a C l_{2} . H_{2} O

4

B a C l_{2} .2 H_{2} O

Explanation:

Weight of hydrated $B a C l_{2}$ = 61 g
Weight of anhydrous $B a C l_{2}$ = 52 g
Loss in mass = 9g
Assuming $B a C l_{2} . x H_{2} O$ as hydrate
Moles of $H_{2} O = \frac{9}{18} = 0.5$
Gram molecular weight of $B a C l_{2}$ = 208 % of $H_{2} O$ in this hydrated
$B a C l_{2} = \frac{9}{61} \times 100 = 14.75 % = \frac{18 x}{208 + 18 x} \times 100$
On solving, x = 2
So, the formula is $B a C l_{2} .2 H_{2} O$

JEE - 2015

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306974 When 22.4 L of $H_{2} (g)$ is mixed with 11.2 L of $C l_{2} (g)$ , each at STP, the moles of $H C l (g)$ formed is equal to

1 1 mole of HCl (g)

2 2 moles of HCl (g)

3 0.5 mole of HCl (g)

4 1.5 mole of HCl (g)

Explanation:

The given problem is related to the concept of stoichiometry of chemical equations. Thus, we have to convert the given volume into their moles and then identify the limiting reagent [possessing minimum number of moles and gets completely used up in the reaction]. The limiting reagent gives the moles of product formed in the reaction.
$H_{2} (g) + C l_{2} (g) \to 2 H C l (g)$
$I n i t i a l v o l . 22 .4 L 11 .2 L 2 m o l$
$∵ 22 .4L$ volume at STP is occupied by $C l_{2} = 1 m o l e$
$∴ 1102 L$ volume will be occupied by, $C l_{2} = \frac{1 \times 11 .2}{22 .4} m o l e = 0 .5 m o l$
Thus, $H_{2} (g) + C l_{2} (g) \to 2 H C l_{2} (g)$
$1 m o l 0 .5 m o l$
Since, $C l_{2}$ possesses minimum number of moles, thus it is the limiting reagent
As per equation,
$1 m o l C l_{2} \equiv 2 m o l H C l$
$∴ 0 .5 m o l C l_{2} = 2 \times 0 .5 m o l H C l = 1 .0 m o l H C l$
Hence 1.0 mole of HCl (g) is produced by 0.5 mole of $C l_{2} [11 .2 L]$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306975 What mass of 95% pure $C a C O_{3}$ will be required to neutralise 50 mL of 0.5 M HCl solution according to the following reaction?
$C a C O_{3 (s)} + 2 H C l_{(a q)} \to C a C L_{2 (g)} + H_{2} O_{(l)}$
[Calculate upto second place of decimal point]

1 1.32 g

2 3.65 g

3 9.50 g

4 1.25 g

Explanation:

$C a C O_{3 (s)} + 2 H C l_{(a q)} \to C a C L_{2 (g)} + C O_{2 (g)} + H_{2} O_{(ι)}$
$N o . o f m o l e s o f C a C O_{3} (p u r e) =$
$\frac{1}{2} \times m o l e o f H C l$
$[M o l e = m o l a r i t y \times v o l u m e (i n l t r .)]$
$= \frac{1}{2} \times 0 .5 \times \frac{50}{1000} = 0 .0125$
$W e i g h t o f C a C O_{3} (p u r e) = m o l e \times m o l . w t$
$= 0 .0125 \times 100 = 1 .25 g$
$% P u r i t y = \frac{w t . o f p u r e s u b s t a n c e}{w t . o f i m p u r e s a m p l e} \times 100$
$95 = \frac{125}{w t . o f i m p u r e s a m p l e} \times 100$
$95 = \frac{125}{w t . o f i m p u r e s a m p l e} \times 100$

NEET - 2022

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306976 The volume of ${Cl}_{2}$ at STP obtained on reacting $4.35 g {MnO}_{2}$ with conc. $HCl$ (At. wt. of $Mn = 55$ )

