228220
Arrange the following in the order increasing mass (atomic mass $\mathrm{O}=16, \mathrm{Cu}=63, \mathrm{~N}=14$ ) I. One molecule of oxygen II. One atom of nitrogen III. $1 \times 10^{-10}$ gram molecule of oxygen IV. $1 \times 10^{-10}$ g of copper
1 II $<$ I $<$ IV $<$ III
2 I $<$ II $<$ III $<$ IV
3 III $<$ II $<$ IV $<$ I
4 IV $<$ II $<$ III $<$ I
5 II $<$ IV $<$ I $<$ III
Explanation:
Comparing the masses, we get correct order of increasing mass is (II) $<$ (I) $<$ (IV) $<$ (III) (I) 1 molecule of oxygen $=\mathrm{O}_2$ $\therefore$ Mass of $\mathrm{O}_2$ $=\frac{16 \times 2}{\mathrm{~N}_{\mathrm{A}}}=\frac{32 \mathrm{~g}}{\mathrm{~N}_{\mathrm{A}}}=\frac{32}{6.22 \times 10^{23}}=5.3 \times 10^{-23} \mathrm{~g}$ (II) Mass of 1 atom of Nitrogen $=1.66 \times 10^{-24} \times 14$ $=23.2 \times 10^{-24} \mathrm{~g}$ (III) $1 \times 10^{-24} \mathrm{gm}$ molecule of oxygen $=1 \times 10^{-10}$ moles of $\mathrm{O}_2$ Mass of $1 \times 10^{-10} \mathrm{gm}$ molecule of oxygen $1 \times 10^{-10} \times 32$ $=3.2 \times 10^{-9} \mathrm{~g}$ (IV) Mass of copper $=1 \times 10^{-10} \mathrm{~g}$ Comparing the masses in (I), (II), (III) and (IV) We get, (II),$<$ (I),$<$ (IV),$<$ (III) Therefore, answer is (II) $<$ (I) $<$ (IV) $<$ (III)
AIIMS-2016 Kerala-CEE-2011
Some Basic Concepts of Chemistry
228221
1.520 $g$ of hydroxide of a metal on ignition gave $0.995 \mathrm{~g}$ of oxide. The equivalent weight of metal is :
1 1.52
2 0.995
3 19.00
4 9.00
Explanation:
Let $\mathrm{E}$ be the equivalent weight of the metal So, $\frac{\mathrm{E}+17}{\mathrm{E}+8}=\frac{1.52}{0.995}$ [17 is equivalent weight of $\mathrm{OH}$ and 8 is equivalent weight of oxygen ] $\begin{aligned} & \Rightarrow 0.995 \mathrm{E}+17 \times 0.995=\mathrm{E} \times 1.52+8 \times 1.52 \\ & \Rightarrow 0.525 \mathrm{E}=16.915-12.16=4.755 \\ & \therefore \mathrm{E}=\frac{4.755}{0.525}=9 \end{aligned}$
BITSAT-2011 BCECE- 2008
Some Basic Concepts of Chemistry
228222
In acidic medium, the equivalent weight of $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ (Mol. wt. $=\mathrm{M}$ ) is
1 $\mathrm{M}$
2 $\frac{\mathrm{M}}{2}$
3 $\frac{M}{3}$
4 $\frac{\mathrm{M}}{6}$
Explanation:
An equivalent weight of a solution is defined as the molecular weight of the solute divided by the valence of the solute. Equivalent weight is used for predicting the mass of substance that react with one atom of hydrogen is acid -base analysis. Balanced chemical reaction of $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ in acidic medium will be- $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7+14 \mathrm{H}^{+}+6 \mathrm{e}^{-} \rightarrow 2 \mathrm{~K}^{+}+2 \mathrm{Cr}^{3+}+7 \mathrm{H}_2 \mathrm{O}$ In the above reaction, oxidation state of chromate ion is changing from +6 to +3 , i.e. the transfer of 6 electrons is taking place. $\therefore \text { Equivalent weight }=\frac{M}{6}$
WBJEE-2012 UPTU/UPSEE-2009
Some Basic Concepts of Chemistry
228223
Assertion : The normality of $0.3 \mathrm{M}$ aqueous solution of $\mathrm{H}_3 \mathrm{PO}_3$ is equal to $0.6 \mathrm{~N}$. Reason: Equivalent weight of $\mathrm{H}_3 \mathrm{PO}_3$ $=\frac{\text { Molecular weight of } \mathrm{H}_3 \mathrm{PO}_3}{3}$
