228256
8.50g of $\mathrm{NH}_3$ is present in $250 \mathrm{~mL}$ volume. Its active mass is
1 $1.0 \mathrm{ML}^{-1}$
2 $0.5 \mathrm{ML}^{-1}$
3 $1.5 \mathrm{ML}^{-1}$
4 $2.0 \mathrm{ML}^{-1}$
Explanation:
Acive mass $=\frac{\text { Given mass of compound }}{\text { molecular mass of compound } \times \text { volume of solution }}$ Active mass is defined as number of $\mathrm{g}$ mol per litre. It is also known as molar concentration molarity. $\left[\begin{array}{l} {\left[\mathrm{NH}_3\right]=\frac{8.50 \mathrm{~g}}{17 \mathrm{~g} / \mathrm{mol} \times 250 \mathrm{~mL}} \times 1000 \mathrm{~mL}} \\ {\left[\mathrm{NH}_3\right]=2.0 \mathrm{~mol} / \mathrm{L}} \end{array}\right.$
UPTU/UPSEE-2016
Some Basic Concepts of Chemistry
228257
Sulphur forms the chlorides $\mathrm{S}_2 \mathrm{Cl}_2$ and $\mathrm{SCl}_2$. The equivalent mass of sulphur in $\mathrm{SCl}_2$ is
1 $8 \mathrm{~g} / \mathrm{mol}$
2 $16 \mathrm{~g} / \mathrm{mol}$
3 $64.8 \mathrm{~g} / \mathrm{mol}$
4 $32 \mathrm{~g} / \mathrm{mol}$
Explanation:
Equivalent mass of sulphur $\begin{aligned} & \Rightarrow \frac{\text { atomic mass of sulphur }}{\text { valency }} \\ & \mathrm{SCl}_2=\mathrm{x}+2(-1)=0 \\ & \mathrm{x}=2 \\ & \Rightarrow \frac{32}{2}=16 \end{aligned}$
AIIMS-2015
Some Basic Concepts of Chemistry
228258
$3.011 \times 10^{22}$ atoms of an element weighs $1.15 \mathrm{~g}$. The atomic mass of the element is :
1 23
2 10
3 16
4 35.5
Explanation:
From Avogadro's law : $\because 3.011 \times 10^{22}$ atoms contain an element weight 1.15 gm. Atomic mass $\rightarrow 1$ mole of atoms $\rightarrow 6.022 \times 10^{23}$ atoms $\begin{aligned} \text { 1atom } & =\frac{1.15}{3.011 \times 10^{22}} \\ 6.022 \times 10^{23} \text { atoms } & =\frac{1.15 \times 6.022 \times 10^{23}}{3.011 \times 10^{22}}=23 \end{aligned}$ Thus, the atomic mass of the element is $=23$.
AP-EAMCET (Engg.)-2015
Some Basic Concepts of Chemistry
228269
A certain metal sulphide, $\mathrm{MS}_2$, is used extensively as a high temperature lubricant, If $\mathrm{MS}_2$ has $40.96 \%$ sulphur by weight, atomic mass of $M$ will be
1 $100.0 \mathrm{amu}$
2 $96.0 \mathrm{amu}$
3 $60.0 \mathrm{amu}$
4 $30.0 \mathrm{amu}$
Explanation:
Weight percentage of sulphur $\begin{aligned} & =\frac{\text { Mass of sulphur }}{\text { Mass of Compound }} \times 100 \\ & \Rightarrow 40.96=\frac{64}{M+64} \times 100 \\ & 40.96(\mathrm{M}+64)=64 \times 100 \\ & 40.96 \mathrm{M}+64 \times 40.96=64 \times 100 \\ & M=96 \mathrm{amu} \end{aligned}$ Where, $M=$ Atomic mass of metal
228256
8.50g of $\mathrm{NH}_3$ is present in $250 \mathrm{~mL}$ volume. Its active mass is
1 $1.0 \mathrm{ML}^{-1}$
2 $0.5 \mathrm{ML}^{-1}$
3 $1.5 \mathrm{ML}^{-1}$
4 $2.0 \mathrm{ML}^{-1}$
Explanation:
Acive mass $=\frac{\text { Given mass of compound }}{\text { molecular mass of compound } \times \text { volume of solution }}$ Active mass is defined as number of $\mathrm{g}$ mol per litre. It is also known as molar concentration molarity. $\left[\begin{array}{l} {\left[\mathrm{NH}_3\right]=\frac{8.50 \mathrm{~g}}{17 \mathrm{~g} / \mathrm{mol} \times 250 \mathrm{~mL}} \times 1000 \mathrm{~mL}} \\ {\left[\mathrm{NH}_3\right]=2.0 \mathrm{~mol} / \mathrm{L}} \end{array}\right.$
UPTU/UPSEE-2016
Some Basic Concepts of Chemistry
228257
Sulphur forms the chlorides $\mathrm{S}_2 \mathrm{Cl}_2$ and $\mathrm{SCl}_2$. The equivalent mass of sulphur in $\mathrm{SCl}_2$ is
1 $8 \mathrm{~g} / \mathrm{mol}$
2 $16 \mathrm{~g} / \mathrm{mol}$
