228235
Assertion: Molecular weight of a compound is 180 , if its vapour density is 90 . Reason: Molecular Weight $=\frac{\text { Vapour density }}{2}$
1 If both Assertion and Reason are correct and 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:
Relation between molecular weight and vapour density is, Molecular weight of compound $=2 \times$ vapour density $\begin{aligned} & =2 \times 90 \\ & =180 \end{aligned}$ Hence, assertion is correct but reason is incorrect.
AIIMS 25 May 2019 (Morning)
Some Basic Concepts of Chemistry
228236
In acid medium $\mathrm{MnO}_4^{-}$is reduced to $\mathrm{Mn}^{2+}$, by a reducing agent. Then the equivalent mass of $\mathrm{KMnO}_4$ is given by : (M = molecular mass)
1 $\mathrm{M} / 2$
2 $\mathrm{M}$
3 $\mathrm{M} / 5$
4 $\mathrm{M} / 3$
Explanation:
In acidic medium, $\mathrm{Mn}^{+7}$ goes to $\mathrm{Mn}^{+2}$ state and hence there is a net gain of 5 electrons. Now, equivalent weight $=\frac{\text { molar mass }}{\text { number of electrons gained or lost }}$ So, equivalent weight $=\frac{M}{5}$
Manipal-2019
Some Basic Concepts of Chemistry
228237
The equivalent weight of oxalic acid in $\mathrm{C}_2 \mathrm{H}_2 \mathrm{O}_4 \cdot 2 \mathrm{H}_2 \mathrm{O}$ is
228239
In the standardization of $\mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3$ using $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ by iodometry, the equivalent weight of $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ is :
1 Molecular weight $/ 2$
2 Molecular weight $/ 6$
3 Molecular weight $/ 3$
4 Same as molecular weight
Explanation:
The reaction between $\mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3$ and $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ is as shown below. $26 \mathrm{H}^{+}+3 \mathrm{~S}_2 \mathrm{O}_3^{2-}+4 \mathrm{Cr}_2 \mathrm{O}_7^{2-} \rightarrow 6 \mathrm{SO}_4^{2-}+8 \mathrm{Cr}^{3+}+13 \mathrm{H}_2 \mathrm{O}$ The oxidation state of chromium in $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ change from +6 to +3 . The net change in oxidation number per formula unit is 6. Hence, The equivalent weight of $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7=\frac{\text { Molecular weight }}{6}$
Manipal-2018
Some Basic Concepts of Chemistry
228240
A certain amount of a metal whose equivalent mass is 28 displaces $0.7 \mathrm{~L}$ of $\mathrm{H}_2$ at STP from an acid. Hence, mass of the element is :
1 $1.75 \mathrm{~g}$
2 $0875 \mathrm{~g}$
3 $3.50 \mathrm{~g}$
4 $7.00 \mathrm{~g}$
Explanation:
Moles of $\mathrm{H}_2=\frac{\text { given volume }}{\text { volume at STP }}$ Equivalent weight $=\frac{\text { molar mass of metal }}{\text { valency }}$ Gram equivalent metal $=$ gram equivalent of $\mathrm{H}_2$ From (i) \& (ii) - Weight of metal $\overline{\text { Equivalent wt. of metal }}=$ Moles of $\mathrm{H}_2 \times$ Valency factor $\begin{aligned} & \Rightarrow \frac{\text { Weight of metal }}{28}=\frac{\text { given volume }}{22.4} \times 2 \\ & \Rightarrow \frac{\text { Weight of metal }}{28}=\frac{0.7}{22.4} \times 2 \\ & \text { Weight }=1.75 \mathrm{gm} \end{aligned}$
228235
Assertion: Molecular weight of a compound is 180 , if its vapour density is 90 . Reason: Molecular Weight $=\frac{\text { Vapour density }}{2}$
1 If both Assertion and Reason are correct and 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:
Relation between molecular weight and vapour density is, Molecular weight of compound $=2 \times$ vapour density $\begin{aligned} & =2 \times 90 \\ & =180 \end{aligned}$ Hence, assertion is correct but reason is incorrect.
