QBManager

Some Basic Concepts of Chemistry

228256 8.50g of ${NH}_{3}$ is present in $250 mL$ volume. Its active mass is

1

1.0 {ML}^{- 1}

2

0.5 {ML}^{- 1}

3

1.5 {ML}^{- 1}

4

2.0 {ML}^{- 1}

Explanation:

Acive mass $= \frac{Given mass of compound}{molecular mass of compound \times volume of solution}$
Active mass is defined as number of $g$ mol per litre. It is also known as molar concentration molarity.
$[\begin{array}{l} [{NH}_{3}] = \frac{8.50 g}{17 g / mol \times 250 mL} \times 1000 mL \\ [{NH}_{3}] = 2.0 mol / L \end{array}$

UPTU/UPSEE-2016

Some Basic Concepts of Chemistry

228257 Sulphur forms the chlorides $S_{2} {Cl}_{2}$ and ${SCl}_{2}$ . The equivalent mass of sulphur in ${SCl}_{2}$ is

1

8 g / mol

2

16 g / mol

3

64.8 g / mol

4

32 g / mol

Explanation:

Equivalent mass of sulphur
$\begin{aligned} \Rightarrow \frac{atomic mass of sulphur}{valency} \\ {SCl}_{2} = x + 2 (- 1) = 0 \\ x = 2 \\ \Rightarrow \frac{32}{2} = 16 \end{aligned}$

AIIMS-2015

Some Basic Concepts of Chemistry

228258 $3.011 \times 10^{22}$ atoms of an element weighs $1.15 g$ . The atomic mass of the element is :

1 23

2 10

3 16

4 35.5

Explanation:

From Avogadro's law :
$∵ 3.011 \times 10^{22}$ atoms contain an element weight 1.15 gm.
Atomic mass $\to 1$ mole of atoms $\to 6.022 \times 10^{23}$ atoms
$\begin{aligned} 1atom & = \frac{1.15}{3.011 \times 10^{22}} \\ 6.022 \times 10^{23} atoms & = \frac{1.15 \times 6.022 \times 10^{23}}{3.011 \times 10^{22}} = 23 \end{aligned}$
Thus, the atomic mass of the element is $= 23$ .

AP-EAMCET (Engg.)-2015

Some Basic Concepts of Chemistry

228269 A certain metal sulphide, ${MS}_{2}$ , is used extensively as a high temperature lubricant, If ${MS}_{2}$ has $40.96 %$ sulphur by weight, atomic mass of $M$ will be

1

100.0 amu

2

96.0 amu

3

60.0 amu

4

30.0 amu

Explanation:

Weight percentage of sulphur
$\begin{aligned} = \frac{Mass of sulphur}{Mass of Compound} \times 100 \\ \Rightarrow 40.96 = \frac{64}{M + 64} \times 100 \\ 40.96 (M + 64) = 64 \times 100 \\ 40.96 M + 64 \times 40.96 = 64 \times 100 \\ M = 96 amu \end{aligned}$
Where, $M =$ Atomic mass of metal

BCECE-2013

Some Basic Concepts of Chemistry

228256 8.50g of ${NH}_{3}$ is present in $250 mL$ volume. Its active mass is

1

1.0 {ML}^{- 1}

2

0.5 {ML}^{- 1}

3

1.5 {ML}^{- 1}

4

2.0 {ML}^{- 1}

Explanation:

Acive mass $= \frac{Given mass of compound}{molecular mass of compound \times volume of solution}$
Active mass is defined as number of $g$ mol per litre. It is also known as molar concentration molarity.
$[\begin{array}{l} [{NH}_{3}] = \frac{8.50 g}{17 g / mol \times 250 mL} \times 1000 mL \\ [{NH}_{3}] = 2.0 mol / L \end{array}$

UPTU/UPSEE-2016

Some Basic Concepts of Chemistry

228257 Sulphur forms the chlorides $S_{2} {Cl}_{2}$ and ${SCl}_{2}$ . The equivalent mass of sulphur in ${SCl}_{2}$ is

1

8 g / mol

2

16 g / mol

3

64.8 g / mol

4

32 g / mol

Explanation:

Equivalent mass of sulphur
$\begin{aligned} \Rightarrow \frac{atomic mass of sulphur}{valency} \\ {SCl}_{2} = x + 2 (- 1) = 0 \\ x = 2 \\ \Rightarrow \frac{32}{2} = 16 \end{aligned}$

AIIMS-2015

Some Basic Concepts of Chemistry

228258 $3.011 \times 10^{22}$ atoms of an element weighs $1.15 g$ . The atomic mass of the element is :

1 23

2 10

3 16

4 35.5

Explanation:

From Avogadro's law :
$∵ 3.011 \times 10^{22}$ atoms contain an element weight 1.15 gm.
Atomic mass $\to 1$ mole of atoms $\to 6.022 \times 10^{23}$ atoms
$\begin{aligned} 1atom & = \frac{1.15}{3.011 \times 10^{22}} \\ 6.022 \times 10^{23} atoms & = \frac{1.15 \times 6.022 \times 10^{23}}{3.011 \times 10^{22}} = 23 \end{aligned}$
Thus, the atomic mass of the element is $= 23$ .

AP-EAMCET (Engg.)-2015

Some Basic Concepts of Chemistry

228269 A certain metal sulphide, ${MS}_{2}$ , is used extensively as a high temperature lubricant, If ${MS}_{2}$ has $40.96 %$ sulphur by weight, atomic mass of $M$ will be

1

100.0 amu

2

96.0 amu

3

60.0 amu

4

30.0 amu

Explanation:

Weight percentage of sulphur
$\begin{aligned} = \frac{Mass of sulphur}{Mass of Compound} \times 100 \\ \Rightarrow 40.96 = \frac{64}{M + 64} \times 100 \\ 40.96 (M + 64) = 64 \times 100 \\ 40.96 M + 64 \times 40.96 = 64 \times 100 \\ M = 96 amu \end{aligned}$
Where, $M =$ Atomic mass of metal

BCECE-2013

Some Basic Concepts of Chemistry

228256 8.50g of ${NH}_{3}$ is present in $250 mL$ volume. Its active mass is

1

1.0 {ML}^{- 1}

2

0.5 {ML}^{- 1}

3

1.5 {ML}^{- 1}

4

2.0 {ML}^{- 1}

Explanation:

Acive mass $= \frac{Given mass of compound}{molecular mass of compound \times volume of solution}$
Active mass is defined as number of $g$ mol per litre. It is also known as molar concentration molarity.
$[\begin{array}{l} [{NH}_{3}] = \frac{8.50 g}{17 g / mol \times 250 mL} \times 1000 mL \\ [{NH}_{3}] = 2.0 mol / L \end{array}$

UPTU/UPSEE-2016

Some Basic Concepts of Chemistry

228257 Sulphur forms the chlorides $S_{2} {Cl}_{2}$ and ${SCl}_{2}$ . The equivalent mass of sulphur in ${SCl}_{2}$ is

1

8 g / mol

2

16 g / mol

3

64.8 g / mol

4

32 g / mol

Explanation:

Equivalent mass of sulphur
$\begin{aligned} \Rightarrow \frac{atomic mass of sulphur}{valency} \\ {SCl}_{2} = x + 2 (- 1) = 0 \\ x = 2 \\ \Rightarrow \frac{32}{2} = 16 \end{aligned}$

AIIMS-2015

Some Basic Concepts of Chemistry

228258 $3.011 \times 10^{22}$ atoms of an element weighs $1.15 g$ . The atomic mass of the element is :

1 23

2 10

3 16

4 35.5

Explanation:

From Avogadro's law :
$∵ 3.011 \times 10^{22}$ atoms contain an element weight 1.15 gm.
Atomic mass $\to 1$ mole of atoms $\to 6.022 \times 10^{23}$ atoms
$\begin{aligned} 1atom & = \frac{1.15}{3.011 \times 10^{22}} \\ 6.022 \times 10^{23} atoms & = \frac{1.15 \times 6.022 \times 10^{23}}{3.011 \times 10^{22}} = 23 \end{aligned}$
Thus, the atomic mass of the element is $= 23$ .

