QBManager

Some Basic Concepts of Chemistry

228224 $0.79 gm$ of a metal oxide is obtained from 0.5 gm of the same metal upon oxidation. Equivalent weight of the metal will be which of the following?

1 10

2 13.8

3 20

4 40

Explanation:

Mass of metal + mass of oxygen $=$ mass of metal oxide
Mass of oxygen $= 0.79 - 0.5$
$= 0.29$
$\frac{Mass of metal}{Mass of oxygen} = \frac{Eq.wt. of metal}{Eq.wt.of oxygen}$
Eq. wt. of metal $= \frac{0.5}{0.29} \times 8$
Eq.wt. of metal $= 13.79 ◻ 13.8$

Tripura JEE-2022

Some Basic Concepts of Chemistry

228225 What amount of conc. $H_{2} {SO}_{4}$ solution should be used to prepare $500 ml$ of $0.5 M H_{2} {SO}_{4}$ ? (The concentration of $H_{2} {SO}_{4}$ solution being used is $90 %$ and molecular mass of $H_{2} {SO}_{4} =$

1

22.06 g

2

24.52 g

3

11.03 g

4

27.24 g

Explanation:

Given that-
$\begin{aligned} V & = 500 ml \\ C & = 0.5 M \end{aligned}$
Concentration of $H_{2} {SO}_{4}$ solution being used is $90 %$
So, $C = \frac{0.5 \times 90}{100} = 0.45 M$
Molecular weight of $H_{2} {SO}_{4} = 98.079 g / mol$
$\begin{aligned} ∴ C = \frac{n (moles)}{V (volume)} \\ or C = \frac{weight}{m.weight} \times \frac{1}{V (Volume)} \\ Putting the value we get- \\ 0.45 = \frac{weight / amount}{98.079 \times 0.5} \\ Amount of H_{2} {SO}_{4} = 0.45 \times 0.5 \times 98.079 = 22.06 gm . \end{aligned}$

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Some Basic Concepts of Chemistry

228227 3.7 gm of a gas at $25^{\circ} C$ occupies some volume. At $17^{\circ} C, 0.184$ gm hydrogen gas occupies same volume when pressures of both gases are same. What will be the molecular weight of the gas?

1 41.98

2 20.67

3 20.94

4 41.34

Explanation:

Given, Amount of gas $= 3.7 g$ Here, volume and pressure of both gases are same. As we know, $PV = nRT$ $P$ and $V$ are same for both gases.
So, $n_{1} T_{1} = n_{2} T_{2}$
$\begin{aligned} \frac{w_{1}}{M_{1}} T_{1} = \frac{W_{2}}{M_{2}} T_{2} \\ \frac{3.7}{M_{1}} \times 298 = \frac{0.184}{2} \times 290 \\ M_{1} = 41.33 g \end{aligned}$

Tripura JEE-2021

Some Basic Concepts of Chemistry

228229 Vapour density of a metal chloride is 83. If equivalent weight of the metal is 6 , its atomic weight will be ....... . 1

1 12

2 24

3 18

4 60

Explanation:

Valency of metal,
$\begin{aligned} (n) & = \frac{2 \times vapour density}{Equivalent weight of metal + 35.5} \\ = \frac{2 \times 83}{6 + 35.5} = 4 \end{aligned}$
The atomic weight of metal is equal to the product of the equivalent weight of metal and valency.
Atomic weight of metal $= n \times$ equivalent weight of metal
$\Rightarrow 4 \times 6 = 24$
Hence, the correct option (b).

Shift-I

Some Basic Concepts of Chemistry

228224 $0.79 gm$ of a metal oxide is obtained from 0.5 gm of the same metal upon oxidation. Equivalent weight of the metal will be which of the following?

1 10

2 13.8

3 20

4 40

Explanation:

Mass of metal + mass of oxygen $=$ mass of metal oxide
Mass of oxygen $= 0.79 - 0.5$
$= 0.29$
$\frac{Mass of metal}{Mass of oxygen} = \frac{Eq.wt. of metal}{Eq.wt.of oxygen}$
Eq. wt. of metal $= \frac{0.5}{0.29} \times 8$
Eq.wt. of metal $= 13.79 ◻ 13.8$

Tripura JEE-2022

Some Basic Concepts of Chemistry

228225 What amount of conc. $H_{2} {SO}_{4}$ solution should be used to prepare $500 ml$ of $0.5 M H_{2} {SO}_{4}$ ? (The concentration of $H_{2} {SO}_{4}$ solution being used is $90 %$ and molecular mass of $H_{2} {SO}_{4} =$

1

22.06 g

2

24.52 g

3

11.03 g

4

27.24 g

Explanation:

