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Some Basic Concepts of Chemistry question

2015 · Q91
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Some Basic Concepts of Chemistry question

2015 · Q91

NEETChemistrySome Basic Concepts of ChemistryMCQ+4 / −1
What is the mass of the precipitate formed when 50 mL of 16.9% solution of AgNO3 is mixed with 50 mL of 5.8% NaCl solution ? (Ag = 107.8, N = 14, O = 16, Na = 23, Cl = 35.5
  1. A
    3.5 g
  2. B
    7 g
  3. C
    14 g
  4. D
    28 g
View written solutionFree

Correct answer: B

50 ml of 16.9% solution of AgNO3

(16.9100×50)\left( {{{16.9} \over {100}} \times 50} \right)(10016.9​×50) = 8.45 g of AgNO3

nmole = 8.45g(107.8+14+16×3)g/mol{{8.45g} \over {(107.8 + 14 + 16 \times 3)g/mol}}(107.8+14+16×3)g/mol8.45g​

= (8.45g169.8g/mol)=0.0497 moles\left( {{{8.45g} \over {169.8g/mol}}} \right) = 0.0497\,moles(169.8g/mol8.45g​)=0.0497moles

50ml of 5.8% solution of NaCl contain

NaCl = (5.8100×50)=2.9g\left( {{{5.8} \over {100}} \times 50} \right) = 2.9g(1005.8​×50)=2.9g

nNaCl = 2.9g(23+35.5)g/mol{{2.9g} \over {(23 + 35.5)g/mol}}(23+35.5)g/mol2.9g​

= 0.0495 moles

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AgNO3+NaCl$ \Rightarrow $AgCl+Na$ \oplus $+Cl$ \ominus $
1 mole1 mole1 mole
0.049 mole0.049 mole0.049 mole of AgCl

n = wM{w \over M}Mw​

⇒\Rightarrow⇒ w = (nAgCl) ×\times× Molecular mass

= (0.049) ×\times× (107.8 + 35.5) = 7.02 g

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