1 If both Assertion and Reason are correct and the Reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.
3 If Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
Explanation:
$\mathrm{H}_3 \mathrm{PO}_3$ is a divalent. This is because it has two ionizable hydrogen atoms bonded to two oxygen atom and one non-ionizable hydrogen atom bonded directly to phosphorus. $\therefore$ Equivalent weight $=\frac{\text { Molecular weight }}{\text { Valency factor }}$ (Where, valency factor $=$ no. of replaceable $\mathrm{H}^{+}$ions) equivalent weight $=\frac{M}{q}$, since, no. of replaceable $\mathrm{H}^{+}$ ions in $\mathrm{H}_3 \mathrm{PO}_3=2$
228220
Arrange the following in the order increasing mass (atomic mass $\mathrm{O}=16, \mathrm{Cu}=63, \mathrm{~N}=14$ ) I. One molecule of oxygen II. One atom of nitrogen III. $1 \times 10^{-10}$ gram molecule of oxygen IV. $1 \times 10^{-10}$ g of copper
1 II $<$ I $<$ IV $<$ III
2 I $<$ II $<$ III $<$ IV
3 III $<$ II $<$ IV $<$ I
4 IV $<$ II $<$ III $<$ I
5 II $<$ IV $<$ I $<$ III
Explanation:
Comparing the masses, we get correct order of increasing mass is (II) $<$ (I) $<$ (IV) $<$ (III) (I) 1 molecule of oxygen $=\mathrm{O}_2$ $\therefore$ Mass of $\mathrm{O}_2$ $=\frac{16 \times 2}{\mathrm{~N}_{\mathrm{A}}}=\frac{32 \mathrm{~g}}{\mathrm{~N}_{\mathrm{A}}}=\frac{32}{6.22 \times 10^{23}}=5.3 \times 10^{-23} \mathrm{~g}$ (II) Mass of 1 atom of Nitrogen $=1.66 \times 10^{-24} \times 14$ $=23.2 \times 10^{-24} \mathrm{~g}$ (III) $1 \times 10^{-24} \mathrm{gm}$ molecule of oxygen $=1 \times 10^{-10}$ moles of $\mathrm{O}_2$ Mass of $1 \times 10^{-10} \mathrm{gm}$ molecule of oxygen $1 \times 10^{-10} \times 32$ $=3.2 \times 10^{-9} \mathrm{~g}$ (IV) Mass of copper $=1 \times 10^{-10} \mathrm{~g}$ Comparing the masses in (I), (II), (III) and (IV) We get, (II),$<$ (I),$<$ (IV),$<$ (III) Therefore, answer is (II) $<$ (I) $<$ (IV) $<$ (III)
AIIMS-2016 Kerala-CEE-2011
Some Basic Concepts of Chemistry
228221
1.520 $g$ of hydroxide of a metal on ignition gave $0.995 \mathrm{~g}$ of oxide. The equivalent weight of metal is :
1 1.52
2 0.995
3 19.00
4 9.00
Explanation:
Let $\mathrm{E}$ be the equivalent weight of the metal So, $\frac{\mathrm{E}+17}{\mathrm{E}+8}=\frac{1.52}{0.995}$ [17 is equivalent weight of $\mathrm{OH}$ and 8 is equivalent weight of oxygen ] $\begin{aligned} & \Rightarrow 0.995 \mathrm{E}+17 \times 0.995=\mathrm{E} \times 1.52+8 \times 1.52 \\ & \Rightarrow 0.525 \mathrm{E}=16.915-12.16=4.755 \\ & \therefore \mathrm{E}=\frac{4.755}{0.525}=9 \end{aligned}$
BITSAT-2011 BCECE- 2008
Some Basic Concepts of Chemistry
228222
In acidic medium, the equivalent weight of $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ (Mol. wt. $=\mathrm{M}$ ) is
1 $\mathrm{M}$
2 $\frac{\mathrm{M}}{2}$
3 $\frac{M}{3}$
4 $\frac{\mathrm{M}}{6}$
Explanation:
An equivalent weight of a solution is defined as the molecular weight of the solute divided by the valence of the solute. Equivalent weight is used for predicting the mass of substance that react with one atom of hydrogen is acid -base analysis. Balanced chemical reaction of $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ in acidic medium will be- $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7+14 \mathrm{H}^{+}+6 \mathrm{e}^{-} \rightarrow 2 \mathrm{~K}^{+}+2 \mathrm{Cr}^{3+}+7 \mathrm{H}_2 \mathrm{O}$ In the above reaction, oxidation state of chromate ion is changing from +6 to +3 , i.e. the transfer of 6 electrons is taking place. $\therefore \text { Equivalent weight }=\frac{M}{6}$
WBJEE-2012 UPTU/UPSEE-2009
Some Basic Concepts of Chemistry
228223
Assertion : The normality of $0.3 \mathrm{M}$ aqueous solution of $\mathrm{H}_3 \mathrm{PO}_3$ is equal to $0.6 \mathrm{~N}$. Reason: Equivalent weight of $\mathrm{H}_3 \mathrm{PO}_3$ $=\frac{\text { Molecular weight of } \mathrm{H}_3 \mathrm{PO}_3}{3}$
