Q3.99P - Chemistry: Molecular Nature Of Matter And Change

Question

Calculate each of the following quantities: (a) Molarity of the solution resulting from dissolving 46.0 g of silver nitrate in enough water to give a final volume of 335 mL (b) Volume in liters of 0.385 M manganese (II) sulfate that contains 63.0 g of solute (c) Volume in milliliters of 6.44×10^-2 M adenosine triphosphate (ATP) that contains 1.68 mmol of ATP.

Step-by-Step Solution

Answer

1. The molarity of the solution resulting from dissolving 46.0 g of silver nitrate in enough water to give a final volume of 335 mL is a
2. The volume in liters of 0.385 M manganese (II) sulfate that contains 63.0 g of solute is a 
3. The volume in milliliters of 6.44×10^-2 M adenosine triphosphate (ATP) that contains 1.68 m mol of ATP is 26.1 mL ATP solution 

Step 1: Finding the molarity

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

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

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

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

Step 2: Finding Volume of solution

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

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

Step 3: Finding the volume in milliliters

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

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