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The strength of an aqueous NaOH solution is most accurately determined by titrating:
(Note: consider that an appropriate indicator is used)
Option: 1 Aq.NaOH in a pipette and aqueous oxalic acid in a burette
Option: 2 Aq.NaOH in a volumetric flask and concentrated H_{2}SO_{4} in a conical flask
Option: 3 Aq.NaOH in a burette and concentrated H_{2}SO_{4} in a conical flask
Option: 4 Aq.NaOH in a burette and aqueous oxalic acid in a conical flask

As we have learnt,

To calculate the strength of NaOH, it is titrated against oxalic acid. For this purpose, NaOH is kept in a burette and oxalic acid in a conical flask.

Therefore, Option(4) is correct.

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Posted by

vishal kumar

Reaction of an inorganic sulphite X with dilute H_{2}SO_{4} generates compound Y.Reaction of Y with \mathrm{NaOH} gives X. Further, the reaction of X with Y and water affords compound Z. Y and Z, respectively, are :
Option: 1 SO_{2}\; \; and \; \; Na_{2}SO_{3}
Option: 2 SO_{3}\; \; and \; \; NaHSO_{3}
Option: 3 SO_{2}\; \; and \; \; NaHSO_{3}
Option: 4 S\; \; and \; \; Na_{2}SO_{3}

The reaction will be-

 

So, Y and Z, respectively, are SO_{2}\; \; and \; \; NaHSO_{3}

Therefore, the correct option is (3).

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Posted by

Kuldeep Maurya

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If you spill a chemical toilet cleaning liquid on your hand, your first aid would be:
Option: 1 aqueous NH3 
Option: 2  gaseous NaHCO3
Option: 3 aqueous NaOH
Option: 4 Vinegar

Chemical toilet cleaning liquid contains acid and hence the first aid would be a weak base like NaHCO3.

Therefore, Option(2) is correct.

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Posted by

Kuldeep Maurya

The addition of dilute \mathrm{NaOH}$ to $\mathrm{Cr}^{3+} salt solution will give :
Option: 1 Precipitate of \mathrm{Cr}_{2} \mathrm{O}_{3} \left(\mathrm{H}_{2} \mathrm{O}\right)_{\mathrm{n}}
Option: 2 Precipitate of \mathrm{Cr}(\mathrm{OH})_{3}
Option: 3 Precipitate of \left[\mathrm{Cr}(\mathrm{OH})_{6}\right]^{3-}
Option: 4 A solution of \left[\mathrm{Cr}(\mathrm{OH})_{4}\right]^{-}

Addition of dilute \mathrm{NaOH} to a salt solution of \mathrm{Cr^{3+}} will lead to the formation of a precipitate of hydrated chromium (III) oxide.

\mathrm{Cr}^{3+}+\underset{\mathrm{dil}}{\mathrm{NaOH}} \longrightarrow \mathrm{Cr}_{2} \mathrm{O}_{3} \cdot \mathrm{nH}_{2} \mathrm{O}

Hence, the correct answer is option (1)

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sudhir.kumar

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Acidic ferric chloride solution on treatment with excess of potassium ferrocyanide gives a Prussian blue coloured colloidal species. It is :
Option: 1 \mathrm{HFe}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]
Option: 2 \mathrm{K}_{5} \mathrm{Fe}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]_{2}
Option: 3 \mathrm{Fe}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]_{3}
Option: 4 \mathrm{KFe}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]

\mathrm{Fe^{3+}} ions react with excess of potassium ferrocyanide to form \mathrm{KFe\left [ Fe\left ( CN \right )_{6} \right ]}.

The reaction is given below:

\mathrm{\underset{excess}{K_{4}\left [ Fe\left ( CN \right ) _{6}\right ]}+Fe^{3+}\rightarrow KFe\left [ Fe\left ( CN \right )_{6} \right ]}

Hence,the correct answer is Option (4)

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Posted by

sudhir.kumar

Match List - I with List - II :
List-I (Metal Ion) List-II (Group in Qualitative analysis)
(a) \mathrm{Mn}^{2+} (i) Group - III
(b) \mathrm{As}^{3+} (ii) Group - IIA
(c) \mathrm{Cu}^{2+} (iii) Group - IV
(d) \mathrm{Al}^{3+} (iv) Group - IIB
Choose the most appropriate answer from the options given below :
Option: 1 (a)-(iii), (b)-(iv), (c)-(ii), (d)-(i)
Option: 2 (a)-(iv), (b)-(ii), (c)-(iii), (d)-(i)
Option: 3 (a)-(i), (b)-(iv), (c)-(ii), (d)-(iii)
Option: 4 (a)-(i), (b)-(ii), (c)-(iii), (d)-(iv)

