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CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306635 Given the following percentage composition:
Na = 29.11%, S = 40.51% and O = 30.38%. If molar mass of the compound is 158, what is its molecular formula?

1

N a_{2} S_{2} O_{3}

2

N a_{2} S_{3} O_{3}

3

N a S_{3} O_{20}

4

N a S_{2} O_{4}

Explanation:

$E m p i r i c a l f o r m u l a = N a_{2} S_{2} O_{3}$
$E m p i r i c a l m a s s = 158$
$n = \frac{M o l a r m a s s}{E m p i r i c a l m a s s} = \frac{158}{158} = 1$
$∴ M o l e c u l a r f o r m u l a = n \times E m p i r i c a l f o r m u l a = N a_{2} S_{2} O_{3}$
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CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306636 0.145 g of hydrocarbon is heated with dry copper (II) oxide and $224 c m^{3}$ of $C O_{2}$ was collected at STP. The empirical formula of the hydrocarbon is

1

C_{2} H_{5}

2

C_{4} H_{10}

3

C_{3} H_{8}

4

C_{2} H_{2}

Explanation:

Sample of hydrocarbon contains carbon and hydrogen
At STP, $22400 c m^{3} o f C O_{2} = 1 m o l e o f C O_{2} = 1 m o l e o f C = 12$
$224 c m^{3} o f C O_{2} = 0 .01 m o l e o f C O_{2} = 0 .01 m o l e o f C = 00 .12 g$
$M a s s o f C = 0 .12 g$
$M a s s o f H = 0 .45 - M a s s o f C = 0 .12 - 0 .145 = 0 .025 g$
Formula of the compound is $C_{2} H_{5}$
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CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306637 A gaseous compound of nitrogen and hydrogen contains 12.5 % by mass of hydrogen. The density of the compound relative to hydrogen is 16. The molecular formula of the compound is:

1

N H_{2}

2

N_{3} H

3

N H_{3}

4

N_{2} H_{4}

Explanation:

Given, % of H = 12.5%
$∴ % o f N = 100 - 12.5 = 87.5 %$ 2 $\times$ vapour density = Mol.wt
$M o l . w t . = 16 \times 2 = 32$
Molecular formula $= n \times e m p i r i c a l f o r m u l a m a s s$
$n = \frac{32}{16} = 2$
$∴$ Molecular formula of the compound will be $= (N H_{2})_{2} = N_{2} H_{4}$
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JEE - 2014

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306638 Statement A :
The empirical mass of ethyne is half of its molecular mass.
Statement B :
The empirical formula represents the simplest whole number ratio of various atoms present in a compound.

1 Statement A is correct but Statement B is incorrect.

2 Statement A is incorrect but Statement B is correct.

3 Both Statements are correct.

4 Both Statements are incorrect.

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306635 Given the following percentage composition:
Na = 29.11%, S = 40.51% and O = 30.38%. If molar mass of the compound is 158, what is its molecular formula?

1

N a_{2} S_{2} O_{3}

2

N a_{2} S_{3} O_{3}

3

N a S_{3} O_{20}

4

N a S_{2} O_{4}

Explanation:

$E m p i r i c a l f o r m u l a = N a_{2} S_{2} O_{3}$
$E m p i r i c a l m a s s = 158$
$n = \frac{M o l a r m a s s}{E m p i r i c a l m a s s} = \frac{158}{158} = 1$
$∴ M o l e c u l a r f o r m u l a = n \times E m p i r i c a l f o r m u l a = N a_{2} S_{2} O_{3}$
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CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306636 0.145 g of hydrocarbon is heated with dry copper (II) oxide and $224 c m^{3}$ of $C O_{2}$ was collected at STP. The empirical formula of the hydrocarbon is

1

C_{2} H_{5}

2

C_{4} H_{10}

3

C_{3} H_{8}

4

C_{2} H_{2}

Explanation:

Sample of hydrocarbon contains carbon and hydrogen
At STP, $22400 c m^{3} o f C O_{2} = 1 m o l e o f C O_{2} = 1 m o l e o f C = 12$
$224 c m^{3} o f C O_{2} = 0 .01 m o l e o f C O_{2} = 0 .01 m o l e o f C = 00 .12 g$
$M a s s o f C = 0 .12 g$
$M a s s o f H = 0 .45 - M a s s o f C = 0 .12 - 0 .145 = 0 .025 g$
Formula of the compound is $C_{2} H_{5}$
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CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306637 A gaseous compound of nitrogen and hydrogen contains 12.5 % by mass of hydrogen. The density of the compound relative to hydrogen is 16. The molecular formula of the compound is:

1

N H_{2}

2

N_{3} H

3

N H_{3}

4

N_{2} H_{4}

Explanation:

Given, % of H = 12.5%
$∴ % o f N = 100 - 12.5 = 87.5 %$ 2 $\times$ vapour density = Mol.wt
$M o l . w t . = 16 \times 2 = 32$
Molecular formula $= n \times e m p i r i c a l f o r m u l a m a s s$
$n = \frac{32}{16} = 2$
$∴$ Molecular formula of the compound will be $= (N H_{2})_{2} = N_{2} H_{4}$
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JEE - 2014

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306638 Statement A :
The empirical mass of ethyne is half of its molecular mass.
Statement B :
The empirical formula represents the simplest whole number ratio of various atoms present in a compound.

1 Statement A is correct but Statement B is incorrect.

2 Statement A is incorrect but Statement B is correct.

3 Both Statements are correct.

4 Both Statements are incorrect.

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306635 Given the following percentage composition:
Na = 29.11%, S = 40.51% and O = 30.38%. If molar mass of the compound is 158, what is its molecular formula?

