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

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

7

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14

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28

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3.5

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Solution

The correct option is B $7$
$A g N O_{3} + N a C l \to A g C l ↓ + N a N O_{3}$

Moles of $A g N O_{3} = \frac{\frac{16.9 \times 50}{100}}{170} = 0.05$ moles

Moles of $N a C l = \frac{\frac{5.8 \times 50}{100}}{58.5} = 0.05$ moles

Therefore, the moles of precipitate $(A g C l) = 0.05$ moles

Mass of precipitate formed $(A g C l) = 0.05 \times 143.5 = 7.165 g$

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Similar questions

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)

50 mL

17 % (w / V)

A g N O_{3}

50 mL

5.85 % (w / V)

N a C l

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

A g N O_{3}

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