The correct match of basic radicalsand their groups in Qualitative analysis is given below

\mathrm{(a)Mn^{2+}:Group\; I V (iii)}

\mathrm{(b)As^{3+}:Group\; IIB(iv)}

\mathrm{(c)Cu^{2+}:Group\; II A(ii)}

\mathrm{(d)Al^{3+}: Group (III)(i )}

Hence the correct answer is option (1)

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sudhir.kumar

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Consider the sulphides \mathrm{HgS}$, $\mathrm{PbS}$, $\mathrm{CuS}_{}, \mathrm{Sb}_{2} \mathrm{~S}_{3},\mathrm{As}_{2} \mathrm{~S}_{3} and \mathrm{CdS}. Number of these sulphides soluble in 50 \% \mathrm{HNO}_{3} is_________.
 

The sulphide which are soluble in \mathrm{50% \ HNO_{3}} are \mathrm{PbS,\ CuS,\ As_{2}S_{3},\ CdS}

Sulphides insoluble in \mathrm{50%\ HNO_{3}} are \mathrm{HgS} and \mathrm{Sb_{2}S_{3}}

Hence, the correct answer is 4

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Posted by

sudhir.kumar

The potassium ferrocyanide solution gives a Prussian blue colour, when added to:
Option: 1 \mathrm{COCl}_{3}
Option: 2 \mathrm{CoCl_{2}}
Option: 3 \mathrm{FeCl_{2}}
Option: 4 \mathrm{FeCl_{3}}

Prussian Blue is formed when \mathrm{K_{4}\left [ Fe\left ( CN \right ) _{6}\right ]}  reacts with ferric ions

\mathrm{K_{4}\left [ Fe\left ( CN \right ) _{6}\right ]+Fe^{3+}\rightarrow \underset{Prussian\ Blue}{Fe_{4}\left [ Fe\left ( CN \right )_{6} \right ]}}

Hence, the correct answer is option (4)

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sudhir.kumar

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Cu^{2+} salt reacts with potassium iodide to give
Option: 1 \mathrm{Cu_{2}I_{2}}
Option: 2 \mathrm{Cu_{2} I_{3}}
Option: 3 \mathrm{C u I}
Option: 4 \mathrm{C u\left(I_{3}\right)_{2}}

Cu^{2+} reacts with I^{-} to form precipitate of Cu_{2}I_{2}.

Cu^{2+} is reduced to Cu^{+} while Iodide ions are oxidised to Iodine as shown below

\mathrm{Cu}^{2+}+\mathrm{I}^{-} \rightarrow \mathrm{Cu}_{2} \mathrm{I}_{2}+\mathrm{I}_{2}

The I_{2} complexes with I^{-} to give tri iodide ions \left(I_{3}^{-}\right) which forms a reddish solution.

Hence, the correct answer is option (1).

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Posted by

sudhir.kumar

An inorganic Compound 'X' on treatment with concentrated \text {H}_{2}\text{SO}_{4} produces brown fumes and gives dark brown ring with \text {FeSO}_{4} in presence of concentrated \text {H}_{2}\text{SO}_{4} . Also Compound 'X' gives precipitate 'Y', when its solution in dilute HCl is treated with \text {H}_{2}\text{S} gas. The precipitate 'Y' on treatment with concentrated \text {HNO}_{3} followed by excess of \text {NH}_{4}\text{OH} further gives deep blue coloured solution, Compound 'X' is:
Option: 1 \mathrm{Co}\left(\mathrm{NO}_{3}\right)_{2}
Option: 2 \mathrm{Pb}\left(\mathrm{NO}_{2}\right)_{2}
Option: 3 \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}
Option: 4 \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}

Acconding to the information given in the question

X\xrightarrow[\text{Conc.}]{\text{H}_{2}\text{SO}_{4}}\text{Brown fumes}

X\xrightarrow[\text{Ring}]{\text{Brown}}\text{(Positive)}

This confirms the presence of Nitrate ions (\text{NO}_{3}^{-}) as the acidic Radical.

X\xrightarrow[\text{dil.HCl}]{\text{H}_{2}\text{S}}\text{Y(ppt)}

X\xrightarrow[\text{Conc.}]{\text{HNO}_{3}}\xrightarrow[\text{excess}]{\text{NH}_{4}\text{OH}}\text{Blue Solution}

The precipitation with \text{H}_{2}\text{S} in presence of dil. HCl tells us that the Basic radical is of Group (2).

Futher reaction of the precipitate with \text{HNO}_{3} and excess \text{NH}_{4}\text{OH} giving a blue solution confirms the presence of \text{Cu}^{2+} ions

\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}: \text { Deep blue color }

Hence, the correct answer is option (3)

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Posted by

vishal kumar

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