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CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306791 When burnt in air, 14.0 g mixture of carbon and sulphur gives a mixture of $C O_{2}$ and $S O_{2}$ in the volume ratio 2 : 1, volume being measured at the same conditions of temperature and pressure. Number of moles carbon in the mixture is

1 0.75

2 0.5

3

0 .40

4 0.25

Explanation:

Let weight of C be x g, then weight of S will be (14 -x) g in a mixture.
$\frac{\frac{x}{12}}{\frac{(14 - x)}{32}} = \frac{2}{1}$ $(∵ V \propto n)$
$\Rightarrow x = 6 g; M o l e s o f C = \frac{6}{12} = 0 .5$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306792 In which case the number of molecules of water is maximum ?

1 0.00224 L of water vapours at 1 atm and 273 K

2

1 0^{- 3} m o l

of water

3 0.18g of water

4 18 mL of water

NEET - 2018

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306793 The ratio of number of oxygen atoms (O) in 16.0 g ozone $(O_{3})$ , 28.0 g carbon monoxide $(C O)$ and 16.0 oxygen $(O_{2})$ is (Atomic mass; C = 12, O = 16 and Avogadro’s constant $N_{A} = 6 .023 \times 1 0^{23} m o l^{- 1}$ )

1

3 : 1 : 2

2

1 : 1 : 2

3

3 : 1 : 1

4

1 : 1 : 1

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306794 If an iodised salt contains 1% of KI and a person takes 2g of the salt every day, the iodide ions going into his body every day would be approx

1

7 .2 \times 1 0^{21}

2

7 .2 \times 1 0^{19}

3

3 .6 \times 1 0^{21}

4

9 .5 \times 1 0^{19}

Explanation:

2 g of the salt will contain KI
$= \frac{2}{100} g = 0 .02 g$
$= \frac{0 .02}{39 + 127} m o l = 1 .2 \times 1 0^{- 4} m o l K I$
$= 1 .2 \times 1 0^{- 4} m o l I^{-} i o n s .$
$= (1 .2 \times 1 0^{- 4}) (6 .02 \times 1 0^{- 23}) I^{-} i o n s .$
$7 .2 \times 1 0^{19} I^{-} i o n s .$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306795 26.8 g of ${Na}_{2} {SO}_{4} \cdot {nH}_{2} O$ contains 12.6 g of water. The value of $n$ is

1 1

2 10

3 6

4 7

Explanation:

${Na}_{2} {SO}_{4}$ molar mass is $142 g . mo l^{-}$
only ${Na}_{2} {SO}_{4}$ mass $= 26.8 - 12.6 = 14.2 g$
$∴ 14.2 g N a_{2} S O_{4}$ has $12.6 g H_{2} O$
$142 g N a_{2} S O_{4} h a s ?$
$= \frac{142 \times 12.6}{14.2} = 126 g H_{2} O$
$n_{H_{2} O} = \frac{126}{18} = 7 m o l e H_{2} O$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306791 When burnt in air, 14.0 g mixture of carbon and sulphur gives a mixture of $C O_{2}$ and $S O_{2}$ in the volume ratio 2 : 1, volume being measured at the same conditions of temperature and pressure. Number of moles carbon in the mixture is

1 0.75

2 0.5

3

0 .40

4 0.25

Explanation:

Let weight of C be x g, then weight of S will be (14 -x) g in a mixture.
$\frac{\frac{x}{12}}{\frac{(14 - x)}{32}} = \frac{2}{1}$ $(∵ V \propto n)$
$\Rightarrow x = 6 g; M o l e s o f C = \frac{6}{12} = 0 .5$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306792 In which case the number of molecules of water is maximum ?

1 0.00224 L of water vapours at 1 atm and 273 K

2

1 0^{- 3} m o l

of water

3 0.18g of water

4 18 mL of water

NEET - 2018

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306793 The ratio of number of oxygen atoms (O) in 16.0 g ozone $(O_{3})$ , 28.0 g carbon monoxide $(C O)$ and 16.0 oxygen $(O_{2})$ is (Atomic mass; C = 12, O = 16 and Avogadro’s constant $N_{A} = 6 .023 \times 1 0^{23} m o l^{- 1}$ )

1

3 : 1 : 2

2

1 : 1 : 2

3

3 : 1 : 1

4

1 : 1 : 1

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306794 If an iodised salt contains 1% of KI and a person takes 2g of the salt every day, the iodide ions going into his body every day would be approx

1

7 .2 \times 1 0^{21}

2

7 .2 \times 1 0^{19}

3

3 .6 \times 1 0^{21}

4

9 .5 \times 1 0^{19}

Explanation:

