What is the mass of precipitate formed when 50 mL of 16.9% (w/v) solution of $A g N O_{3}$ is mixed with 50 mL of 5.8% (w/v) NaCl solution?

28 g

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3.5 g

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7 g

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14 g

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Solution

The correct option is C 7 g
16.9% solution of $A g N O_{3}$ means 16.9 g $A g N O_{3}$ is present in 100 mL of solution
$∴$ 8.45 g $A g N O_{3}$ – 50 mL of solution
$A g N O_{3} + N a C l \to A g C l ↓ + N a N O_{3}$
Moles of $A g N O_{3}$ = Moles of NaCl = Moles of AgCl $= 0.049 (= \frac{8.45}{169.8})$
Mass of AgCl precipitated = 0.049 $\times$ 143.5 = 7g

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Similar questions

What is the mass of the precipitate formed when $50 m L$ of $16.9 %$ solution of ${AgNO}_{3}$ is mixed with $50 m L$ of $5.8 %$ w/v $NaCl$ solution?

$[A g = 107.8, N = 14, O = 16, N a = 23, C l = 35.5]$

50 mL

16.9 %

A g N O_{3}

50 mL

5.8 %

N a C l

(A g = 107.8, N = 14, O = 16, N a = 23, C l = 35.5)

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