AIIMS 25 May 2019 (Morning)
Some Basic Concepts of Chemistry
228236
In acid medium $\mathrm{MnO}_4^{-}$is reduced to $\mathrm{Mn}^{2+}$, by a reducing agent. Then the equivalent mass of $\mathrm{KMnO}_4$ is given by : (M = molecular mass)
1 $\mathrm{M} / 2$
2 $\mathrm{M}$
3 $\mathrm{M} / 5$
4 $\mathrm{M} / 3$
Explanation:
In acidic medium, $\mathrm{Mn}^{+7}$ goes to $\mathrm{Mn}^{+2}$ state and hence there is a net gain of 5 electrons. Now, equivalent weight $=\frac{\text { molar mass }}{\text { number of electrons gained or lost }}$ So, equivalent weight $=\frac{M}{5}$
Manipal-2019
Some Basic Concepts of Chemistry
228237
The equivalent weight of oxalic acid in $\mathrm{C}_2 \mathrm{H}_2 \mathrm{O}_4 \cdot 2 \mathrm{H}_2 \mathrm{O}$ is
228239
In the standardization of $\mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3$ using $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ by iodometry, the equivalent weight of $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ is :
1 Molecular weight $/ 2$
2 Molecular weight $/ 6$
3 Molecular weight $/ 3$
4 Same as molecular weight
Explanation:
The reaction between $\mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3$ and $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ is as shown below. $26 \mathrm{H}^{+}+3 \mathrm{~S}_2 \mathrm{O}_3^{2-}+4 \mathrm{Cr}_2 \mathrm{O}_7^{2-} \rightarrow 6 \mathrm{SO}_4^{2-}+8 \mathrm{Cr}^{3+}+13 \mathrm{H}_2 \mathrm{O}$ The oxidation state of chromium in $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ change from +6 to +3 . The net change in oxidation number per formula unit is 6. Hence, The equivalent weight of $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7=\frac{\text { Molecular weight }}{6}$
Manipal-2018
Some Basic Concepts of Chemistry
228240
A certain amount of a metal whose equivalent mass is 28 displaces $0.7 \mathrm{~L}$ of $\mathrm{H}_2$ at STP from an acid. Hence, mass of the element is :
1 $1.75 \mathrm{~g}$
2 $0875 \mathrm{~g}$
3 $3.50 \mathrm{~g}$
4 $7.00 \mathrm{~g}$
Explanation:
Moles of $\mathrm{H}_2=\frac{\text { given volume }}{\text { volume at STP }}$ Equivalent weight $=\frac{\text { molar mass of metal }}{\text { valency }}$ Gram equivalent metal $=$ gram equivalent of $\mathrm{H}_2$ From (i) \& (ii) - Weight of metal $\overline{\text { Equivalent wt. of metal }}=$ Moles of $\mathrm{H}_2 \times$ Valency factor $\begin{aligned} & \Rightarrow \frac{\text { Weight of metal }}{28}=\frac{\text { given volume }}{22.4} \times 2 \\ & \Rightarrow \frac{\text { Weight of metal }}{28}=\frac{0.7}{22.4} \times 2 \\ & \text { Weight }=1.75 \mathrm{gm} \end{aligned}$
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Some Basic Concepts of Chemistry
228235
Assertion: Molecular weight of a compound is 180 , if its vapour density is 90 . Reason: Molecular Weight $=\frac{\text { Vapour density }}{2}$
1 If both Assertion and Reason are correct and 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:
Relation between molecular weight and vapour density is, Molecular weight of compound $=2 \times$ vapour density $\begin{aligned} & =2 \times 90 \\ & =180 \end{aligned}$ Hence, assertion is correct but reason is incorrect.
