Question

1. Which acid is used as ink-stain remover?

(a) Oxalic acid (b) Tartaric acid

(c) Acetic acid (d) Boric acid



2. 2.12g of an impure mixture containing anhydrous sodium sulphate dissolved in water. An excess of barium chloride solution is added and 1.74 gram of barium sulphate is obtained as a dry precipitate.

Calculate the percentage purity of the impure sample.

(a) 70% (b) 36.7% (c) 50% (d) 82%



3. A compound contains carbon 14.4 %, hydrogen 1.2 % and chlorine 84.5 %. Work correct to 1 decimal place. The relative molecular mass of the compound is 168. Find its molecular formula.

(a) CHCI2 (b) C2H2CI2 (c) C2H2CI4 (d) CHCI4​

Answer 1

Acid is used as ink-stain remover?, Percentage purity of the impure sample. The relative molecular mass of the compound is 168. ,Find its molecular formula

Explanation:

1. Oxalic acid is the right answer.

2. 70%

3. C2H2CL14

• Oxalic acid is the acid that is used in laundry factories. to rinse the color. It is effective on stain and removing the rust. It converts the insoluble compounds into soluble compounds.
• Na2SO4+BaCL2 = 2Nacl +BaSO4(S)
• 140G = 233G
• x ?  =  1.74g
• Now, there is x/142  1.74g/233g
• x = 1.74*142/233
• = 1.06g
• therefore purity is  = 1.06/1.5*100
• = 70.7%

• Molecular formula is  = (CHCL2) * 2 = C2H2CL4

Learn More: Oxalic acid, sulphate

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Answer 2

Oxalic acid is widely used as an acid rinse in laundries, where it is effective in removing rust and ink stains because it converts most insoluble iron compounds into a soluble complex ion.

% age of purity = 70.67 %

Explanation:

equation

Na2SO4 +  BaCl2 → 2NaCl + BaSO4

here it's clear that

1 mol of Na2SO4 = 1 mol of BaSO4

i.e

142g of Na2SO4 produces 233 g of BaSO4

so , x g of Na2SO4 produces 1.74 g BaSO4

But given mass of Na2SO4 = 1.5 g

x = 1.74 X 142/233 = 1.06g

∴ impurity present = 1.5 - 1.06

                              = 0.44 g

∴ Purity % age = 1.06/1.5 X 100 = 70.67 %

Element    % composition     Atomic weight        Relative No. of atoms         Simplest ratio

Carbon                14.4            12           14.4/12 = 1.2          1.2/1.2 = 1

Hydrogen         1.2          1           1.2/1 = 1.2          1.2/1.2 = 1

Chlorine         84.5       35.5           84.5/35.5 = 2.38          2.38/1.2 = 2

Hence, empirical formula is CHCl  

2

​  

.  

n = Molecular mass/Empirical formula mass  

= 168/(12 + 1 + 70)  

= 168/83  

= 2.  

Therefore, molecular formula =(CHCl  

2

​  

)×2=C  

2

​  

H  

2

​  

Cl  

4

​  

.