What is the mass of precipitate formed when 50 mL of 16.9% solution of $A g N O_{3}$ is mixed with 50 mL of 5.8% NaCl solution ?
(Ag = 107.8, N = 14, O = 16, Na = 23, Cl = 35.5)

18g

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28g

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3.5g

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Solution

The correct option is B 7g
Moles of $A g N O_{3} = \frac{50 \times 16.9}{100 \times 169.8} = 0.05$ moles
Moles of $N a C l = \frac{50 \times 5.8}{100 \times 58.5} = 0.05$ moles
$A g N o_{3} + N a C l \to A g C l + N a N O_{3}$
Mass of AgCl $=$ mole x molar mass
$= 0.05 \times 143.5 = 7.16$ g

Hence option C is correct.

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Similar questions

50

16.9

(w / v)

A g N O_{3}

50

5.8

(w / v)

N a C l

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

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17 % (w / V)

A g N O_{3}

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5.85 % (w / V)

N a C l

(A g = 108, N = 14, O = 16, N a = 23, C l = 35.5)

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