AIIMS 25 May 2019 (Morning)
Some Basic Concepts of Chemistry
228236
In acid medium $\mathrm{MnO}_4^{-}$is reduced to $\mathrm{Mn}^{2+}$, by a reducing agent. Then the equivalent mass of $\mathrm{KMnO}_4$ is given by : (M = molecular mass)
1 $\mathrm{M} / 2$
2 $\mathrm{M}$
3 $\mathrm{M} / 5$
4 $\mathrm{M} / 3$
Explanation:
In acidic medium, $\mathrm{Mn}^{+7}$ goes to $\mathrm{Mn}^{+2}$ state and hence there is a net gain of 5 electrons. Now, equivalent weight $=\frac{\text { molar mass }}{\text { number of electrons gained or lost }}$ So, equivalent weight $=\frac{M}{5}$
Manipal-2019
Some Basic Concepts of Chemistry
228237
The equivalent weight of oxalic acid in $\mathrm{C}_2 \mathrm{H}_2 \mathrm{O}_4 \cdot 2 \mathrm{H}_2 \mathrm{O}$ is
228239
In the standardization of $\mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3$ using $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ by iodometry, the equivalent weight of $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ is :
1 Molecular weight $/ 2$
2 Molecular weight $/ 6$
3 Molecular weight $/ 3$
4 Same as molecular weight
Explanation:
The reaction between $\mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3$ and $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ is as shown below. $26 \mathrm{H}^{+}+3 \mathrm{~S}_2 \mathrm{O}_3^{2-}+4 \mathrm{Cr}_2 \mathrm{O}_7^{2-} \rightarrow 6 \mathrm{SO}_4^{2-}+8 \mathrm{Cr}^{3+}+13 \mathrm{H}_2 \mathrm{O}$ The oxidation state of chromium in $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ change from +6 to +3 . The net change in oxidation number per formula unit is 6. Hence, The equivalent weight of $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7=\frac{\text { Molecular weight }}{6}$
Manipal-2018
Some Basic Concepts of Chemistry
228240
A certain amount of a metal whose equivalent mass is 28 displaces $0.7 \mathrm{~L}$ of $\mathrm{H}_2$ at STP from an acid. Hence, mass of the element is :
1 $1.75 \mathrm{~g}$
2 $0875 \mathrm{~g}$
3 $3.50 \mathrm{~g}$
4 $7.00 \mathrm{~g}$
Explanation:
Moles of $\mathrm{H}_2=\frac{\text { given volume }}{\text { volume at STP }}$ Equivalent weight $=\frac{\text { molar mass of metal }}{\text { valency }}$ Gram equivalent metal $=$ gram equivalent of $\mathrm{H}_2$ From (i) \& (ii) - Weight of metal $\overline{\text { Equivalent wt. of metal }}=$ Moles of $\mathrm{H}_2 \times$ Valency factor $\begin{aligned} & \Rightarrow \frac{\text { Weight of metal }}{28}=\frac{\text { given volume }}{22.4} \times 2 \\ & \Rightarrow \frac{\text { Weight of metal }}{28}=\frac{0.7}{22.4} \times 2 \\ & \text { Weight }=1.75 \mathrm{gm} \end{aligned}$
228235
Assertion: Molecular weight of a compound is 180 , if its vapour density is 90 . Reason: Molecular Weight $=\frac{\text { Vapour density }}{2}$
1 If both Assertion and Reason are correct and 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:
Relation between molecular weight and vapour density is, Molecular weight of compound $=2 \times$ vapour density $\begin{aligned} & =2 \times 90 \\ & =180 \end{aligned}$ Hence, assertion is correct but reason is incorrect.
AIIMS 25 May 2019 (Morning)
Some Basic Concepts of Chemistry
228236
In acid medium $\mathrm{MnO}_4^{-}$is reduced to $\mathrm{Mn}^{2+}$, by a reducing agent. Then the equivalent mass of $\mathrm{KMnO}_4$ is given by : (M = molecular mass)
1 $\mathrm{M} / 2$
2 $\mathrm{M}$
3 $\mathrm{M} / 5$
4 $\mathrm{M} / 3$
Explanation:
In acidic medium, $\mathrm{Mn}^{+7}$ goes to $\mathrm{Mn}^{+2}$ state and hence there is a net gain of 5 electrons. Now, equivalent weight $=\frac{\text { molar mass }}{\text { number of electrons gained or lost }}$ So, equivalent weight $=\frac{M}{5}$
Manipal-2019
Some Basic Concepts of Chemistry
228237
The equivalent weight of oxalic acid in $\mathrm{C}_2 \mathrm{H}_2 \mathrm{O}_4 \cdot 2 \mathrm{H}_2 \mathrm{O}$ is
228239
In the standardization of $\mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3$ using $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ by iodometry, the equivalent weight of $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ is :
1 Molecular weight $/ 2$
2 Molecular weight $/ 6$
3 Molecular weight $/ 3$
4 Same as molecular weight
Explanation:
The reaction between $\mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3$ and $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ is as shown below. $26 \mathrm{H}^{+}+3 \mathrm{~S}_2 \mathrm{O}_3^{2-}+4 \mathrm{Cr}_2 \mathrm{O}_7^{2-} \rightarrow 6 \mathrm{SO}_4^{2-}+8 \mathrm{Cr}^{3+}+13 \mathrm{H}_2 \mathrm{O}$ The oxidation state of chromium in $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ change from +6 to +3 . The net change in oxidation number per formula unit is 6. Hence, The equivalent weight of $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7=\frac{\text { Molecular weight }}{6}$
Manipal-2018
Some Basic Concepts of Chemistry
228240
A certain amount of a metal whose equivalent mass is 28 displaces $0.7 \mathrm{~L}$ of $\mathrm{H}_2$ at STP from an acid. Hence, mass of the element is :
1 $1.75 \mathrm{~g}$
2 $0875 \mathrm{~g}$
3 $3.50 \mathrm{~g}$
4 $7.00 \mathrm{~g}$
Explanation:
Moles of $\mathrm{H}_2=\frac{\text { given volume }}{\text { volume at STP }}$ Equivalent weight $=\frac{\text { molar mass of metal }}{\text { valency }}$ Gram equivalent metal $=$ gram equivalent of $\mathrm{H}_2$ From (i) \& (ii) - Weight of metal $\overline{\text { Equivalent wt. of metal }}=$ Moles of $\mathrm{H}_2 \times$ Valency factor $\begin{aligned} & \Rightarrow \frac{\text { Weight of metal }}{28}=\frac{\text { given volume }}{22.4} \times 2 \\ & \Rightarrow \frac{\text { Weight of metal }}{28}=\frac{0.7}{22.4} \times 2 \\ & \text { Weight }=1.75 \mathrm{gm} \end{aligned}$
228235
Assertion: Molecular weight of a compound is 180 , if its vapour density is 90 . Reason: Molecular Weight $=\frac{\text { Vapour density }}{2}$
1 If both Assertion and Reason are correct and 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:
Relation between molecular weight and vapour density is, Molecular weight of compound $=2 \times$ vapour density $\begin{aligned} & =2 \times 90 \\ & =180 \end{aligned}$ Hence, assertion is correct but reason is incorrect.
