Question

how many grams of nacl are required to precipitate most of the ag+ ions from 1.50×10² ml of 0.0287 m agno₃ solution? round your answer to 3 significant digits.

Explanation:

Step1: Calculate the moles of \(AgNO_3\)

The formula for moles \(n = C\times V\). Given \(C = 0.0287\space M\) and \(V=1.50\times10^{2}\space mL=0.150\space L\).
\(n(AgNO_3)=0.0287\space mol/L\times0.150\space L = 0.004305\space mol\)
Since \(AgNO_3\) dissociates as \(AgNO_3=Ag^{+}+NO_3^{-}\), the moles of \(Ag^{+}\) is \(n(Ag^{+}) = n(AgNO_3)=0.004305\space mol\)
The reaction between \(NaCl\) and \(AgNO_3\) is \(NaCl + AgNO_3=AgCl\downarrow+NaNO_3\). The mole - ratio of \(NaCl\) to \(AgNO_3\) is \(1:1\), so \(n(NaCl)=n(AgNO_3) = 0.004305\space mol\)

Step2: Calculate the mass of \(NaCl\)

The molar mass of \(NaCl\), \(M(NaCl)=(22.99 + 35.45)\space g/mol=58.44\space g/mol\)
The formula for mass \(m=n\times M\).
\(m(NaCl)=0.004305\space mol\times58.44\space g/mol\)
\(m(NaCl)=0.004305\times58.44 = 0.2516\space g\approx0.252\space g\)

Answer:

\(0.252\space g\)