3 $64.8 \mathrm{~g} / \mathrm{mol}$
4 $32 \mathrm{~g} / \mathrm{mol}$
Explanation:
Equivalent mass of sulphur $\begin{aligned} & \Rightarrow \frac{\text { atomic mass of sulphur }}{\text { valency }} \\ & \mathrm{SCl}_2=\mathrm{x}+2(-1)=0 \\ & \mathrm{x}=2 \\ & \Rightarrow \frac{32}{2}=16 \end{aligned}$
AIIMS-2015
Some Basic Concepts of Chemistry
228258
$3.011 \times 10^{22}$ atoms of an element weighs $1.15 \mathrm{~g}$. The atomic mass of the element is :
1 23
2 10
3 16
4 35.5
Explanation:
From Avogadro's law : $\because 3.011 \times 10^{22}$ atoms contain an element weight 1.15 gm. Atomic mass $\rightarrow 1$ mole of atoms $\rightarrow 6.022 \times 10^{23}$ atoms $\begin{aligned} \text { 1atom } & =\frac{1.15}{3.011 \times 10^{22}} \\ 6.022 \times 10^{23} \text { atoms } & =\frac{1.15 \times 6.022 \times 10^{23}}{3.011 \times 10^{22}}=23 \end{aligned}$ Thus, the atomic mass of the element is $=23$.
AP-EAMCET (Engg.)-2015
Some Basic Concepts of Chemistry
228269
A certain metal sulphide, $\mathrm{MS}_2$, is used extensively as a high temperature lubricant, If $\mathrm{MS}_2$ has $40.96 \%$ sulphur by weight, atomic mass of $M$ will be
1 $100.0 \mathrm{amu}$
2 $96.0 \mathrm{amu}$
3 $60.0 \mathrm{amu}$
4 $30.0 \mathrm{amu}$
Explanation:
Weight percentage of sulphur $\begin{aligned} & =\frac{\text { Mass of sulphur }}{\text { Mass of Compound }} \times 100 \\ & \Rightarrow 40.96=\frac{64}{M+64} \times 100 \\ & 40.96(\mathrm{M}+64)=64 \times 100 \\ & 40.96 \mathrm{M}+64 \times 40.96=64 \times 100 \\ & M=96 \mathrm{amu} \end{aligned}$ Where, $M=$ Atomic mass of metal
228256
8.50g of $\mathrm{NH}_3$ is present in $250 \mathrm{~mL}$ volume. Its active mass is
1 $1.0 \mathrm{ML}^{-1}$
2 $0.5 \mathrm{ML}^{-1}$
3 $1.5 \mathrm{ML}^{-1}$
4 $2.0 \mathrm{ML}^{-1}$
Explanation:
Acive mass $=\frac{\text { Given mass of compound }}{\text { molecular mass of compound } \times \text { volume of solution }}$ Active mass is defined as number of $\mathrm{g}$ mol per litre. It is also known as molar concentration molarity. $\left[\begin{array}{l} {\left[\mathrm{NH}_3\right]=\frac{8.50 \mathrm{~g}}{17 \mathrm{~g} / \mathrm{mol} \times 250 \mathrm{~mL}} \times 1000 \mathrm{~mL}} \\ {\left[\mathrm{NH}_3\right]=2.0 \mathrm{~mol} / \mathrm{L}} \end{array}\right.$
UPTU/UPSEE-2016
Some Basic Concepts of Chemistry
228257
Sulphur forms the chlorides $\mathrm{S}_2 \mathrm{Cl}_2$ and $\mathrm{SCl}_2$. The equivalent mass of sulphur in $\mathrm{SCl}_2$ is
1 $8 \mathrm{~g} / \mathrm{mol}$
2 $16 \mathrm{~g} / \mathrm{mol}$
3 $64.8 \mathrm{~g} / \mathrm{mol}$
4 $32 \mathrm{~g} / \mathrm{mol}$
Explanation:
Equivalent mass of sulphur $\begin{aligned} & \Rightarrow \frac{\text { atomic mass of sulphur }}{\text { valency }} \\ & \mathrm{SCl}_2=\mathrm{x}+2(-1)=0 \\ & \mathrm{x}=2 \\ & \Rightarrow \frac{32}{2}=16 \end{aligned}$
AIIMS-2015
Some Basic Concepts of Chemistry
228258
$3.011 \times 10^{22}$ atoms of an element weighs $1.15 \mathrm{~g}$. The atomic mass of the element is :
1 23
2 10
3 16
4 35.5
Explanation:
From Avogadro's law : $\because 3.011 \times 10^{22}$ atoms contain an element weight 1.15 gm. Atomic mass $\rightarrow 1$ mole of atoms $\rightarrow 6.022 \times 10^{23}$ atoms $\begin{aligned} \text { 1atom } & =\frac{1.15}{3.011 \times 10^{22}} \\ 6.022 \times 10^{23} \text { atoms } & =\frac{1.15 \times 6.022 \times 10^{23}}{3.011 \times 10^{22}}=23 \end{aligned}$ Thus, the atomic mass of the element is $=23$.
