91Ó°ÊÓ

a.Multiply the mass of AgNO₃ by the molar mass reciprocal to find mole number.

$\begin{array}{rcl}Molesof AgN{O}_3& =& 46 g AgN{O}_3×\frac{mol AgN{O}_3}{169.87 g N{O}_3}\\                                      & =& 0.271 mol AgN{O}_3.\end{array}$

Now, divide the mole number calculated above by the given solution volume to find the molarity solution.

$\begin{array}{rcl}Molarity& =& \frac{0.271 mol AgN{O}_3}{335×{10}^{-3} L so\ln }\\                      & =& 0.809  M.\end{array}$

b. Multiply the given MnSO₄ mass by the molar mass reciprocal to find the mole number present in the solution.

$\begin{array}{rcl}Moles of MnS{O}_4& =& 63 g MnS{O}_4×\frac{mol MnS{O}_4}{151 g MnS{O}_4}\backslash \mathrm{hfill}\\                                         & =& 0.117 mol  MnS{O}_4.\\ Volume of the solution& =& \frac{0.417 mol MnS{O}_4}{0.385 \frac{mol MnS{O}_4}{L so\ln }}\\                                                      & =& 1.08  L.\end{array}$

c.The given mole number of ATP can be divided by the given molarity of the solution to find the solution’s volume.

$\begin{array}{rcl}Volume of the solution& =& \frac{1.68×{10}^{-3} mol ATP}{6.44×{10}^{-2} \frac{mol ATP}{L so\ln }}\\                                     & =& 0.0261  L\\                                    & =& 26.1  mL.\end{array}$