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CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306763 Two elements $X$ and $Y$ combine in gaseous state to form $XY$ in the ratio 1:35.5 by mass. The mass of $Y$ that will be required to react with $2 g$ of $X$ is

1

7.1 g

2

3.55 g

3

71 g

4

35.5 g

Explanation:

$1 g$ of $X$ combines with $35.5 g$ ' $y$ '
$∴ 2 g$ of $X$ combines with $\frac{2 \times 35.5}{1} = 71 g Y$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306764 The vapour density of a mixture containing ${NO}_{2}$ and $N_{2} O_{4}$ is 38.3 at $27^{\circ} C$ . Calculate the moles of ${NO}_{2}$ in $500 g$ of the mixture(approx.)

1 1

2 2

3 3

4 4

Explanation:

Mol. mass of mixture $= 2 \times 38.3 = 76.6$
No. of moles in $500 g$ of mixture $= \frac{500}{76.6}$
Let a g of ${NO}_{2}$ is present in mixture.
Moles of ${NO}_{2} +$ Moles of $N_{2} O_{4} =$ Moles of mixture
$\frac{a}{46} + \frac{500 - a}{76.6} = \frac{500}{76.6}$ or $a = 100.52 g$
Moles of ${NO}_{2}$ in mixture $= \frac{100.52}{46} = 2.19$ $\approx 2$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306765 Number of moles of electrons in $0.5 m o l N^{3 -}$ ions will be _____.

1 4

2 5

3 7

4 10

Explanation:

$N^{3 -}$ ion has 10 electrons. Thus, $0.5 mol$ of $N^{3 -}$ ions will have $5 mol$ electrons.

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306766 How many molecules are present in $22400 {cm}^{3}$ of a gas at STP?

1

22.4 \times 10^{20}

2

6.022 \times 10^{23}

3

6.022 \times 10^{20}

4

22.4 \times 10^{23}

Explanation:

1 mole of gas at STP occupies $22400 {cm}^{3}$ volume and contains $6.022 \times 10^{23}$ molecules

MHTCET - 2021

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306767 Arrange the following (I to IV) in the order of increasing masses.
(I) 0.5 mole of $O_{3}$
(II) 0.5 gm molecule of Nitrogen
(III) $3 .011 \times 1 0^{23}$ molecules of $O_{2}$
(IV) 11.35 L of $C O_{2}$ at STP

1 IV < III < II < I

2 II < III < IV < I

3 III < II < I < IV

4 I < II < III < IV

Explanation:

Convert all the value in mass and the increasing order is $(I I < I I I < I V < I)$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306763 Two elements $X$ and $Y$ combine in gaseous state to form $XY$ in the ratio 1:35.5 by mass. The mass of $Y$ that will be required to react with $2 g$ of $X$ is

1

7.1 g

2

3.55 g

3

71 g

4

35.5 g

Explanation:

$1 g$ of $X$ combines with $35.5 g$ ' $y$ '
$∴ 2 g$ of $X$ combines with $\frac{2 \times 35.5}{1} = 71 g Y$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306764 The vapour density of a mixture containing ${NO}_{2}$ and $N_{2} O_{4}$ is 38.3 at $27^{\circ} C$ . Calculate the moles of ${NO}_{2}$ in $500 g$ of the mixture(approx.)

1 1

2 2

3 3

4 4

Explanation:

Mol. mass of mixture $= 2 \times 38.3 = 76.6$
No. of moles in $500 g$ of mixture $= \frac{500}{76.6}$
Let a g of ${NO}_{2}$ is present in mixture.
Moles of ${NO}_{2} +$ Moles of $N_{2} O_{4} =$ Moles of mixture
$\frac{a}{46} + \frac{500 - a}{76.6} = \frac{500}{76.6}$ or $a = 100.52 g$
Moles of ${NO}_{2}$ in mixture $= \frac{100.52}{46} = 2.19$ $\approx 2$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306765 Number of moles of electrons in $0.5 m o l N^{3 -}$ ions will be _____.

1 4

2 5

3 7

4 10

Explanation:

$N^{3 -}$ ion has 10 electrons. Thus, $0.5 mol$ of $N^{3 -}$ ions will have $5 mol$ electrons.

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306766 How many molecules are present in $22400 {cm}^{3}$ of a gas at STP?

