How many milliliters of \(0.750 \quad M \mathrm{H}_{3} \mathrm{PO}_{4}\) will contain the following? (a) \(0.15 \mathrm{~mol} \mathrm{H}_{3} \mathrm{PO}_{4}\) (b) \(35.5 \mathrm{~g} \mathrm{H}_{3} \mathrm{PO}_{4}\) (c) \(7.34 \times 10^{22}\) molecules of \(\mathrm{H}_{3} \mathrm{PO}_{4}\)

Molarity is an essential concept in chemistry, defining the concentration of a solution. It represents the number of moles of solute (the substance dissolved) per liter of solution. This is widely expressed using the symbol \( M \). The formula to calculate molarity is straightforward: \[ M = \frac{n}{V} \]- \( M \) = Molarity (moles per liter)- \( n \) = Number of moles of solute- \( V \) = Volume of the solution in litersUnderstanding molarity is critical when performing volume and concentration calculations for various chemical reactions and solutions. To calculate the volume required to obtain a certain amount of solute, you can rearrange the formula as follows: \[ V = \frac{n}{M} \]This rearranged formula is handy when you need to dilute or concentrate solutions to achieve the desired molarity.

Molar mass is the mass of one mole of a given substance and is usually expressed in grams per mole (g/mol). It is a crucial parameter because it helps in converting between the mass of a substance and the number of moles.To find the molar mass of a compound, sum the atomic masses of all the atoms in its molecular formula. Let's take phosphoric acid (\( \mathrm{H}_{3} \mathrm{PO}_{4} \)) as an example:

• Hydrogen (H): 1 g/mol
• Phosphorus (P): 31 g/mol
• Oxygen (O): 16 g/mol

Now, calculate the molar mass of \( \mathrm{H}_{3} \mathrm{PO}_{4} \): \[ M_{\mathrm{H}_{3} \mathrm{PO}_{4}} = 3(1) + 31 + 4(16) = 98 \ g/mol \]This value means that one mole of \( \mathrm{H}_{3} \mathrm{PO}_{4} \) weighs 98 grams. Molar mass is foundational for converting between the mass of a substance and the moles of that substance in stoichiometric calculations.

Avogadro's number, \( 6.022 \times 10^{23} \), is a fundamental constant in chemistry, representing the number of representative particles (atoms, molecules, ions, etc.) in one mole of a substance. This number bridges the macroscopic and atomic worlds, allowing chemists to count atoms by weighing them.To convert between molecules and moles, you use Avogadro's number. For example, to find the number of moles in \( 7.34 \times 10^{22} \) molecules of \( \mathrm{H}_{3} \mathrm{PO}_{4} \):\[ n = \frac{7.34 \times 10^{22} \ \mathrm{molecules}}{6.022 \times 10^{23} \ \mathrm{molecules/mol}} = 0.1219 \ mol \]By understanding Avogadro's number, you can easily perform conversions between the microscopic number of particles and macroscopic quantities of matter in mole-related calculations.

Converting between moles and volume is a common task in chemistry. Knowing the volume of a solution required to get a specific number of moles of solute is essential for preparing solutions of desired concentrations.This conversion relies on the molarity formula rearranged as: \[ V = \frac{n}{M} \]Where:- \( V \) = Volume of the solution in liters- \( n \) = Number of moles of solute- \( M \) = Molarity of the solutionFor instance, to prepare a solution containing 0.362 moles of \( \mathrm{H}_{3} \mathrm{PO}_{4} \) at a molarity of 0.750 M, the volume needed would be: \[ V = \frac{0.362 \ mol}{0.750 \ M} = 0.48267 \ L \approx 483 \ mL \]This conversion technique is indispensable in laboratory settings and industrial applications where precise control of solution concentrations is necessary.