AP-EAMCET (Engg.)-2015
Some Basic Concepts of Chemistry
228269
A certain metal sulphide, $\mathrm{MS}_2$, is used extensively as a high temperature lubricant, If $\mathrm{MS}_2$ has $40.96 \%$ sulphur by weight, atomic mass of $M$ will be
1 $100.0 \mathrm{amu}$
2 $96.0 \mathrm{amu}$
3 $60.0 \mathrm{amu}$
4 $30.0 \mathrm{amu}$
Explanation:
Weight percentage of sulphur $\begin{aligned} & =\frac{\text { Mass of sulphur }}{\text { Mass of Compound }} \times 100 \\ & \Rightarrow 40.96=\frac{64}{M+64} \times 100 \\ & 40.96(\mathrm{M}+64)=64 \times 100 \\ & 40.96 \mathrm{M}+64 \times 40.96=64 \times 100 \\ & M=96 \mathrm{amu} \end{aligned}$ Where, $M=$ Atomic mass of metal
228256
8.50g of $\mathrm{NH}_3$ is present in $250 \mathrm{~mL}$ volume. Its active mass is
1 $1.0 \mathrm{ML}^{-1}$
2 $0.5 \mathrm{ML}^{-1}$
3 $1.5 \mathrm{ML}^{-1}$
4 $2.0 \mathrm{ML}^{-1}$
Explanation:
Acive mass $=\frac{\text { Given mass of compound }}{\text { molecular mass of compound } \times \text { volume of solution }}$ Active mass is defined as number of $\mathrm{g}$ mol per litre. It is also known as molar concentration molarity. $\left[\begin{array}{l} {\left[\mathrm{NH}_3\right]=\frac{8.50 \mathrm{~g}}{17 \mathrm{~g} / \mathrm{mol} \times 250 \mathrm{~mL}} \times 1000 \mathrm{~mL}} \\ {\left[\mathrm{NH}_3\right]=2.0 \mathrm{~mol} / \mathrm{L}} \end{array}\right.$
UPTU/UPSEE-2016
Some Basic Concepts of Chemistry
228257
Sulphur forms the chlorides $\mathrm{S}_2 \mathrm{Cl}_2$ and $\mathrm{SCl}_2$. The equivalent mass of sulphur in $\mathrm{SCl}_2$ is
1 $8 \mathrm{~g} / \mathrm{mol}$
2 $16 \mathrm{~g} / \mathrm{mol}$
3 $64.8 \mathrm{~g} / \mathrm{mol}$
4 $32 \mathrm{~g} / \mathrm{mol}$
Explanation:
Equivalent mass of sulphur $\begin{aligned} & \Rightarrow \frac{\text { atomic mass of sulphur }}{\text { valency }} \\ & \mathrm{SCl}_2=\mathrm{x}+2(-1)=0 \\ & \mathrm{x}=2 \\ & \Rightarrow \frac{32}{2}=16 \end{aligned}$
AIIMS-2015
Some Basic Concepts of Chemistry
228258
$3.011 \times 10^{22}$ atoms of an element weighs $1.15 \mathrm{~g}$. The atomic mass of the element is :
1 23
2 10
3 16
4 35.5
Explanation:
From Avogadro's law : $\because 3.011 \times 10^{22}$ atoms contain an element weight 1.15 gm. Atomic mass $\rightarrow 1$ mole of atoms $\rightarrow 6.022 \times 10^{23}$ atoms $\begin{aligned} \text { 1atom } & =\frac{1.15}{3.011 \times 10^{22}} \\ 6.022 \times 10^{23} \text { atoms } & =\frac{1.15 \times 6.022 \times 10^{23}}{3.011 \times 10^{22}}=23 \end{aligned}$ Thus, the atomic mass of the element is $=23$.
AP-EAMCET (Engg.)-2015
Some Basic Concepts of Chemistry
228269
A certain metal sulphide, $\mathrm{MS}_2$, is used extensively as a high temperature lubricant, If $\mathrm{MS}_2$ has $40.96 \%$ sulphur by weight, atomic mass of $M$ will be
1 $100.0 \mathrm{amu}$
2 $96.0 \mathrm{amu}$
3 $60.0 \mathrm{amu}$
4 $30.0 \mathrm{amu}$
Explanation:
Weight percentage of sulphur $\begin{aligned} & =\frac{\text { Mass of sulphur }}{\text { Mass of Compound }} \times 100 \\ & \Rightarrow 40.96=\frac{64}{M+64} \times 100 \\ & 40.96(\mathrm{M}+64)=64 \times 100 \\ & 40.96 \mathrm{M}+64 \times 40.96=64 \times 100 \\ & M=96 \mathrm{amu} \end{aligned}$ Where, $M=$ Atomic mass of metal
