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CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

307062 Match the column I with column II and mark the appropriate choice.

Column I	Column II
A - Mass of $H_{2}$ produced when 0.5 mole of zinc reacts with excess of HCl	P - $3.01 \times 10^{23}$ molecules
B - Mass of all atoms of a compound with formula $C_{70} H_{22}$	Q - $6.023 \times 10^{23}$ molecules
C - Number of molecules in $35.5$ g of $C l_{2}$	R - $1.43 \times 10^{- 21}$ g
D - Number of molecules in $64$ g of $S O_{2}$	S - $1$ g

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

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

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

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

Explanation:

(A) $Z n + 2 H C l \to Z n C l_{2} + H_{2}$
1 mole of Zn produces 2g of $H_{2}$
0.5 mole of Zn will produce 1g of $H_{2}$
(B) molar mass of $C_{70} H_{22} = 862$
Mass of atoms = $862 / 6 .023 \times 1 0^{23} = 1 .43 \times 1 0^{- 21} g$
(C) 71g of $C l_{2} = 6 .023 \times 1 0^{23}$ molecules
(D) molar mass of $S O_{2} = 64 = 1 m o l e$
64g of $S O_{2} = 6 .023 \times 1 0^{23}$ molecules

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

307013 The amount of zinc required to produce 224 mL of $H_{2}$ at STP on treatment with dil. $H_{2} S O_{4}$ will be

1 0.65 g

2 6.5 g

3 65 g

4 0.065 g

Explanation:

$Z n + H_{2} S O_{4} \to Z n S O_{4} + H_{2}$
At STP, 65 g of Zn produce 22,400 mL of $H_{2}$ 22.4 mL of $H_{2}$ obtained from 0.65 g of Zn.

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306968 10 g of a mixture of BaO and CaO requires $100 c m^{3}$ of 2.5 M HCl to react completely. The percentage of calcium oxide in the mixture is approximately
(Given: molar mass of BaO = 153)

1 52.6

2 55.1

3 44.9

4 47.4

Explanation:

$B a O + 2 H C l \to B a C l_{2} + H_{2} O$
$C a O + 2 H C l \to C a C l_{2} + H_{2} O$
Reactions show that 2 moles of mix. of BaO and CaO requires 4 moles of HCl to react completely.
Hence, no. of moles of HCl
$= \frac{2 .5 \times 100}{1000} = 0 .25 m o l e s$
Now, HCl BaO + CaO
Initial 4 moles 2 moles
Final 0.25 moles ?
No. of moles of mixture of BaO and CaO
$= \frac{0 .25 \times 2}{4} = 0 .125 m o l e s$
Let the mass of CaO be x g and BaO is $(10 - x) g$
Then, $\frac{x}{56} + \frac{10 - x}{153} = 0 .125$
[Mol. mass of CaO = 56, BaO = 153]
$153 x + 56 (10 - x) = 0 .125 \times 56 \times 153$
$153 x + 560 = 56 x = 0 .125 \times 56 \times 153$
$97 x = 1071 - 560$
$\Rightarrow 97 x = 511 \Rightarrow x = 5 .268$
$∴ % o f C a O = \frac{5 .268}{10} \times 100 = 52 .68$

KCET - 2014

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306969 A sample of $N a C l O_{3}$ is converted by heat to NaCl with a loss of 0.16 g of oxygen. The residue is dissolved in water and precipitated as AgCl . The mass of AgCl (in g) obtained will be: (Given: Molar mass of $A g C l = 143 .5 m o l^{- 1}$ )

1 0.35

2 0.54

3 0.41

4 0.48

Explanation:

$2 N a C l O_{3} \to 2 N a C l + 3 O_{2}$
$0 .16$
$\frac{n_{N a C l}}{2} = \frac{n_{O_{2}}}{3}$ [n = number of moles]
$n_{N a C l} = \frac{0 .16}{32} \times \frac{2}{3} = \frac{1}{200} \times \frac{2}{3} = \frac{1}{300}$
$N a C l \to A g C l$
$P O A C o f C l$
$1 \times n_{N a C l} = 1 \times n_{A g C l}$
$\frac{1}{300} = n_{A g C l}$
Weight of $A g C l = \frac{143 .5}{300}$
$= 0 .48 g$

JEE - 2018

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306970 20 mL of methane is completely burnt using 50 mL of oxygen. The volume of gas left after cooling to room temperature is

1 80 mL

2 40 mL

3 60 mL

4 30 mL

Explanation:

$C H_{4} + 2 O_{2} \to C O_{2} + 2 H_{2} O$
According to the Balenced chemical equation:
$∴$ 20 mL of methane will require = 40 mL of oxygen.
This means that 20 mL of methane will burn completley using 40 mL of oxygen, producing 20 mL of $C O_{2}$ .