1

N a_{2} S_{2} O_{3}

2

N a_{2} S_{3} O_{3}

3

N a S_{3} O_{20}

4

N a S_{2} O_{4}

Explanation:

$E m p i r i c a l f o r m u l a = N a_{2} S_{2} O_{3}$
$E m p i r i c a l m a s s = 158$
$n = \frac{M o l a r m a s s}{E m p i r i c a l m a s s} = \frac{158}{158} = 1$
$∴ M o l e c u l a r f o r m u l a = n \times E m p i r i c a l f o r m u l a = N a_{2} S_{2} O_{3}$
supporting img

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306636 0.145 g of hydrocarbon is heated with dry copper (II) oxide and $224 c m^{3}$ of $C O_{2}$ was collected at STP. The empirical formula of the hydrocarbon is

1

C_{2} H_{5}

2

C_{4} H_{10}

3

C_{3} H_{8}

4

C_{2} H_{2}

Explanation:

Sample of hydrocarbon contains carbon and hydrogen
At STP, $22400 c m^{3} o f C O_{2} = 1 m o l e o f C O_{2} = 1 m o l e o f C = 12$
$224 c m^{3} o f C O_{2} = 0 .01 m o l e o f C O_{2} = 0 .01 m o l e o f C = 00 .12 g$
$M a s s o f C = 0 .12 g$
$M a s s o f H = 0 .45 - M a s s o f C = 0 .12 - 0 .145 = 0 .025 g$
Formula of the compound is $C_{2} H_{5}$
supporting img

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306637 A gaseous compound of nitrogen and hydrogen contains 12.5 % by mass of hydrogen. The density of the compound relative to hydrogen is 16. The molecular formula of the compound is:

1

N H_{2}

2

N_{3} H

3

N H_{3}

4

N_{2} H_{4}

Explanation:

Given, % of H = 12.5%
$∴ % o f N = 100 - 12.5 = 87.5 %$ 2 $\times$ vapour density = Mol.wt
$M o l . w t . = 16 \times 2 = 32$
Molecular formula $= n \times e m p i r i c a l f o r m u l a m a s s$
$n = \frac{32}{16} = 2$
$∴$ Molecular formula of the compound will be $= (N H_{2})_{2} = N_{2} H_{4}$
supporting img

JEE - 2014

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306638 Statement A :
The empirical mass of ethyne is half of its molecular mass.
Statement B :
The empirical formula represents the simplest whole number ratio of various atoms present in a compound.

1 Statement A is correct but Statement B is incorrect.

2 Statement A is incorrect but Statement B is correct.

3 Both Statements are correct.

4 Both Statements are incorrect.

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306635 Given the following percentage composition:
Na = 29.11%, S = 40.51% and O = 30.38%. If molar mass of the compound is 158, what is its molecular formula?

1

N a_{2} S_{2} O_{3}

2

N a_{2} S_{3} O_{3}

3

N a S_{3} O_{20}

4

N a S_{2} O_{4}

Explanation:

$E m p i r i c a l f o r m u l a = N a_{2} S_{2} O_{3}$
$E m p i r i c a l m a s s = 158$
$n = \frac{M o l a r m a s s}{E m p i r i c a l m a s s} = \frac{158}{158} = 1$
$∴ M o l e c u l a r f o r m u l a = n \times E m p i r i c a l f o r m u l a = N a_{2} S_{2} O_{3}$
supporting img

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306636 0.145 g of hydrocarbon is heated with dry copper (II) oxide and $224 c m^{3}$ of $C O_{2}$ was collected at STP. The empirical formula of the hydrocarbon is

1

C_{2} H_{5}

2

C_{4} H_{10}

3

C_{3} H_{8}

4

C_{2} H_{2}

Explanation:

Sample of hydrocarbon contains carbon and hydrogen
At STP, $22400 c m^{3} o f C O_{2} = 1 m o l e o f C O_{2} = 1 m o l e o f C = 12$
$224 c m^{3} o f C O_{2} = 0 .01 m o l e o f C O_{2} = 0 .01 m o l e o f C = 00 .12 g$
$M a s s o f C = 0 .12 g$
$M a s s o f H = 0 .45 - M a s s o f C = 0 .12 - 0 .145 = 0 .025 g$
Formula of the compound is $C_{2} H_{5}$
supporting img

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306637 A gaseous compound of nitrogen and hydrogen contains 12.5 % by mass of hydrogen. The density of the compound relative to hydrogen is 16. The molecular formula of the compound is:

1

N H_{2}

2

N_{3} H

3

N H_{3}

4

N_{2} H_{4}

Explanation:

Given, % of H = 12.5%
$∴ % o f N = 100 - 12.5 = 87.5 %$ 2 $\times$ vapour density = Mol.wt
$M o l . w t . = 16 \times 2 = 32$
Molecular formula $= n \times e m p i r i c a l f o r m u l a m a s s$
$n = \frac{32}{16} = 2$
$∴$ Molecular formula of the compound will be $= (N H_{2})_{2} = N_{2} H_{4}$
supporting img

JEE - 2014

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306638 Statement A :
The empirical mass of ethyne is half of its molecular mass.
Statement B :
The empirical formula represents the simplest whole number ratio of various atoms present in a compound.

1 Statement A is correct but Statement B is incorrect.

2 Statement A is incorrect but Statement B is correct.

3 Both Statements are correct.

4 Both Statements are incorrect.

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