1 4.48 litre

2 2.24 litres

3 1.12 litre

4 0.56 litre

Explanation:

$\underset{87 g}{M n O_{2}} + 4 H C l \to M n C l_{2} + 2 H_{2} O + \underset{\begin{matrix} 22 .4 l i t r e s \\ a t S T P \end{matrix}}{C l_{2}}$
$1 g m M n O_{2} g i v e s = \frac{22 .4}{87} l i t r e s o f C l_{2}$
$4 .35 g M n O_{2} g i v e s = \frac{22 .4 \times 4 .35}{87}$
$= 1 .12 l i t . o f C l_{2} .$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306977 A sample of a hydrate of barium chloride weighing 61 g was heated until all the water of hydration is removed. The dried sample weighed 52 g. The formula of the hydrated salt is:
(atomic mass : Ba = 137 amu , Cl = 35.5 amu)

1

B a C l_{2} .4 H_{2} O

2

B a C l_{2} .3 H_{2} O

3

B a C l_{2} . H_{2} O

4

B a C l_{2} .2 H_{2} O

Explanation:

Weight of hydrated $B a C l_{2}$ = 61 g
Weight of anhydrous $B a C l_{2}$ = 52 g
Loss in mass = 9g
Assuming $B a C l_{2} . x H_{2} O$ as hydrate
Moles of $H_{2} O = \frac{9}{18} = 0.5$
Gram molecular weight of $B a C l_{2}$ = 208 % of $H_{2} O$ in this hydrated
$B a C l_{2} = \frac{9}{61} \times 100 = 14.75 % = \frac{18 x}{208 + 18 x} \times 100$
On solving, x = 2
So, the formula is $B a C l_{2} .2 H_{2} O$

JEE - 2015

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306974 When 22.4 L of $H_{2} (g)$ is mixed with 11.2 L of $C l_{2} (g)$ , each at STP, the moles of $H C l (g)$ formed is equal to

1 1 mole of HCl (g)

2 2 moles of HCl (g)

3 0.5 mole of HCl (g)

4 1.5 mole of HCl (g)

Explanation:

The given problem is related to the concept of stoichiometry of chemical equations. Thus, we have to convert the given volume into their moles and then identify the limiting reagent [possessing minimum number of moles and gets completely used up in the reaction]. The limiting reagent gives the moles of product formed in the reaction.
$H_{2} (g) + C l_{2} (g) \to 2 H C l (g)$
$I n i t i a l v o l . 22 .4 L 11 .2 L 2 m o l$
$∵ 22 .4L$ volume at STP is occupied by $C l_{2} = 1 m o l e$
$∴ 1102 L$ volume will be occupied by, $C l_{2} = \frac{1 \times 11 .2}{22 .4} m o l e = 0 .5 m o l$
Thus, $H_{2} (g) + C l_{2} (g) \to 2 H C l_{2} (g)$
$1 m o l 0 .5 m o l$
Since, $C l_{2}$ possesses minimum number of moles, thus it is the limiting reagent
As per equation,
$1 m o l C l_{2} \equiv 2 m o l H C l$
$∴ 0 .5 m o l C l_{2} = 2 \times 0 .5 m o l H C l = 1 .0 m o l H C l$
Hence 1.0 mole of HCl (g) is produced by 0.5 mole of $C l_{2} [11 .2 L]$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306975 What mass of 95% pure $C a C O_{3}$ will be required to neutralise 50 mL of 0.5 M HCl solution according to the following reaction?
$C a C O_{3 (s)} + 2 H C l_{(a q)} \to C a C L_{2 (g)} + H_{2} O_{(l)}$
[Calculate upto second place of decimal point]

1 1.32 g

2 3.65 g

3 9.50 g

4 1.25 g

Explanation:

$C a C O_{3 (s)} + 2 H C l_{(a q)} \to C a C L_{2 (g)} + C O_{2 (g)} + H_{2} O_{(ι)}$
$N o . o f m o l e s o f C a C O_{3} (p u r e) =$
$\frac{1}{2} \times m o l e o f H C l$
$[M o l e = m o l a r i t y \times v o l u m e (i n l t r .)]$
$= \frac{1}{2} \times 0 .5 \times \frac{50}{1000} = 0 .0125$
$W e i g h t o f C a C O_{3} (p u r e) = m o l e \times m o l . w t$
$= 0 .0125 \times 100 = 1 .25 g$
$% P u r i t y = \frac{w t . o f p u r e s u b s t a n c e}{w t . o f i m p u r e s a m p l e} \times 100$
$95 = \frac{125}{w t . o f i m p u r e s a m p l e} \times 100$
$95 = \frac{125}{w t . o f i m p u r e s a m p l e} \times 100$

NEET - 2022

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306976 The volume of ${Cl}_{2}$ at STP obtained on reacting $4.35 g {MnO}_{2}$ with conc. $HCl$ (At. wt. of $Mn = 55$ )

1 4.48 litre

2 2.24 litres

3 1.12 litre

4 0.56 litre

Explanation:

$\underset{87 g}{M n O_{2}} + 4 H C l \to M n C l_{2} + 2 H_{2} O + \underset{\begin{matrix} 22 .4 l i t r e s \\ a t S T P \end{matrix}}{C l_{2}}$
$1 g m M n O_{2} g i v e s = \frac{22 .4}{87} l i t r e s o f C l_{2}$
$4 .35 g M n O_{2} g i v e s = \frac{22 .4 \times 4 .35}{87}$
$= 1 .12 l i t . o f C l_{2} .$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306977 A sample of a hydrate of barium chloride weighing 61 g was heated until all the water of hydration is removed. The dried sample weighed 52 g. The formula of the hydrated salt is:
(atomic mass : Ba = 137 amu , Cl = 35.5 amu)

1

B a C l_{2} .4 H_{2} O

2

B a C l_{2} .3 H_{2} O

3

B a C l_{2} . H_{2} O

4

B a C l_{2} .2 H_{2} O

Explanation:

Weight of hydrated $B a C l_{2}$ = 61 g
Weight of anhydrous $B a C l_{2}$ = 52 g
Loss in mass = 9g
Assuming $B a C l_{2} . x H_{2} O$ as hydrate
Moles of $H_{2} O = \frac{9}{18} = 0.5$
Gram molecular weight of $B a C l_{2}$ = 208 % of $H_{2} O$ in this hydrated
$B a C l_{2} = \frac{9}{61} \times 100 = 14.75 % = \frac{18 x}{208 + 18 x} \times 100$
On solving, x = 2
So, the formula is $B a C l_{2} .2 H_{2} O$

JEE - 2015

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306974 When 22.4 L of $H_{2} (g)$ is mixed with 11.2 L of $C l_{2} (g)$ , each at STP, the moles of $H C l (g)$ formed is equal to

1 1 mole of HCl (g)

2 2 moles of HCl (g)

3 0.5 mole of HCl (g)

4 1.5 mole of HCl (g)

Explanation:

The given problem is related to the concept of stoichiometry of chemical equations. Thus, we have to convert the given volume into their moles and then identify the limiting reagent [possessing minimum number of moles and gets completely used up in the reaction]. The limiting reagent gives the moles of product formed in the reaction.
$H_{2} (g) + C l_{2} (g) \to 2 H C l (g)$
$I n i t i a l v o l . 22 .4 L 11 .2 L 2 m o l$
$∵ 22 .4L$ volume at STP is occupied by $C l_{2} = 1 m o l e$
$∴ 1102 L$ volume will be occupied by, $C l_{2} = \frac{1 \times 11 .2}{22 .4} m o l e = 0 .5 m o l$
Thus, $H_{2} (g) + C l_{2} (g) \to 2 H C l_{2} (g)$
$1 m o l 0 .5 m o l$
Since, $C l_{2}$ possesses minimum number of moles, thus it is the limiting reagent
As per equation,
$1 m o l C l_{2} \equiv 2 m o l H C l$
$∴ 0 .5 m o l C l_{2} = 2 \times 0 .5 m o l H C l = 1 .0 m o l H C l$
Hence 1.0 mole of HCl (g) is produced by 0.5 mole of $C l_{2} [11 .2 L]$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306975 What mass of 95% pure $C a C O_{3}$ will be required to neutralise 50 mL of 0.5 M HCl solution according to the following reaction?
$C a C O_{3 (s)} + 2 H C l_{(a q)} \to C a C L_{2 (g)} + H_{2} O_{(l)}$
[Calculate upto second place of decimal point]

1 1.32 g

2 3.65 g

3 9.50 g

4 1.25 g

Explanation:

$C a C O_{3 (s)} + 2 H C l_{(a q)} \to C a C L_{2 (g)} + C O_{2 (g)} + H_{2} O_{(ι)}$
$N o . o f m o l e s o f C a C O_{3} (p u r e) =$
$\frac{1}{2} \times m o l e o f H C l$
$[M o l e = m o l a r i t y \times v o l u m e (i n l t r .)]$
$= \frac{1}{2} \times 0 .5 \times \frac{50}{1000} = 0 .0125$
$W e i g h t o f C a C O_{3} (p u r e) = m o l e \times m o l . w t$
$= 0 .0125 \times 100 = 1 .25 g$
$% P u r i t y = \frac{w t . o f p u r e s u b s t a n c e}{w t . o f i m p u r e s a m p l e} \times 100$
$95 = \frac{125}{w t . o f i m p u r e s a m p l e} \times 100$
$95 = \frac{125}{w t . o f i m p u r e s a m p l e} \times 100$

NEET - 2022

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306976 The volume of ${Cl}_{2}$ at STP obtained on reacting $4.35 g {MnO}_{2}$ with conc. $HCl$ (At. wt. of $Mn = 55$ )

1 4.48 litre

2 2.24 litres

3 1.12 litre

4 0.56 litre

Explanation:

$\underset{87 g}{M n O_{2}} + 4 H C l \to M n C l_{2} + 2 H_{2} O + \underset{\begin{matrix} 22 .4 l i t r e s \\ a t S T P \end{matrix}}{C l_{2}}$
$1 g m M n O_{2} g i v e s = \frac{22 .4}{87} l i t r e s o f C l_{2}$
$4 .35 g M n O_{2} g i v e s = \frac{22 .4 \times 4 .35}{87}$
$= 1 .12 l i t . o f C l_{2} .$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306977 A sample of a hydrate of barium chloride weighing 61 g was heated until all the water of hydration is removed. The dried sample weighed 52 g. The formula of the hydrated salt is:
(atomic mass : Ba = 137 amu , Cl = 35.5 amu)

1

B a C l_{2} .4 H_{2} O

2

B a C l_{2} .3 H_{2} O

3

B a C l_{2} . H_{2} O

4

B a C l_{2} .2 H_{2} O

Explanation:

Weight of hydrated $B a C l_{2}$ = 61 g
Weight of anhydrous $B a C l_{2}$ = 52 g
Loss in mass = 9g
Assuming $B a C l_{2} . x H_{2} O$ as hydrate
Moles of $H_{2} O = \frac{9}{18} = 0.5$
Gram molecular weight of $B a C l_{2}$ = 208 % of $H_{2} O$ in this hydrated
$B a C l_{2} = \frac{9}{61} \times 100 = 14.75 % = \frac{18 x}{208 + 18 x} \times 100$
On solving, x = 2
So, the formula is $B a C l_{2} .2 H_{2} O$

JEE - 2015

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