A colourless (A) when place into water a heavy white turbidly of (B) solid (A) gives a close solution in conesolution in cone `HCI` when `HCI` solution is added to clear solution water ,(B) forms again (B) dissolves in dilute `HCI`. When `H_(2)S` is passed through a sespension of (A) or (B), a black precipitate (C ) forms , (C ) is insolves in yellow ammonium sulphide `(NH_(4))_(2)S` , cone `H_(2)SO_(4)` added to solid (A) liberates gas (D) gas (D) is water soluble and gives white precipitate with mercuric salts (E ) and not mercuric salt .The black precipitate (C ) dissolves in `HNO_(3), (1,1)` to give a solution to which `H_(2)SO_(4)` is added followed by addition of `NH_(4)OH` when a white precipitate (F) is formed (E ) gives a black ppt , (G) with solution of sodium stannite.
Compound (B ) is not soluble in

Options

A. Tartaric ric acid
B. HCI
C. `HNO_(3)`
D. `H_(2)SO_(4)`

Explanation

To solve the problem step by step, let's analyze the information given in the question and the video transcript.

### Step 1: Identify Compound (A)
- **Given Information**: A colorless compound (A) produces a heavy white turbidity (B) when placed in water.
- **Conclusion**: Compound (A) is likely to be Bismuth chloride (BiCl3), which forms a white precipitate of Bismuth oxychloride (BiOCl) when it reacts with water.

### Step 2: Identify Compound (B)
- **Given Information**: The heavy white turbidity (B) is formed when (A) is added to water. This compound (B) dissolves in concentrated HCl.
- **Conclusion**: Compound (B) is Bismuth oxychloride (BiOCl), which is known to dissolve in concentrated HCl to regenerate Bismuth chloride (BiCl3).

### Step 3: Behavior of Compound (B) with H2S
- **Given Information**: When H2S is passed through a suspension of (A) or (B), a black precipitate (C) forms.
- **Conclusion**: The black precipitate (C) is likely Bismuth sulfide (Bi2S3), which is formed when H2S reacts with bismuth compounds.

### Step 4: Solubility of Compound (C)
- **Given Information**: The black precipitate (C) is insoluble in yellow ammonium sulfide ((NH4)2S).
- **Conclusion**: This confirms that (C) is indeed Bismuth sulfide (Bi2S3), as it is known to be insoluble in ammonium sulfide.

### Step 5: Reaction of Compound (A) with H2SO4
- **Given Information**: Concentrated H2SO4 added to solid (A) liberates gas (D), which is water-soluble and gives a white precipitate with mercuric salts (E).
- **Conclusion**: The gas (D) is likely sulfur dioxide (SO2), which is produced when Bismuth chloride reacts with sulfuric acid. The white precipitate (E) is likely mercurous chloride (Hg2Cl2).

### Step 6: Formation of Precipitate (F)
- **Given Information**: The black precipitate (C) dissolves in dilute HNO3 to give a solution. When H2SO4 is added, followed by NH4OH, a white precipitate (F) is formed.
- **Conclusion**: The white precipitate (F) is likely Bismuth hydroxide (Bi(OH)3), which forms when bismuth ions react with ammonium hydroxide.

### Step 7: Reaction with Sodium Stannite
- **Given Information**: The compound (E) gives a black precipitate (G) with sodium stannite.
- **Conclusion**: This indicates that the mercuric salt (E) reacts with sodium stannite to form a black precipitate, likely mercuric stannate.

### Step 8: Identify the Solubility of Compound (B)
- **Given Information**: The question asks which solution compound (B) is not soluble in.
- **Conclusion**: Based on the reactions and solubility discussed, compound (B) (BiOCl) is not soluble in tartaric acid.

### Final Answer
- **Compound (B) is not soluble in tartaric acid.**