1 If both Assertion and Reason are correct and the Reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.
3 If Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
Explanation:
$\mathrm{H}_3 \mathrm{PO}_3$ is a divalent. This is because it has two ionizable hydrogen atoms bonded to two oxygen atom and one non-ionizable hydrogen atom bonded directly to phosphorus. $\therefore$ Equivalent weight $=\frac{\text { Molecular weight }}{\text { Valency factor }}$ (Where, valency factor $=$ no. of replaceable $\mathrm{H}^{+}$ions) equivalent weight $=\frac{M}{q}$, since, no. of replaceable $\mathrm{H}^{+}$ ions in $\mathrm{H}_3 \mathrm{PO}_3=2$
228220
Arrange the following in the order increasing mass (atomic mass $\mathrm{O}=16, \mathrm{Cu}=63, \mathrm{~N}=14$ ) I. One molecule of oxygen II. One atom of nitrogen III. $1 \times 10^{-10}$ gram molecule of oxygen IV. $1 \times 10^{-10}$ g of copper
1 II $<$ I $<$ IV $<$ III
2 I $<$ II $<$ III $<$ IV
3 III $<$ II $<$ IV $<$ I
4 IV $<$ II $<$ III $<$ I
5 II $<$ IV $<$ I $<$ III
Explanation:
Comparing the masses, we get correct order of increasing mass is (II) $<$ (I) $<$ (IV) $<$ (III) (I) 1 molecule of oxygen $=\mathrm{O}_2$ $\therefore$ Mass of $\mathrm{O}_2$ $=\frac{16 \times 2}{\mathrm{~N}_{\mathrm{A}}}=\frac{32 \mathrm{~g}}{\mathrm{~N}_{\mathrm{A}}}=\frac{32}{6.22 \times 10^{23}}=5.3 \times 10^{-23} \mathrm{~g}$ (II) Mass of 1 atom of Nitrogen $=1.66 \times 10^{-24} \times 14$ $=23.2 \times 10^{-24} \mathrm{~g}$ (III) $1 \times 10^{-24} \mathrm{gm}$ molecule of oxygen $=1 \times 10^{-10}$ moles of $\mathrm{O}_2$ Mass of $1 \times 10^{-10} \mathrm{gm}$ molecule of oxygen $1 \times 10^{-10} \times 32$ $=3.2 \times 10^{-9} \mathrm{~g}$ (IV) Mass of copper $=1 \times 10^{-10} \mathrm{~g}$ Comparing the masses in (I), (II), (III) and (IV) We get, (II),$<$ (I),$<$ (IV),$<$ (III) Therefore, answer is (II) $<$ (I) $<$ (IV) $<$ (III)
AIIMS-2016 Kerala-CEE-2011
Some Basic Concepts of Chemistry
228221
1.520 $g$ of hydroxide of a metal on ignition gave $0.995 \mathrm{~g}$ of oxide. The equivalent weight of metal is :
1 1.52
2 0.995
3 19.00
4 9.00
Explanation:
Let $\mathrm{E}$ be the equivalent weight of the metal So, $\frac{\mathrm{E}+17}{\mathrm{E}+8}=\frac{1.52}{0.995}$ [17 is equivalent weight of $\mathrm{OH}$ and 8 is equivalent weight of oxygen ] $\begin{aligned} & \Rightarrow 0.995 \mathrm{E}+17 \times 0.995=\mathrm{E} \times 1.52+8 \times 1.52 \\ & \Rightarrow 0.525 \mathrm{E}=16.915-12.16=4.755 \\ & \therefore \mathrm{E}=\frac{4.755}{0.525}=9 \end{aligned}$
BITSAT-2011 BCECE- 2008
Some Basic Concepts of Chemistry
228222
In acidic medium, the equivalent weight of $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ (Mol. wt. $=\mathrm{M}$ ) is
1 $\mathrm{M}$
2 $\frac{\mathrm{M}}{2}$
3 $\frac{M}{3}$
4 $\frac{\mathrm{M}}{6}$
Explanation:
An equivalent weight of a solution is defined as the molecular weight of the solute divided by the valence of the solute. Equivalent weight is used for predicting the mass of substance that react with one atom of hydrogen is acid -base analysis. Balanced chemical reaction of $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ in acidic medium will be- $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7+14 \mathrm{H}^{+}+6 \mathrm{e}^{-} \rightarrow 2 \mathrm{~K}^{+}+2 \mathrm{Cr}^{3+}+7 \mathrm{H}_2 \mathrm{O}$ In the above reaction, oxidation state of chromate ion is changing from +6 to +3 , i.e. the transfer of 6 electrons is taking place. $\therefore \text { Equivalent weight }=\frac{M}{6}$
WBJEE-2012 UPTU/UPSEE-2009
Some Basic Concepts of Chemistry
228223
Assertion : The normality of $0.3 \mathrm{M}$ aqueous solution of $\mathrm{H}_3 \mathrm{PO}_3$ is equal to $0.6 \mathrm{~N}$. Reason: Equivalent weight of $\mathrm{H}_3 \mathrm{PO}_3$ $=\frac{\text { Molecular weight of } \mathrm{H}_3 \mathrm{PO}_3}{3}$
1 If both Assertion and Reason are correct and the Reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.
3 If Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
Explanation:
$\mathrm{H}_3 \mathrm{PO}_3$ is a divalent. This is because it has two ionizable hydrogen atoms bonded to two oxygen atom and one non-ionizable hydrogen atom bonded directly to phosphorus. $\therefore$ Equivalent weight $=\frac{\text { Molecular weight }}{\text { Valency factor }}$ (Where, valency factor $=$ no. of replaceable $\mathrm{H}^{+}$ions) equivalent weight $=\frac{M}{q}$, since, no. of replaceable $\mathrm{H}^{+}$ ions in $\mathrm{H}_3 \mathrm{PO}_3=2$
228220
Arrange the following in the order increasing mass (atomic mass $\mathrm{O}=16, \mathrm{Cu}=63, \mathrm{~N}=14$ ) I. One molecule of oxygen II. One atom of nitrogen III. $1 \times 10^{-10}$ gram molecule of oxygen IV. $1 \times 10^{-10}$ g of copper
1 II $<$ I $<$ IV $<$ III
2 I $<$ II $<$ III $<$ IV
3 III $<$ II $<$ IV $<$ I
4 IV $<$ II $<$ III $<$ I
5 II $<$ IV $<$ I $<$ III
Explanation:
Comparing the masses, we get correct order of increasing mass is (II) $<$ (I) $<$ (IV) $<$ (III) (I) 1 molecule of oxygen $=\mathrm{O}_2$ $\therefore$ Mass of $\mathrm{O}_2$ $=\frac{16 \times 2}{\mathrm{~N}_{\mathrm{A}}}=\frac{32 \mathrm{~g}}{\mathrm{~N}_{\mathrm{A}}}=\frac{32}{6.22 \times 10^{23}}=5.3 \times 10^{-23} \mathrm{~g}$ (II) Mass of 1 atom of Nitrogen $=1.66 \times 10^{-24} \times 14$ $=23.2 \times 10^{-24} \mathrm{~g}$ (III) $1 \times 10^{-24} \mathrm{gm}$ molecule of oxygen $=1 \times 10^{-10}$ moles of $\mathrm{O}_2$ Mass of $1 \times 10^{-10} \mathrm{gm}$ molecule of oxygen $1 \times 10^{-10} \times 32$ $=3.2 \times 10^{-9} \mathrm{~g}$ (IV) Mass of copper $=1 \times 10^{-10} \mathrm{~g}$ Comparing the masses in (I), (II), (III) and (IV) We get, (II),$<$ (I),$<$ (IV),$<$ (III) Therefore, answer is (II) $<$ (I) $<$ (IV) $<$ (III)
AIIMS-2016 Kerala-CEE-2011
Some Basic Concepts of Chemistry
228221
1.520 $g$ of hydroxide of a metal on ignition gave $0.995 \mathrm{~g}$ of oxide. The equivalent weight of metal is :
1 1.52
2 0.995
3 19.00
4 9.00
Explanation:
Let $\mathrm{E}$ be the equivalent weight of the metal So, $\frac{\mathrm{E}+17}{\mathrm{E}+8}=\frac{1.52}{0.995}$ [17 is equivalent weight of $\mathrm{OH}$ and 8 is equivalent weight of oxygen ] $\begin{aligned} & \Rightarrow 0.995 \mathrm{E}+17 \times 0.995=\mathrm{E} \times 1.52+8 \times 1.52 \\ & \Rightarrow 0.525 \mathrm{E}=16.915-12.16=4.755 \\ & \therefore \mathrm{E}=\frac{4.755}{0.525}=9 \end{aligned}$
BITSAT-2011 BCECE- 2008
Some Basic Concepts of Chemistry
228222
In acidic medium, the equivalent weight of $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ (Mol. wt. $=\mathrm{M}$ ) is
1 $\mathrm{M}$
2 $\frac{\mathrm{M}}{2}$
3 $\frac{M}{3}$
4 $\frac{\mathrm{M}}{6}$
Explanation:
An equivalent weight of a solution is defined as the molecular weight of the solute divided by the valence of the solute. Equivalent weight is used for predicting the mass of substance that react with one atom of hydrogen is acid -base analysis. Balanced chemical reaction of $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ in acidic medium will be- $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7+14 \mathrm{H}^{+}+6 \mathrm{e}^{-} \rightarrow 2 \mathrm{~K}^{+}+2 \mathrm{Cr}^{3+}+7 \mathrm{H}_2 \mathrm{O}$ In the above reaction, oxidation state of chromate ion is changing from +6 to +3 , i.e. the transfer of 6 electrons is taking place. $\therefore \text { Equivalent weight }=\frac{M}{6}$
WBJEE-2012 UPTU/UPSEE-2009
Some Basic Concepts of Chemistry
228223
Assertion : The normality of $0.3 \mathrm{M}$ aqueous solution of $\mathrm{H}_3 \mathrm{PO}_3$ is equal to $0.6 \mathrm{~N}$. Reason: Equivalent weight of $\mathrm{H}_3 \mathrm{PO}_3$ $=\frac{\text { Molecular weight of } \mathrm{H}_3 \mathrm{PO}_3}{3}$
1 If both Assertion and Reason are correct and the Reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.
3 If Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
Explanation:
$\mathrm{H}_3 \mathrm{PO}_3$ is a divalent. This is because it has two ionizable hydrogen atoms bonded to two oxygen atom and one non-ionizable hydrogen atom bonded directly to phosphorus. $\therefore$ Equivalent weight $=\frac{\text { Molecular weight }}{\text { Valency factor }}$ (Where, valency factor $=$ no. of replaceable $\mathrm{H}^{+}$ions) equivalent weight $=\frac{M}{q}$, since, no. of replaceable $\mathrm{H}^{+}$ ions in $\mathrm{H}_3 \mathrm{PO}_3=2$
228220
Arrange the following in the order increasing mass (atomic mass $\mathrm{O}=16, \mathrm{Cu}=63, \mathrm{~N}=14$ ) I. One molecule of oxygen II. One atom of nitrogen III. $1 \times 10^{-10}$ gram molecule of oxygen IV. $1 \times 10^{-10}$ g of copper
1 II $<$ I $<$ IV $<$ III
2 I $<$ II $<$ III $<$ IV
3 III $<$ II $<$ IV $<$ I
4 IV $<$ II $<$ III $<$ I
5 II $<$ IV $<$ I $<$ III
Explanation:
Comparing the masses, we get correct order of increasing mass is (II) $<$ (I) $<$ (IV) $<$ (III) (I) 1 molecule of oxygen $=\mathrm{O}_2$ $\therefore$ Mass of $\mathrm{O}_2$ $=\frac{16 \times 2}{\mathrm{~N}_{\mathrm{A}}}=\frac{32 \mathrm{~g}}{\mathrm{~N}_{\mathrm{A}}}=\frac{32}{6.22 \times 10^{23}}=5.3 \times 10^{-23} \mathrm{~g}$ (II) Mass of 1 atom of Nitrogen $=1.66 \times 10^{-24} \times 14$ $=23.2 \times 10^{-24} \mathrm{~g}$ (III) $1 \times 10^{-24} \mathrm{gm}$ molecule of oxygen $=1 \times 10^{-10}$ moles of $\mathrm{O}_2$ Mass of $1 \times 10^{-10} \mathrm{gm}$ molecule of oxygen $1 \times 10^{-10} \times 32$ $=3.2 \times 10^{-9} \mathrm{~g}$ (IV) Mass of copper $=1 \times 10^{-10} \mathrm{~g}$ Comparing the masses in (I), (II), (III) and (IV) We get, (II),$<$ (I),$<$ (IV),$<$ (III) Therefore, answer is (II) $<$ (I) $<$ (IV) $<$ (III)
AIIMS-2016 Kerala-CEE-2011
Some Basic Concepts of Chemistry
228221
1.520 $g$ of hydroxide of a metal on ignition gave $0.995 \mathrm{~g}$ of oxide. The equivalent weight of metal is :
1 1.52
2 0.995
3 19.00
4 9.00
Explanation:
Let $\mathrm{E}$ be the equivalent weight of the metal So, $\frac{\mathrm{E}+17}{\mathrm{E}+8}=\frac{1.52}{0.995}$ [17 is equivalent weight of $\mathrm{OH}$ and 8 is equivalent weight of oxygen ] $\begin{aligned} & \Rightarrow 0.995 \mathrm{E}+17 \times 0.995=\mathrm{E} \times 1.52+8 \times 1.52 \\ & \Rightarrow 0.525 \mathrm{E}=16.915-12.16=4.755 \\ & \therefore \mathrm{E}=\frac{4.755}{0.525}=9 \end{aligned}$
BITSAT-2011 BCECE- 2008
Some Basic Concepts of Chemistry
228222
In acidic medium, the equivalent weight of $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ (Mol. wt. $=\mathrm{M}$ ) is
1 $\mathrm{M}$
2 $\frac{\mathrm{M}}{2}$
3 $\frac{M}{3}$
4 $\frac{\mathrm{M}}{6}$
Explanation:
An equivalent weight of a solution is defined as the molecular weight of the solute divided by the valence of the solute. Equivalent weight is used for predicting the mass of substance that react with one atom of hydrogen is acid -base analysis. Balanced chemical reaction of $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ in acidic medium will be- $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7+14 \mathrm{H}^{+}+6 \mathrm{e}^{-} \rightarrow 2 \mathrm{~K}^{+}+2 \mathrm{Cr}^{3+}+7 \mathrm{H}_2 \mathrm{O}$ In the above reaction, oxidation state of chromate ion is changing from +6 to +3 , i.e. the transfer of 6 electrons is taking place. $\therefore \text { Equivalent weight }=\frac{M}{6}$
WBJEE-2012 UPTU/UPSEE-2009
Some Basic Concepts of Chemistry
228223
Assertion : The normality of $0.3 \mathrm{M}$ aqueous solution of $\mathrm{H}_3 \mathrm{PO}_3$ is equal to $0.6 \mathrm{~N}$. Reason: Equivalent weight of $\mathrm{H}_3 \mathrm{PO}_3$ $=\frac{\text { Molecular weight of } \mathrm{H}_3 \mathrm{PO}_3}{3}$
1 If both Assertion and Reason are correct and the Reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.
3 If Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
Explanation:
$\mathrm{H}_3 \mathrm{PO}_3$ is a divalent. This is because it has two ionizable hydrogen atoms bonded to two oxygen atom and one non-ionizable hydrogen atom bonded directly to phosphorus. $\therefore$ Equivalent weight $=\frac{\text { Molecular weight }}{\text { Valency factor }}$ (Where, valency factor $=$ no. of replaceable $\mathrm{H}^{+}$ions) equivalent weight $=\frac{M}{q}$, since, no. of replaceable $\mathrm{H}^{+}$ ions in $\mathrm{H}_3 \mathrm{PO}_3=2$