2 g of the salt will contain KI
$= \frac{2}{100} g = 0 .02 g$
$= \frac{0 .02}{39 + 127} m o l = 1 .2 \times 1 0^{- 4} m o l K I$
$= 1 .2 \times 1 0^{- 4} m o l I^{-} i o n s .$
$= (1 .2 \times 1 0^{- 4}) (6 .02 \times 1 0^{- 23}) I^{-} i o n s .$
$7 .2 \times 1 0^{19} I^{-} i o n s .$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306795 26.8 g of ${Na}_{2} {SO}_{4} \cdot {nH}_{2} O$ contains 12.6 g of water. The value of $n$ is

1 1

2 10

3 6

4 7

Explanation:

${Na}_{2} {SO}_{4}$ molar mass is $142 g . mo l^{-}$
only ${Na}_{2} {SO}_{4}$ mass $= 26.8 - 12.6 = 14.2 g$
$∴ 14.2 g N a_{2} S O_{4}$ has $12.6 g H_{2} O$
$142 g N a_{2} S O_{4} h a s ?$
$= \frac{142 \times 12.6}{14.2} = 126 g H_{2} O$
$n_{H_{2} O} = \frac{126}{18} = 7 m o l e H_{2} O$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306791 When burnt in air, 14.0 g mixture of carbon and sulphur gives a mixture of $C O_{2}$ and $S O_{2}$ in the volume ratio 2 : 1, volume being measured at the same conditions of temperature and pressure. Number of moles carbon in the mixture is

1 0.75

2 0.5

3

0 .40

4 0.25

Explanation:

Let weight of C be x g, then weight of S will be (14 -x) g in a mixture.
$\frac{\frac{x}{12}}{\frac{(14 - x)}{32}} = \frac{2}{1}$ $(∵ V \propto n)$
$\Rightarrow x = 6 g; M o l e s o f C = \frac{6}{12} = 0 .5$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306792 In which case the number of molecules of water is maximum ?

1 0.00224 L of water vapours at 1 atm and 273 K

2

1 0^{- 3} m o l

of water

3 0.18g of water

4 18 mL of water

NEET - 2018

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306793 The ratio of number of oxygen atoms (O) in 16.0 g ozone $(O_{3})$ , 28.0 g carbon monoxide $(C O)$ and 16.0 oxygen $(O_{2})$ is (Atomic mass; C = 12, O = 16 and Avogadro’s constant $N_{A} = 6 .023 \times 1 0^{23} m o l^{- 1}$ )

1

3 : 1 : 2

2

1 : 1 : 2

3

3 : 1 : 1

4

1 : 1 : 1

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306794 If an iodised salt contains 1% of KI and a person takes 2g of the salt every day, the iodide ions going into his body every day would be approx

1

7 .2 \times 1 0^{21}

2

7 .2 \times 1 0^{19}

3

3 .6 \times 1 0^{21}

4

9 .5 \times 1 0^{19}

Explanation:

2 g of the salt will contain KI
$= \frac{2}{100} g = 0 .02 g$
$= \frac{0 .02}{39 + 127} m o l = 1 .2 \times 1 0^{- 4} m o l K I$
$= 1 .2 \times 1 0^{- 4} m o l I^{-} i o n s .$
$= (1 .2 \times 1 0^{- 4}) (6 .02 \times 1 0^{- 23}) I^{-} i o n s .$
$7 .2 \times 1 0^{19} I^{-} i o n s .$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306795 26.8 g of ${Na}_{2} {SO}_{4} \cdot {nH}_{2} O$ contains 12.6 g of water. The value of $n$ is

1 1

2 10

3 6

4 7

Explanation:

${Na}_{2} {SO}_{4}$ molar mass is $142 g . mo l^{-}$
only ${Na}_{2} {SO}_{4}$ mass $= 26.8 - 12.6 = 14.2 g$
$∴ 14.2 g N a_{2} S O_{4}$ has $12.6 g H_{2} O$
$142 g N a_{2} S O_{4} h a s ?$
$= \frac{142 \times 12.6}{14.2} = 126 g H_{2} O$
$n_{H_{2} O} = \frac{126}{18} = 7 m o l e H_{2} O$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306791 When burnt in air, 14.0 g mixture of carbon and sulphur gives a mixture of $C O_{2}$ and $S O_{2}$ in the volume ratio 2 : 1, volume being measured at the same conditions of temperature and pressure. Number of moles carbon in the mixture is

1 0.75

2 0.5

3

0 .40

4 0.25

Explanation:

Let weight of C be x g, then weight of S will be (14 -x) g in a mixture.
$\frac{\frac{x}{12}}{\frac{(14 - x)}{32}} = \frac{2}{1}$ $(∵ V \propto n)$
$\Rightarrow x = 6 g; M o l e s o f C = \frac{6}{12} = 0 .5$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306792 In which case the number of molecules of water is maximum ?

