Question

100 mL of Na\(_3\)PO\(_4\) solution contains 3.45 g of sodium. The molarity of the solution is __________ \(\times\)10\(^{-2}\) mol L\(^{-1}\). (Nearest integer)
[Atomic Masses - Na: 23.0 u, O: 16.0 u, P: 31.0 u]

Hint:

In problems involving stoichiometry of solutions, always use the mole concept. Find the moles of the substance for which data is given (here, sodium) and then use the stoichiometric ratio from the chemical formula to find the moles of the required substance (here, Na\(_3\)PO\(_4\)). Finally, apply the definition of concentration (molarity, molality, etc.).

Answer 1

Correct Answer: 50

To find the molarity of the Na\(_3\)PO\(_4\) solution, we first need to determine the number of moles of Na\(_3\)PO\(_4\) present. Here are the steps:

Step 1: Determine moles of Sodium (Na)

The atomic mass of Na is 23.0 u. Given that 3.45 g of sodium is present:

Moles of Na = \(\frac{\text{Mass}}{\text{Molar Mass}} = \frac{3.45\, \text{g}}{23.0\, \text{g/mol}}\)

Moles of Na = 0.15 mol

Step 2: Determine moles of Na\(_3\)PO\(_4\)

Each molecule of Na\(_3\)PO\(_4\) contains 3 moles of Na:

Moles of Na\(_3\)PO\(_4\) = \(\frac{\text{Moles of Na}}{3} = \frac{0.15}{3} = 0.05\) mol

Step 3: Calculate molarity of Na\(_3\)PO\(_4\)

Given that the volume of the solution is 100 mL (0.1 L):

Molarity (M) = \(\frac{\text{Moles of Na}_3\text{PO}_4}{\text{Volume in L}}\) = \(\frac{0.05 \, \text{mol}}{0.1\, \text{L}} = 0.5\) mol/L

To express this as \(\times 10^{-2}\) mol L\(^{-1}\):

Molarity = 0.5 × 10\(^{2-2}\) mol L\(^{-1}\) = 50 × 10\(^{-2}\) mol L\(^{-1}\)

Verification:

The final value, 50, fits within the expected range (50,50).