AP-EAMCET (Engg.)-2015

Some Basic Concepts of Chemistry

228269 A certain metal sulphide, ${MS}_{2}$ , is used extensively as a high temperature lubricant, If ${MS}_{2}$ has $40.96 %$ sulphur by weight, atomic mass of $M$ will be

1

100.0 amu

2

96.0 amu

3

60.0 amu

4

30.0 amu

Explanation:

Weight percentage of sulphur
$\begin{aligned} = \frac{Mass of sulphur}{Mass of Compound} \times 100 \\ \Rightarrow 40.96 = \frac{64}{M + 64} \times 100 \\ 40.96 (M + 64) = 64 \times 100 \\ 40.96 M + 64 \times 40.96 = 64 \times 100 \\ M = 96 amu \end{aligned}$
Where, $M =$ Atomic mass of metal

BCECE-2013

Some Basic Concepts of Chemistry

228256 8.50g of ${NH}_{3}$ is present in $250 mL$ volume. Its active mass is

1

1.0 {ML}^{- 1}

2

0.5 {ML}^{- 1}

3

1.5 {ML}^{- 1}

4

2.0 {ML}^{- 1}

Explanation:

Acive mass $= \frac{Given mass of compound}{molecular mass of compound \times volume of solution}$
Active mass is defined as number of $g$ mol per litre. It is also known as molar concentration molarity.
$[\begin{array}{l} [{NH}_{3}] = \frac{8.50 g}{17 g / mol \times 250 mL} \times 1000 mL \\ [{NH}_{3}] = 2.0 mol / L \end{array}$

UPTU/UPSEE-2016

Some Basic Concepts of Chemistry

228257 Sulphur forms the chlorides $S_{2} {Cl}_{2}$ and ${SCl}_{2}$ . The equivalent mass of sulphur in ${SCl}_{2}$ is

1

8 g / mol

2

16 g / mol

3

64.8 g / mol

4

32 g / mol

Explanation:

Equivalent mass of sulphur
$\begin{aligned} \Rightarrow \frac{atomic mass of sulphur}{valency} \\ {SCl}_{2} = x + 2 (- 1) = 0 \\ x = 2 \\ \Rightarrow \frac{32}{2} = 16 \end{aligned}$

AIIMS-2015

Some Basic Concepts of Chemistry

228258 $3.011 \times 10^{22}$ atoms of an element weighs $1.15 g$ . The atomic mass of the element is :

1 23

2 10

3 16

4 35.5

Explanation:

From Avogadro's law :
$∵ 3.011 \times 10^{22}$ atoms contain an element weight 1.15 gm.
Atomic mass $\to 1$ mole of atoms $\to 6.022 \times 10^{23}$ atoms
$\begin{aligned} 1atom & = \frac{1.15}{3.011 \times 10^{22}} \\ 6.022 \times 10^{23} atoms & = \frac{1.15 \times 6.022 \times 10^{23}}{3.011 \times 10^{22}} = 23 \end{aligned}$
Thus, the atomic mass of the element is $= 23$ .

AP-EAMCET (Engg.)-2015

Some Basic Concepts of Chemistry

228269 A certain metal sulphide, ${MS}_{2}$ , is used extensively as a high temperature lubricant, If ${MS}_{2}$ has $40.96 %$ sulphur by weight, atomic mass of $M$ will be

1

100.0 amu

2

96.0 amu

3

60.0 amu

4

30.0 amu

Explanation:

Weight percentage of sulphur
$\begin{aligned} = \frac{Mass of sulphur}{Mass of Compound} \times 100 \\ \Rightarrow 40.96 = \frac{64}{M + 64} \times 100 \\ 40.96 (M + 64) = 64 \times 100 \\ 40.96 M + 64 \times 40.96 = 64 \times 100 \\ M = 96 amu \end{aligned}$
Where, $M =$ Atomic mass of metal

BCECE-2013

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