1 0.00224 L of water vapours at 1 atm and 273 K

2

1 0^{- 3} m o l

of water

3 0.18g of water

4 18 mL of water

NEET - 2018

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306793 The ratio of number of oxygen atoms (O) in 16.0 g ozone $(O_{3})$ , 28.0 g carbon monoxide $(C O)$ and 16.0 oxygen $(O_{2})$ is (Atomic mass; C = 12, O = 16 and Avogadro’s constant $N_{A} = 6 .023 \times 1 0^{23} m o l^{- 1}$ )

1

3 : 1 : 2

2

1 : 1 : 2

3

3 : 1 : 1

4

1 : 1 : 1

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306794 If an iodised salt contains 1% of KI and a person takes 2g of the salt every day, the iodide ions going into his body every day would be approx

1

7 .2 \times 1 0^{21}

2

7 .2 \times 1 0^{19}

3

3 .6 \times 1 0^{21}

4

9 .5 \times 1 0^{19}

Explanation:

2 g of the salt will contain KI
$= \frac{2}{100} g = 0 .02 g$
$= \frac{0 .02}{39 + 127} m o l = 1 .2 \times 1 0^{- 4} m o l K I$
$= 1 .2 \times 1 0^{- 4} m o l I^{-} i o n s .$
$= (1 .2 \times 1 0^{- 4}) (6 .02 \times 1 0^{- 23}) I^{-} i o n s .$
$7 .2 \times 1 0^{19} I^{-} i o n s .$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306795 26.8 g of ${Na}_{2} {SO}_{4} \cdot {nH}_{2} O$ contains 12.6 g of water. The value of $n$ is

1 1

2 10

3 6

4 7

Explanation:

${Na}_{2} {SO}_{4}$ molar mass is $142 g . mo l^{-}$
only ${Na}_{2} {SO}_{4}$ mass $= 26.8 - 12.6 = 14.2 g$
$∴ 14.2 g N a_{2} S O_{4}$ has $12.6 g H_{2} O$
$142 g N a_{2} S O_{4} h a s ?$
$= \frac{142 \times 12.6}{14.2} = 126 g H_{2} O$
$n_{H_{2} O} = \frac{126}{18} = 7 m o l e H_{2} O$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306791 When burnt in air, 14.0 g mixture of carbon and sulphur gives a mixture of $C O_{2}$ and $S O_{2}$ in the volume ratio 2 : 1, volume being measured at the same conditions of temperature and pressure. Number of moles carbon in the mixture is

1 0.75

2 0.5

3

0 .40

4 0.25

Explanation:

Let weight of C be x g, then weight of S will be (14 -x) g in a mixture.
$\frac{\frac{x}{12}}{\frac{(14 - x)}{32}} = \frac{2}{1}$ $(∵ V \propto n)$
$\Rightarrow x = 6 g; M o l e s o f C = \frac{6}{12} = 0 .5$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306792 In which case the number of molecules of water is maximum ?

1 0.00224 L of water vapours at 1 atm and 273 K

2

1 0^{- 3} m o l

of water

3 0.18g of water

4 18 mL of water

NEET - 2018

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306793 The ratio of number of oxygen atoms (O) in 16.0 g ozone $(O_{3})$ , 28.0 g carbon monoxide $(C O)$ and 16.0 oxygen $(O_{2})$ is (Atomic mass; C = 12, O = 16 and Avogadro’s constant $N_{A} = 6 .023 \times 1 0^{23} m o l^{- 1}$ )

1

3 : 1 : 2

2

1 : 1 : 2

3

3 : 1 : 1

4

1 : 1 : 1

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306794 If an iodised salt contains 1% of KI and a person takes 2g of the salt every day, the iodide ions going into his body every day would be approx

1

7 .2 \times 1 0^{21}

2

7 .2 \times 1 0^{19}

3

3 .6 \times 1 0^{21}

4

9 .5 \times 1 0^{19}

Explanation:

2 g of the salt will contain KI
$= \frac{2}{100} g = 0 .02 g$
$= \frac{0 .02}{39 + 127} m o l = 1 .2 \times 1 0^{- 4} m o l K I$
$= 1 .2 \times 1 0^{- 4} m o l I^{-} i o n s .$
$= (1 .2 \times 1 0^{- 4}) (6 .02 \times 1 0^{- 23}) I^{-} i o n s .$
$7 .2 \times 1 0^{19} I^{-} i o n s .$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306795 26.8 g of ${Na}_{2} {SO}_{4} \cdot {nH}_{2} O$ contains 12.6 g of water. The value of $n$ is

1 1

2 10

3 6

4 7

Explanation:

${Na}_{2} {SO}_{4}$ molar mass is $142 g . mo l^{-}$
only ${Na}_{2} {SO}_{4}$ mass $= 26.8 - 12.6 = 14.2 g$
$∴ 14.2 g N a_{2} S O_{4}$ has $12.6 g H_{2} O$
$142 g N a_{2} S O_{4} h a s ?$
$= \frac{142 \times 12.6}{14.2} = 126 g H_{2} O$
$n_{H_{2} O} = \frac{126}{18} = 7 m o l e H_{2} O$

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