Three different samples of a salt solution give the following test results: One sample gives a white precipitate with ammonia solution. One sample gives a white precipitate with dilute $NaCl$ solution. One sample gives a black precipitate with $H_2S$. Identify the salt.
- A$AgNO_3$
- B$Pb(NO_3)_2$
- C$Hg(NO_3)_2$
- D$MnSO_4$
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Similar Questions
Match column-$I$ and column-$II$ with correct relation:
Column-$I$ Column-$II$ $(i)$ Blood like red colour $(a)$ $Fe_4[Fe(CN)_6]_3$ $(ii)$ Sodium nitroprusside $(b)$ $(NH_4)_2MoO_4$ $(iii)$ Ammonium molybdate $(c)$ $[Fe(SCN)]^{2+}$ $(iv)$ Ferri-ferro cyanide $(d)$ $Na_2[Fe(CN)_5NO]$
| Column-$I$ | Column-$II$ |
|---|---|
| $(i)$ Blood like red colour | $(a)$ $Fe_4[Fe(CN)_6]_3$ |
| $(ii)$ Sodium nitroprusside | $(b)$ $(NH_4)_2MoO_4$ |
| $(iii)$ Ammonium molybdate | $(c)$ $[Fe(SCN)]^{2+}$ |
| $(iv)$ Ferri-ferro cyanide | $(d)$ $Na_2[Fe(CN)_5NO]$ |
Easy
View SolutionWhen $MgSO_4$ is treated with $NH_4OH$ and $Na_2HPO_4$,a white crystalline solid is formed. What is its formula?
Medium
View SolutionWhich ion will not be precipitated by aqueous $(NH_4)_2S$?
Medium
View SolutionMatch List-$I$ with List-$II$:
List-$I$ (Ion) List-$II$ (Group Number in Cation Analysis) $A. Co^{2+}$ $I. Group-I$ $B. Mg^{2+}$ $II. Group-III$ $C. Pb^{2+}$ $III. Group-IV$ $D. Al^{3+}$ $IV. Group-VI$
Choose the correct answer from the options given below:
| List-$I$ (Ion) | List-$II$ (Group Number in Cation Analysis) |
| $A. Co^{2+}$ | $I. Group-I$ |
| $B. Mg^{2+}$ | $II. Group-III$ |
| $C. Pb^{2+}$ | $III. Group-IV$ |
| $D. Al^{3+}$ | $IV. Group-VI$ |
Choose the correct answer from the options given below:
MediumNEET 2025
View SolutionOn treating a solid salt $A$ with dil. $H_2SO_4$ and warming,reddish-brown fumes of gas $B$ are evolved. Addition of $C$ to the salt solution of $A$ followed by freshly prepared starch solution and acidification with acetic acid produces a blue colour. $C$ may contain:
Difficult
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