$∴$ Volume of the $O_{2}$ gas left = (50 – 40) = 10 mL.
Total volume of gas left after reaction $= 10 m L O_{2} + 20 m L C O_{2} = 30 m L$ gas.

KCET - 2012

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

307062 Match the column I with column II and mark the appropriate choice.

Column I	Column II
A - Mass of $H_{2}$ produced when 0.5 mole of zinc reacts with excess of HCl	P - $3.01 \times 10^{23}$ molecules
B - Mass of all atoms of a compound with formula $C_{70} H_{22}$	Q - $6.023 \times 10^{23}$ molecules
C - Number of molecules in $35.5$ g of $C l_{2}$	R - $1.43 \times 10^{- 21}$ g
D - Number of molecules in $64$ g of $S O_{2}$	S - $1$ g

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

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

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

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

Explanation:

(A) $Z n + 2 H C l \to Z n C l_{2} + H_{2}$
1 mole of Zn produces 2g of $H_{2}$
0.5 mole of Zn will produce 1g of $H_{2}$
(B) molar mass of $C_{70} H_{22} = 862$
Mass of atoms = $862 / 6 .023 \times 1 0^{23} = 1 .43 \times 1 0^{- 21} g$
(C) 71g of $C l_{2} = 6 .023 \times 1 0^{23}$ molecules
(D) molar mass of $S O_{2} = 64 = 1 m o l e$
64g of $S O_{2} = 6 .023 \times 1 0^{23}$ molecules

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

307013 The amount of zinc required to produce 224 mL of $H_{2}$ at STP on treatment with dil. $H_{2} S O_{4}$ will be

1 0.65 g

2 6.5 g

3 65 g

4 0.065 g

Explanation:

$Z n + H_{2} S O_{4} \to Z n S O_{4} + H_{2}$
At STP, 65 g of Zn produce 22,400 mL of $H_{2}$ 22.4 mL of $H_{2}$ obtained from 0.65 g of Zn.

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306968 10 g of a mixture of BaO and CaO requires $100 c m^{3}$ of 2.5 M HCl to react completely. The percentage of calcium oxide in the mixture is approximately
(Given: molar mass of BaO = 153)

1 52.6

2 55.1

3 44.9

4 47.4

Explanation:

$B a O + 2 H C l \to B a C l_{2} + H_{2} O$
$C a O + 2 H C l \to C a C l_{2} + H_{2} O$
Reactions show that 2 moles of mix. of BaO and CaO requires 4 moles of HCl to react completely.
Hence, no. of moles of HCl
$= \frac{2 .5 \times 100}{1000} = 0 .25 m o l e s$
Now, HCl BaO + CaO
Initial 4 moles 2 moles
Final 0.25 moles ?
No. of moles of mixture of BaO and CaO
$= \frac{0 .25 \times 2}{4} = 0 .125 m o l e s$
Let the mass of CaO be x g and BaO is $(10 - x) g$
Then, $\frac{x}{56} + \frac{10 - x}{153} = 0 .125$
[Mol. mass of CaO = 56, BaO = 153]
$153 x + 56 (10 - x) = 0 .125 \times 56 \times 153$
$153 x + 560 = 56 x = 0 .125 \times 56 \times 153$
$97 x = 1071 - 560$
$\Rightarrow 97 x = 511 \Rightarrow x = 5 .268$
$∴ % o f C a O = \frac{5 .268}{10} \times 100 = 52 .68$

