Filter By

All Questions

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 in a conical flask Option: 3 Aq.NaOH in a burette and concentrated 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.

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

The reaction will be-

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

Therefore, the correct option is (3).

Crack CUET with india's "Best Teachers"

• HD Video Lectures
• Unlimited Mock Tests
• Faculty Support

If you spill a chemical toilet cleaning liquid on your hand, your first aid would be:Option: 1 aqueous NH3Option: 2  gaseous NaHCO3Option: 3 aqueous NaOHOption: 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.

The addition of dilute salt solution will give :Option: 1 Precipitate of Option: 2 Precipitate of Option: 3 Precipitate of Option: 4 A solution of

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)

Crack NEET with "AI Coach"

• HD Video Lectures
• Unlimited Mock Tests
• Faculty Support

Acidic ferric chloride solution on treatment with excess of potassium ferrocyanide gives a Prussian blue coloured colloidal species. It is :Option: 1Option: 2Option: 3Option: 4

$\mathrm{Fe^{3+}}$ 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)

Match List - I with List - II : List-I (Metal Ion) List-II (Group in Qualitative analysis) (i) Group - III (ii) Group - IIA (iii) Group - IV (iv) Group - IIB Choose the most appropriate answer from the options given below :Option: 1Option: 2Option: 3Option: 4

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)

Crack JEE Main with "AI Coach"

• HD Video Lectures
• Unlimited Mock Tests
• Faculty Support

Consider the sulphides and . Number of these sulphides soluble in 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

The potassium ferrocyanide solution gives a Prussian blue colour, when added to:Option: 1Option: 2Option: 3Option: 4

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)

Crack CUET with india's "Best Teachers"

• HD Video Lectures
• Unlimited Mock Tests
• Faculty Support

salt reacts with potassium iodide to giveOption: 1Option: 2Option: 3Option: 4

$Cu^{2+}$ 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).

An inorganic Compound 'X' on treatment with concentrated  produces brown fumes and gives dark brown ring with  in presence of concentrated  . Also Compound 'X' gives precipitate 'Y', when its solution in dilute HCl is treated with  gas. The precipitate 'Y' on treatment with concentrated  followed by excess of further gives deep blue coloured solution, Compound 'X' is: Option: 1 Option: 2 Option: 3 Option: 4

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)