Given that-
$\begin{aligned} V & = 500 ml \\ C & = 0.5 M \end{aligned}$
Concentration of $H_{2} {SO}_{4}$ solution being used is $90 %$
So, $C = \frac{0.5 \times 90}{100} = 0.45 M$
Molecular weight of $H_{2} {SO}_{4} = 98.079 g / mol$
$\begin{aligned} ∴ C = \frac{n (moles)}{V (volume)} \\ or C = \frac{weight}{m.weight} \times \frac{1}{V (Volume)} \\ Putting the value we get- \\ 0.45 = \frac{weight / amount}{98.079 \times 0.5} \\ Amount of H_{2} {SO}_{4} = 0.45 \times 0.5 \times 98.079 = 22.06 gm . \end{aligned}$

Shift-I

Some Basic Concepts of Chemistry

228227 3.7 gm of a gas at $25^{\circ} C$ occupies some volume. At $17^{\circ} C, 0.184$ gm hydrogen gas occupies same volume when pressures of both gases are same. What will be the molecular weight of the gas?

1 41.98

2 20.67

3 20.94

4 41.34

Explanation:

Given, Amount of gas $= 3.7 g$ Here, volume and pressure of both gases are same. As we know, $PV = nRT$ $P$ and $V$ are same for both gases.
So, $n_{1} T_{1} = n_{2} T_{2}$
$\begin{aligned} \frac{w_{1}}{M_{1}} T_{1} = \frac{W_{2}}{M_{2}} T_{2} \\ \frac{3.7}{M_{1}} \times 298 = \frac{0.184}{2} \times 290 \\ M_{1} = 41.33 g \end{aligned}$

Tripura JEE-2021

Some Basic Concepts of Chemistry

228229 Vapour density of a metal chloride is 83. If equivalent weight of the metal is 6 , its atomic weight will be ....... . 1

1 12

2 24

3 18

4 60

Explanation:

Valency of metal,
$\begin{aligned} (n) & = \frac{2 \times vapour density}{Equivalent weight of metal + 35.5} \\ = \frac{2 \times 83}{6 + 35.5} = 4 \end{aligned}$
The atomic weight of metal is equal to the product of the equivalent weight of metal and valency.
Atomic weight of metal $= n \times$ equivalent weight of metal
$\Rightarrow 4 \times 6 = 24$
Hence, the correct option (b).

Shift-I

Some Basic Concepts of Chemistry

228224 $0.79 gm$ of a metal oxide is obtained from 0.5 gm of the same metal upon oxidation. Equivalent weight of the metal will be which of the following?

1 10

2 13.8

3 20

4 40

Explanation:

Mass of metal + mass of oxygen $=$ mass of metal oxide
Mass of oxygen $= 0.79 - 0.5$
$= 0.29$
$\frac{Mass of metal}{Mass of oxygen} = \frac{Eq.wt. of metal}{Eq.wt.of oxygen}$
Eq. wt. of metal $= \frac{0.5}{0.29} \times 8$
Eq.wt. of metal $= 13.79 ◻ 13.8$

Tripura JEE-2022

Some Basic Concepts of Chemistry

228225 What amount of conc. $H_{2} {SO}_{4}$ solution should be used to prepare $500 ml$ of $0.5 M H_{2} {SO}_{4}$ ? (The concentration of $H_{2} {SO}_{4}$ solution being used is $90 %$ and molecular mass of $H_{2} {SO}_{4} =$

1

22.06 g

2

24.52 g

3

11.03 g

4

27.24 g

Explanation:

Given that-
$\begin{aligned} V & = 500 ml \\ C & = 0.5 M \end{aligned}$
Concentration of $H_{2} {SO}_{4}$ solution being used is $90 %$
So, $C = \frac{0.5 \times 90}{100} = 0.45 M$
Molecular weight of $H_{2} {SO}_{4} = 98.079 g / mol$
$\begin{aligned} ∴ C = \frac{n (moles)}{V (volume)} \\ or C = \frac{weight}{m.weight} \times \frac{1}{V (Volume)} \\ Putting the value we get- \\ 0.45 = \frac{weight / amount}{98.079 \times 0.5} \\ Amount of H_{2} {SO}_{4} = 0.45 \times 0.5 \times 98.079 = 22.06 gm . \end{aligned}$

Shift-I

Some Basic Concepts of Chemistry

228227 3.7 gm of a gas at $25^{\circ} C$ occupies some volume. At $17^{\circ} C, 0.184$ gm hydrogen gas occupies same volume when pressures of both gases are same. What will be the molecular weight of the gas?