KCET - 2014

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306969 A sample of $N a C l O_{3}$ is converted by heat to NaCl with a loss of 0.16 g of oxygen. The residue is dissolved in water and precipitated as AgCl . The mass of AgCl (in g) obtained will be: (Given: Molar mass of $A g C l = 143 .5 m o l^{- 1}$ )

1 0.35

2 0.54

3 0.41

4 0.48

Explanation:

$2 N a C l O_{3} \to 2 N a C l + 3 O_{2}$
$0 .16$
$\frac{n_{N a C l}}{2} = \frac{n_{O_{2}}}{3}$ [n = number of moles]
$n_{N a C l} = \frac{0 .16}{32} \times \frac{2}{3} = \frac{1}{200} \times \frac{2}{3} = \frac{1}{300}$
$N a C l \to A g C l$
$P O A C o f C l$
$1 \times n_{N a C l} = 1 \times n_{A g C l}$
$\frac{1}{300} = n_{A g C l}$
Weight of $A g C l = \frac{143 .5}{300}$
$= 0 .48 g$

JEE - 2018

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306970 20 mL of methane is completely burnt using 50 mL of oxygen. The volume of gas left after cooling to room temperature is

1 80 mL

2 40 mL

3 60 mL

4 30 mL

Explanation:

$C H_{4} + 2 O_{2} \to C O_{2} + 2 H_{2} O$
According to the Balenced chemical equation:
$∴$ 20 mL of methane will require = 40 mL of oxygen.
This means that 20 mL of methane will burn completley using 40 mL of oxygen, producing 20 mL of $C O_{2}$ .

$∴$ Volume of the $O_{2}$ gas left = (50 – 40) = 10 mL.
Total volume of gas left after reaction $= 10 m L O_{2} + 20 m L C O_{2} = 30 m L$ gas.

KCET - 2012

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

307062 Match the column I with column II and mark the appropriate choice.

Column I	Column II
A - Mass of $H_{2}$ produced when 0.5 mole of zinc reacts with excess of HCl	P - $3.01 \times 10^{23}$ molecules
B - Mass of all atoms of a compound with formula $C_{70} H_{22}$	Q - $6.023 \times 10^{23}$ molecules
C - Number of molecules in $35.5$ g of $C l_{2}$	R - $1.43 \times 10^{- 21}$ g
D - Number of molecules in $64$ g of $S O_{2}$	S - $1$ g

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

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

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

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

Explanation:

(A) $Z n + 2 H C l \to Z n C l_{2} + H_{2}$
1 mole of Zn produces 2g of $H_{2}$
0.5 mole of Zn will produce 1g of $H_{2}$
(B) molar mass of $C_{70} H_{22} = 862$
Mass of atoms = $862 / 6 .023 \times 1 0^{23} = 1 .43 \times 1 0^{- 21} g$
(C) 71g of $C l_{2} = 6 .023 \times 1 0^{23}$ molecules
(D) molar mass of $S O_{2} = 64 = 1 m o l e$
64g of $S O_{2} = 6 .023 \times 1 0^{23}$ molecules

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

307013 The amount of zinc required to produce 224 mL of $H_{2}$ at STP on treatment with dil. $H_{2} S O_{4}$ will be

1 0.65 g

2 6.5 g

3 65 g

4 0.065 g

Explanation:

$Z n + H_{2} S O_{4} \to Z n S O_{4} + H_{2}$
At STP, 65 g of Zn produce 22,400 mL of $H_{2}$ 22.4 mL of $H_{2}$ obtained from 0.65 g of Zn.

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306968 10 g of a mixture of BaO and CaO requires $100 c m^{3}$ of 2.5 M HCl to react completely. The percentage of calcium oxide in the mixture is approximately
(Given: molar mass of BaO = 153)

1 52.6

2 55.1

3 44.9

4 47.4

Explanation:

$B a O + 2 H C l \to B a C l_{2} + H_{2} O$
$C a O + 2 H C l \to C a C l_{2} + H_{2} O$
Reactions show that 2 moles of mix. of BaO and CaO requires 4 moles of HCl to react completely.
Hence, no. of moles of HCl
$= \frac{2 .5 \times 100}{1000} = 0 .25 m o l e s$
Now, HCl BaO + CaO
Initial 4 moles 2 moles
Final 0.25 moles ?
No. of moles of mixture of BaO and CaO
$= \frac{0 .25 \times 2}{4} = 0 .125 m o l e s$
Let the mass of CaO be x g and BaO is $(10 - x) g$
Then, $\frac{x}{56} + \frac{10 - x}{153} = 0 .125$
[Mol. mass of CaO = 56, BaO = 153]
$153 x + 56 (10 - x) = 0 .125 \times 56 \times 153$
$153 x + 560 = 56 x = 0 .125 \times 56 \times 153$
$97 x = 1071 - 560$
$\Rightarrow 97 x = 511 \Rightarrow x = 5 .268$
$∴ % o f C a O = \frac{5 .268}{10} \times 100 = 52 .68$

KCET - 2014

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306969 A sample of $N a C l O_{3}$ is converted by heat to NaCl with a loss of 0.16 g of oxygen. The residue is dissolved in water and precipitated as AgCl . The mass of AgCl (in g) obtained will be: (Given: Molar mass of $A g C l = 143 .5 m o l^{- 1}$ )

1 0.35

2 0.54

3 0.41

4 0.48

Explanation:

$2 N a C l O_{3} \to 2 N a C l + 3 O_{2}$
$0 .16$
$\frac{n_{N a C l}}{2} = \frac{n_{O_{2}}}{3}$ [n = number of moles]
$n_{N a C l} = \frac{0 .16}{32} \times \frac{2}{3} = \frac{1}{200} \times \frac{2}{3} = \frac{1}{300}$
$N a C l \to A g C l$
$P O A C o f C l$
$1 \times n_{N a C l} = 1 \times n_{A g C l}$
$\frac{1}{300} = n_{A g C l}$
Weight of $A g C l = \frac{143 .5}{300}$
$= 0 .48 g$

JEE - 2018

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306970 20 mL of methane is completely burnt using 50 mL of oxygen. The volume of gas left after cooling to room temperature is

1 80 mL

2 40 mL

3 60 mL

4 30 mL

Explanation:

$C H_{4} + 2 O_{2} \to C O_{2} + 2 H_{2} O$
According to the Balenced chemical equation:
$∴$ 20 mL of methane will require = 40 mL of oxygen.
This means that 20 mL of methane will burn completley using 40 mL of oxygen, producing 20 mL of $C O_{2}$ .

$∴$ Volume of the $O_{2}$ gas left = (50 – 40) = 10 mL.
Total volume of gas left after reaction $= 10 m L O_{2} + 20 m L C O_{2} = 30 m L$ gas.

KCET - 2012

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

307062 Match the column I with column II and mark the appropriate choice.

Column I	Column II
A - Mass of $H_{2}$ produced when 0.5 mole of zinc reacts with excess of HCl	P - $3.01 \times 10^{23}$ molecules
B - Mass of all atoms of a compound with formula $C_{70} H_{22}$	Q - $6.023 \times 10^{23}$ molecules
C - Number of molecules in $35.5$ g of $C l_{2}$	R - $1.43 \times 10^{- 21}$ g
D - Number of molecules in $64$ g of $S O_{2}$	S - $1$ g

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

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

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

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

Explanation:

(A) $Z n + 2 H C l \to Z n C l_{2} + H_{2}$
1 mole of Zn produces 2g of $H_{2}$
0.5 mole of Zn will produce 1g of $H_{2}$
(B) molar mass of $C_{70} H_{22} = 862$
Mass of atoms = $862 / 6 .023 \times 1 0^{23} = 1 .43 \times 1 0^{- 21} g$
(C) 71g of $C l_{2} = 6 .023 \times 1 0^{23}$ molecules
(D) molar mass of $S O_{2} = 64 = 1 m o l e$
64g of $S O_{2} = 6 .023 \times 1 0^{23}$ molecules

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

307013 The amount of zinc required to produce 224 mL of $H_{2}$ at STP on treatment with dil. $H_{2} S O_{4}$ will be

1 0.65 g

2 6.5 g

3 65 g

4 0.065 g

Explanation:

$Z n + H_{2} S O_{4} \to Z n S O_{4} + H_{2}$
At STP, 65 g of Zn produce 22,400 mL of $H_{2}$ 22.4 mL of $H_{2}$ obtained from 0.65 g of Zn.