1

22.4 \times 10^{20}

2

6.022 \times 10^{23}

3

6.022 \times 10^{20}

4

22.4 \times 10^{23}

Explanation:

1 mole of gas at STP occupies $22400 {cm}^{3}$ volume and contains $6.022 \times 10^{23}$ molecules

MHTCET - 2021

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306767 Arrange the following (I to IV) in the order of increasing masses.
(I) 0.5 mole of $O_{3}$
(II) 0.5 gm molecule of Nitrogen
(III) $3 .011 \times 1 0^{23}$ molecules of $O_{2}$
(IV) 11.35 L of $C O_{2}$ at STP

1 IV < III < II < I

2 II < III < IV < I

3 III < II < I < IV

4 I < II < III < IV

Explanation:

Convert all the value in mass and the increasing order is $(I I < I I I < I V < I)$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306763 Two elements $X$ and $Y$ combine in gaseous state to form $XY$ in the ratio 1:35.5 by mass. The mass of $Y$ that will be required to react with $2 g$ of $X$ is

1

7.1 g

2

3.55 g

3

71 g

4

35.5 g

Explanation:

$1 g$ of $X$ combines with $35.5 g$ ' $y$ '
$∴ 2 g$ of $X$ combines with $\frac{2 \times 35.5}{1} = 71 g Y$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306764 The vapour density of a mixture containing ${NO}_{2}$ and $N_{2} O_{4}$ is 38.3 at $27^{\circ} C$ . Calculate the moles of ${NO}_{2}$ in $500 g$ of the mixture(approx.)

1 1

2 2

3 3

4 4

Explanation:

Mol. mass of mixture $= 2 \times 38.3 = 76.6$
No. of moles in $500 g$ of mixture $= \frac{500}{76.6}$
Let a g of ${NO}_{2}$ is present in mixture.
Moles of ${NO}_{2} +$ Moles of $N_{2} O_{4} =$ Moles of mixture
$\frac{a}{46} + \frac{500 - a}{76.6} = \frac{500}{76.6}$ or $a = 100.52 g$
Moles of ${NO}_{2}$ in mixture $= \frac{100.52}{46} = 2.19$ $\approx 2$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306765 Number of moles of electrons in $0.5 m o l N^{3 -}$ ions will be _____.

1 4

2 5

3 7

4 10

Explanation:

$N^{3 -}$ ion has 10 electrons. Thus, $0.5 mol$ of $N^{3 -}$ ions will have $5 mol$ electrons.

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306766 How many molecules are present in $22400 {cm}^{3}$ of a gas at STP?

1

22.4 \times 10^{20}

2

6.022 \times 10^{23}

3

6.022 \times 10^{20}

4

22.4 \times 10^{23}

Explanation:

1 mole of gas at STP occupies $22400 {cm}^{3}$ volume and contains $6.022 \times 10^{23}$ molecules

MHTCET - 2021

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306767 Arrange the following (I to IV) in the order of increasing masses.
(I) 0.5 mole of $O_{3}$
(II) 0.5 gm molecule of Nitrogen
(III) $3 .011 \times 1 0^{23}$ molecules of $O_{2}$
(IV) 11.35 L of $C O_{2}$ at STP

1 IV < III < II < I

2 II < III < IV < I

3 III < II < I < IV

4 I < II < III < IV

Explanation:

Convert all the value in mass and the increasing order is $(I I < I I I < I V < I)$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306763 Two elements $X$ and $Y$ combine in gaseous state to form $XY$ in the ratio 1:35.5 by mass. The mass of $Y$ that will be required to react with $2 g$ of $X$ is

1

7.1 g

2

3.55 g

3

71 g

4

35.5 g

Explanation:

$1 g$ of $X$ combines with $35.5 g$ ' $y$ '
$∴ 2 g$ of $X$ combines with $\frac{2 \times 35.5}{1} = 71 g Y$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306764 The vapour density of a mixture containing ${NO}_{2}$ and $N_{2} O_{4}$ is 38.3 at $27^{\circ} C$ . Calculate the moles of ${NO}_{2}$ in $500 g$ of the mixture(approx.)

