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

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Solution

Moles of $A g N O_{3} = \frac{50 \times 16.9}{100 \times 169.8} = 0.05$ mole

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

$A g N O_{3} + N a C l ⟶ A g C l + N a N O_{3}$

Mass of $A g C l$ precipitate= Mole $\times$ Molar mass

$= 0.05 \times 143.5$

$= 7.16 g$

Hence, 7.16 g AgCl will be precipitated out.

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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)

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