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306968 10 g of a mixture of BaO and CaO requires $100 c m^{3}$ of 2.5 M HCl to react completely. The percentage of calcium oxide in the mixture is approximately
(Given: molar mass of BaO = 153)

1 52.6

2 55.1

3 44.9

4 47.4

Explanation:

$B a O + 2 H C l \to B a C l_{2} + H_{2} O$
$C a O + 2 H C l \to C a C l_{2} + H_{2} O$
Reactions show that 2 moles of mix. of BaO and CaO requires 4 moles of HCl to react completely.
Hence, no. of moles of HCl
$= \frac{2 .5 \times 100}{1000} = 0 .25 m o l e s$
Now, HCl BaO + CaO
Initial 4 moles 2 moles
Final 0.25 moles ?
No. of moles of mixture of BaO and CaO
$= \frac{0 .25 \times 2}{4} = 0 .125 m o l e s$
Let the mass of CaO be x g and BaO is $(10 - x) g$
Then, $\frac{x}{56} + \frac{10 - x}{153} = 0 .125$
[Mol. mass of CaO = 56, BaO = 153]
$153 x + 56 (10 - x) = 0 .125 \times 56 \times 153$
$153 x + 560 = 56 x = 0 .125 \times 56 \times 153$
$97 x = 1071 - 560$
$\Rightarrow 97 x = 511 \Rightarrow x = 5 .268$
$∴ % o f C a O = \frac{5 .268}{10} \times 100 = 52 .68$

KCET - 2014

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306969 A sample of $N a C l O_{3}$ is converted by heat to NaCl with a loss of 0.16 g of oxygen. The residue is dissolved in water and precipitated as AgCl . The mass of AgCl (in g) obtained will be: (Given: Molar mass of $A g C l = 143 .5 m o l^{- 1}$ )

1 0.35

2 0.54

3 0.41

4 0.48

Explanation:

$2 N a C l O_{3} \to 2 N a C l + 3 O_{2}$
$0 .16$
$\frac{n_{N a C l}}{2} = \frac{n_{O_{2}}}{3}$ [n = number of moles]
$n_{N a C l} = \frac{0 .16}{32} \times \frac{2}{3} = \frac{1}{200} \times \frac{2}{3} = \frac{1}{300}$
$N a C l \to A g C l$
$P O A C o f C l$
$1 \times n_{N a C l} = 1 \times n_{A g C l}$
$\frac{1}{300} = n_{A g C l}$
Weight of $A g C l = \frac{143 .5}{300}$
$= 0 .48 g$

JEE - 2018

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306970 20 mL of methane is completely burnt using 50 mL of oxygen. The volume of gas left after cooling to room temperature is

1 80 mL

2 40 mL

3 60 mL

4 30 mL

Explanation:

$C H_{4} + 2 O_{2} \to C O_{2} + 2 H_{2} O$
According to the Balenced chemical equation:
$∴$ 20 mL of methane will require = 40 mL of oxygen.
This means that 20 mL of methane will burn completley using 40 mL of oxygen, producing 20 mL of $C O_{2}$ .

$∴$ Volume of the $O_{2}$ gas left = (50 – 40) = 10 mL.
Total volume of gas left after reaction $= 10 m L O_{2} + 20 m L C O_{2} = 30 m L$ gas.

KCET - 2012

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

307062 Match the column I with column II and mark the appropriate choice.

Column I	Column II
A - Mass of $H_{2}$ produced when 0.5 mole of zinc reacts with excess of HCl	P - $3.01 \times 10^{23}$ molecules
B - Mass of all atoms of a compound with formula $C_{70} H_{22}$	Q - $6.023 \times 10^{23}$ molecules
C - Number of molecules in $35.5$ g of $C l_{2}$	R - $1.43 \times 10^{- 21}$ g
D - Number of molecules in $64$ g of $S O_{2}$	S - $1$ g

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

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

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

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

Explanation:

(A) $Z n + 2 H C l \to Z n C l_{2} + H_{2}$
1 mole of Zn produces 2g of $H_{2}$
0.5 mole of Zn will produce 1g of $H_{2}$
(B) molar mass of $C_{70} H_{22} = 862$
Mass of atoms = $862 / 6 .023 \times 1 0^{23} = 1 .43 \times 1 0^{- 21} g$
(C) 71g of $C l_{2} = 6 .023 \times 1 0^{23}$ molecules
(D) molar mass of $S O_{2} = 64 = 1 m o l e$
64g of $S O_{2} = 6 .023 \times 1 0^{23}$ molecules

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

307013 The amount of zinc required to produce 224 mL of $H_{2}$ at STP on treatment with dil. $H_{2} S O_{4}$ will be

1 0.65 g

2 6.5 g

3 65 g

4 0.065 g

Explanation:

$Z n + H_{2} S O_{4} \to Z n S O_{4} + H_{2}$
At STP, 65 g of Zn produce 22,400 mL of $H_{2}$ 22.4 mL of $H_{2}$ obtained from 0.65 g of Zn.

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306968 10 g of a mixture of BaO and CaO requires $100 c m^{3}$ of 2.5 M HCl to react completely. The percentage of calcium oxide in the mixture is approximately
(Given: molar mass of BaO = 153)

1 52.6

2 55.1

3 44.9

4 47.4

Explanation:

$B a O + 2 H C l \to B a C l_{2} + H_{2} O$
$C a O + 2 H C l \to C a C l_{2} + H_{2} O$
Reactions show that 2 moles of mix. of BaO and CaO requires 4 moles of HCl to react completely.
Hence, no. of moles of HCl
$= \frac{2 .5 \times 100}{1000} = 0 .25 m o l e s$
Now, HCl BaO + CaO
Initial 4 moles 2 moles
Final 0.25 moles ?
No. of moles of mixture of BaO and CaO
$= \frac{0 .25 \times 2}{4} = 0 .125 m o l e s$
Let the mass of CaO be x g and BaO is $(10 - x) g$
Then, $\frac{x}{56} + \frac{10 - x}{153} = 0 .125$
[Mol. mass of CaO = 56, BaO = 153]
$153 x + 56 (10 - x) = 0 .125 \times 56 \times 153$
$153 x + 560 = 56 x = 0 .125 \times 56 \times 153$
$97 x = 1071 - 560$
$\Rightarrow 97 x = 511 \Rightarrow x = 5 .268$
$∴ % o f C a O = \frac{5 .268}{10} \times 100 = 52 .68$

KCET - 2014

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306969 A sample of $N a C l O_{3}$ is converted by heat to NaCl with a loss of 0.16 g of oxygen. The residue is dissolved in water and precipitated as AgCl . The mass of AgCl (in g) obtained will be: (Given: Molar mass of $A g C l = 143 .5 m o l^{- 1}$ )

1 0.35

2 0.54

3 0.41

4 0.48

Explanation:

$2 N a C l O_{3} \to 2 N a C l + 3 O_{2}$
$0 .16$
$\frac{n_{N a C l}}{2} = \frac{n_{O_{2}}}{3}$ [n = number of moles]
$n_{N a C l} = \frac{0 .16}{32} \times \frac{2}{3} = \frac{1}{200} \times \frac{2}{3} = \frac{1}{300}$
$N a C l \to A g C l$
$P O A C o f C l$
$1 \times n_{N a C l} = 1 \times n_{A g C l}$
$\frac{1}{300} = n_{A g C l}$
Weight of $A g C l = \frac{143 .5}{300}$
$= 0 .48 g$

JEE - 2018

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306970 20 mL of methane is completely burnt using 50 mL of oxygen. The volume of gas left after cooling to room temperature is

1 80 mL

2 40 mL

3 60 mL

4 30 mL

Explanation:

$C H_{4} + 2 O_{2} \to C O_{2} + 2 H_{2} O$
According to the Balenced chemical equation:
$∴$ 20 mL of methane will require = 40 mL of oxygen.
This means that 20 mL of methane will burn completley using 40 mL of oxygen, producing 20 mL of $C O_{2}$ .

$∴$ Volume of the $O_{2}$ gas left = (50 – 40) = 10 mL.
Total volume of gas left after reaction $= 10 m L O_{2} + 20 m L C O_{2} = 30 m L$ gas.

KCET - 2012