What is the mass of precipitate formed, when 50 mL of 16.9 % solution of AgNO3 is mixed with 50 mL of 5.87 % NaCl solution(Ag = 107.8, N = 14, O = l6, Na = 23, Cl = 35.5)

Moles of AgNO₃= $\frac{50 \times 16.9}{100 \times 169.8}$ = 0.05 moles

Moles of NaCl = $\frac{50 \times 5.8}{100 \times 58.5}$ = 0.05 moles

AgNO₃+ NaCl ⟶ AgCl + NaNO₃

Mass of AgCl precipitate= Mole × Molar mass

= 0.05 × 143.5

= 7.18 g

Hence, 7.18 g AgCl will be precipitated out.

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What is the mass of precipitate formed, when 50 mL of 16.9 % solution of AgNO₃ is mixed with 50 mL of 5.87 % NaCl solution
(Ag = 107.8, N = 14, O = l6, Na = 23, Cl = 35.5)