1 1

2 2

3 3

4 4

Explanation:

Mol. mass of mixture $= 2 \times 38.3 = 76.6$
No. of moles in $500 g$ of mixture $= \frac{500}{76.6}$
Let a g of ${NO}_{2}$ is present in mixture.
Moles of ${NO}_{2} +$ Moles of $N_{2} O_{4} =$ Moles of mixture
$\frac{a}{46} + \frac{500 - a}{76.6} = \frac{500}{76.6}$ or $a = 100.52 g$
Moles of ${NO}_{2}$ in mixture $= \frac{100.52}{46} = 2.19$ $\approx 2$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306765 Number of moles of electrons in $0.5 m o l N^{3 -}$ ions will be _____.

1 4

2 5

3 7

4 10

Explanation:

$N^{3 -}$ ion has 10 electrons. Thus, $0.5 mol$ of $N^{3 -}$ ions will have $5 mol$ electrons.

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306766 How many molecules are present in $22400 {cm}^{3}$ of a gas at STP?

1

22.4 \times 10^{20}

2

6.022 \times 10^{23}

3

6.022 \times 10^{20}

4

22.4 \times 10^{23}

Explanation:

1 mole of gas at STP occupies $22400 {cm}^{3}$ volume and contains $6.022 \times 10^{23}$ molecules

MHTCET - 2021

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306767 Arrange the following (I to IV) in the order of increasing masses.
(I) 0.5 mole of $O_{3}$
(II) 0.5 gm molecule of Nitrogen
(III) $3 .011 \times 1 0^{23}$ molecules of $O_{2}$
(IV) 11.35 L of $C O_{2}$ at STP

1 IV < III < II < I

2 II < III < IV < I

3 III < II < I < IV

4 I < II < III < IV

Explanation:

Convert all the value in mass and the increasing order is $(I I < I I I < I V < I)$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306763 Two elements $X$ and $Y$ combine in gaseous state to form $XY$ in the ratio 1:35.5 by mass. The mass of $Y$ that will be required to react with $2 g$ of $X$ is

1

7.1 g

2

3.55 g

3

71 g

4

35.5 g

Explanation:

$1 g$ of $X$ combines with $35.5 g$ ' $y$ '
$∴ 2 g$ of $X$ combines with $\frac{2 \times 35.5}{1} = 71 g Y$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306764 The vapour density of a mixture containing ${NO}_{2}$ and $N_{2} O_{4}$ is 38.3 at $27^{\circ} C$ . Calculate the moles of ${NO}_{2}$ in $500 g$ of the mixture(approx.)

1 1

2 2

3 3

4 4

Explanation:

Mol. mass of mixture $= 2 \times 38.3 = 76.6$
No. of moles in $500 g$ of mixture $= \frac{500}{76.6}$
Let a g of ${NO}_{2}$ is present in mixture.
Moles of ${NO}_{2} +$ Moles of $N_{2} O_{4} =$ Moles of mixture
$\frac{a}{46} + \frac{500 - a}{76.6} = \frac{500}{76.6}$ or $a = 100.52 g$
Moles of ${NO}_{2}$ in mixture $= \frac{100.52}{46} = 2.19$ $\approx 2$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306765 Number of moles of electrons in $0.5 m o l N^{3 -}$ ions will be _____.

1 4

2 5

3 7

4 10

Explanation:

$N^{3 -}$ ion has 10 electrons. Thus, $0.5 mol$ of $N^{3 -}$ ions will have $5 mol$ electrons.

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306766 How many molecules are present in $22400 {cm}^{3}$ of a gas at STP?

1

22.4 \times 10^{20}

2

6.022 \times 10^{23}

3

6.022 \times 10^{20}

4

22.4 \times 10^{23}

Explanation:

1 mole of gas at STP occupies $22400 {cm}^{3}$ volume and contains $6.022 \times 10^{23}$ molecules

MHTCET - 2021

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306767 Arrange the following (I to IV) in the order of increasing masses.
(I) 0.5 mole of $O_{3}$
(II) 0.5 gm molecule of Nitrogen
(III) $3 .011 \times 1 0^{23}$ molecules of $O_{2}$
(IV) 11.35 L of $C O_{2}$ at STP

1 IV < III < II < I

2 II < III < IV < I

3 III < II < I < IV

4 I < II < III < IV

Explanation:

Convert all the value in mass and the increasing order is $(I I < I I I < I V < I)$

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