AIIMS 25 May 2019 (Morning)
Some Basic Concepts of Chemistry
228236
In acid medium $\mathrm{MnO}_4^{-}$is reduced to $\mathrm{Mn}^{2+}$, by a reducing agent. Then the equivalent mass of $\mathrm{KMnO}_4$ is given by : (M = molecular mass)
1 $\mathrm{M} / 2$
2 $\mathrm{M}$
3 $\mathrm{M} / 5$
4 $\mathrm{M} / 3$
Explanation:
In acidic medium, $\mathrm{Mn}^{+7}$ goes to $\mathrm{Mn}^{+2}$ state and hence there is a net gain of 5 electrons. Now, equivalent weight $=\frac{\text { molar mass }}{\text { number of electrons gained or lost }}$ So, equivalent weight $=\frac{M}{5}$
Manipal-2019
Some Basic Concepts of Chemistry
228237
The equivalent weight of oxalic acid in $\mathrm{C}_2 \mathrm{H}_2 \mathrm{O}_4 \cdot 2 \mathrm{H}_2 \mathrm{O}$ is
228239
In the standardization of $\mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3$ using $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ by iodometry, the equivalent weight of $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ is :
1 Molecular weight $/ 2$
2 Molecular weight $/ 6$
3 Molecular weight $/ 3$
4 Same as molecular weight
Explanation:
The reaction between $\mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3$ and $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ is as shown below. $26 \mathrm{H}^{+}+3 \mathrm{~S}_2 \mathrm{O}_3^{2-}+4 \mathrm{Cr}_2 \mathrm{O}_7^{2-} \rightarrow 6 \mathrm{SO}_4^{2-}+8 \mathrm{Cr}^{3+}+13 \mathrm{H}_2 \mathrm{O}$ The oxidation state of chromium in $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ change from +6 to +3 . The net change in oxidation number per formula unit is 6. Hence, The equivalent weight of $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7=\frac{\text { Molecular weight }}{6}$
Manipal-2018
Some Basic Concepts of Chemistry
228240
A certain amount of a metal whose equivalent mass is 28 displaces $0.7 \mathrm{~L}$ of $\mathrm{H}_2$ at STP from an acid. Hence, mass of the element is :
1 $1.75 \mathrm{~g}$
2 $0875 \mathrm{~g}$
3 $3.50 \mathrm{~g}$
4 $7.00 \mathrm{~g}$
Explanation:
Moles of $\mathrm{H}_2=\frac{\text { given volume }}{\text { volume at STP }}$ Equivalent weight $=\frac{\text { molar mass of metal }}{\text { valency }}$ Gram equivalent metal $=$ gram equivalent of $\mathrm{H}_2$ From (i) \& (ii) - Weight of metal $\overline{\text { Equivalent wt. of metal }}=$ Moles of $\mathrm{H}_2 \times$ Valency factor $\begin{aligned} & \Rightarrow \frac{\text { Weight of metal }}{28}=\frac{\text { given volume }}{22.4} \times 2 \\ & \Rightarrow \frac{\text { Weight of metal }}{28}=\frac{0.7}{22.4} \times 2 \\ & \text { Weight }=1.75 \mathrm{gm} \end{aligned}$