1 41.98

2 20.67

3 20.94

4 41.34

Explanation:

Given, Amount of gas $= 3.7 g$ Here, volume and pressure of both gases are same. As we know, $PV = nRT$ $P$ and $V$ are same for both gases.
So, $n_{1} T_{1} = n_{2} T_{2}$
$\begin{aligned} \frac{w_{1}}{M_{1}} T_{1} = \frac{W_{2}}{M_{2}} T_{2} \\ \frac{3.7}{M_{1}} \times 298 = \frac{0.184}{2} \times 290 \\ M_{1} = 41.33 g \end{aligned}$

Tripura JEE-2021

Some Basic Concepts of Chemistry

228229 Vapour density of a metal chloride is 83. If equivalent weight of the metal is 6 , its atomic weight will be ....... . 1

1 12

2 24

3 18

4 60

Explanation:

Valency of metal,
$\begin{aligned} (n) & = \frac{2 \times vapour density}{Equivalent weight of metal + 35.5} \\ = \frac{2 \times 83}{6 + 35.5} = 4 \end{aligned}$
The atomic weight of metal is equal to the product of the equivalent weight of metal and valency.
Atomic weight of metal $= n \times$ equivalent weight of metal
$\Rightarrow 4 \times 6 = 24$
Hence, the correct option (b).

Shift-I

Some Basic Concepts of Chemistry

228224 $0.79 gm$ of a metal oxide is obtained from 0.5 gm of the same metal upon oxidation. Equivalent weight of the metal will be which of the following?

1 10

2 13.8

3 20

4 40

Explanation:

Mass of metal + mass of oxygen $=$ mass of metal oxide
Mass of oxygen $= 0.79 - 0.5$
$= 0.29$
$\frac{Mass of metal}{Mass of oxygen} = \frac{Eq.wt. of metal}{Eq.wt.of oxygen}$
Eq. wt. of metal $= \frac{0.5}{0.29} \times 8$
Eq.wt. of metal $= 13.79 ◻ 13.8$

Tripura JEE-2022

Some Basic Concepts of Chemistry

228225 What amount of conc. $H_{2} {SO}_{4}$ solution should be used to prepare $500 ml$ of $0.5 M H_{2} {SO}_{4}$ ? (The concentration of $H_{2} {SO}_{4}$ solution being used is $90 %$ and molecular mass of $H_{2} {SO}_{4} =$

1

22.06 g

2

24.52 g

3

11.03 g

4

27.24 g

Explanation:

Given that-
$\begin{aligned} V & = 500 ml \\ C & = 0.5 M \end{aligned}$
Concentration of $H_{2} {SO}_{4}$ solution being used is $90 %$
So, $C = \frac{0.5 \times 90}{100} = 0.45 M$
Molecular weight of $H_{2} {SO}_{4} = 98.079 g / mol$
$\begin{aligned} ∴ C = \frac{n (moles)}{V (volume)} \\ or C = \frac{weight}{m.weight} \times \frac{1}{V (Volume)} \\ Putting the value we get- \\ 0.45 = \frac{weight / amount}{98.079 \times 0.5} \\ Amount of H_{2} {SO}_{4} = 0.45 \times 0.5 \times 98.079 = 22.06 gm . \end{aligned}$

Shift-I

Some Basic Concepts of Chemistry

228227 3.7 gm of a gas at $25^{\circ} C$ occupies some volume. At $17^{\circ} C, 0.184$ gm hydrogen gas occupies same volume when pressures of both gases are same. What will be the molecular weight of the gas?

1 41.98

2 20.67

3 20.94

4 41.34

Explanation:

Given, Amount of gas $= 3.7 g$ Here, volume and pressure of both gases are same. As we know, $PV = nRT$ $P$ and $V$ are same for both gases.
So, $n_{1} T_{1} = n_{2} T_{2}$
$\begin{aligned} \frac{w_{1}}{M_{1}} T_{1} = \frac{W_{2}}{M_{2}} T_{2} \\ \frac{3.7}{M_{1}} \times 298 = \frac{0.184}{2} \times 290 \\ M_{1} = 41.33 g \end{aligned}$

Tripura JEE-2021

Some Basic Concepts of Chemistry

228229 Vapour density of a metal chloride is 83. If equivalent weight of the metal is 6 , its atomic weight will be ....... . 1

1 12

2 24

3 18

4 60

Explanation:

Valency of metal,
$\begin{aligned} (n) & = \frac{2 \times vapour density}{Equivalent weight of metal + 35.5} \\ = \frac{2 \times 83}{6 + 35.5} = 4 \end{aligned}$
The atomic weight of metal is equal to the product of the equivalent weight of metal and valency.
Atomic weight of metal $= n \times$ equivalent weight of metal
$\Rightarrow 4 \times 6 = 24$
Hence, the correct option (b).

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