Molarity and Solution Calculations in GOB Chemistry

Calculating Molarity

Definition and Formula

Molarity (M) is a measure of concentration, representing the number of moles of solute per liter of solution. It is a fundamental concept in solution chemistry, used to quantify how much solute is present in a given volume of solvent.

• Key Term: Molarity (M) – the concentration of a solution expressed as moles of solute per liter of solution.

• Formula:

$ext{Molarity} = \frac{\text{moles solute}}{\text{liters solution}}$

Example Calculation

To calculate the molarity of a solution prepared by dissolving a known mass of solute in a specific volume of solvent:

• Convert mass of solute to moles using molar mass.

• Divide moles of solute by volume of solution in liters.

Example: Calculate the molarity of a solution prepared by dissolving 2.37 g NaOH in enough water to make 2.50 L of solution.

• Moles of NaOH: $\frac{2.37\,\text{g}}{39.998\,\text{g/mol}} = 0.05925\,\text{mol}$

• Molarity: $\frac{0.05925\,\text{mol}}{2.50\,\text{L}} = 0.0237\,\text{M}$

Additional info: The molar mass of NaOH is calculated as Na (22.99 g/mol) + O (16.00 g/mol) + H (1.008 g/mol) = 39.998 g/mol.

Using Molarity to Calculate Unknowns

Conversion Factors and End Amounts

Molarity values can be used as conversion factors to determine unknown quantities in solution chemistry. Problems may require isolating an end amount using given amounts and conversion factors.

• Given amount: The known quantity (mass, volume, or moles).

• Conversion factor: Molarity or other relationships (e.g., molar mass).

• End amount: The unknown to be solved for (e.g., mass, moles, volume).

Example: How many grams of Na3PO4 (MW: 163.94 g/mol) are present in 300.0 mL of a 0.550 M Na3PO4 solution?

• Convert volume to liters: $300.0\,\text{mL} = 0.300\,\text{L}$

• Calculate moles: $0.550\,\text{mol/L} \times 0.300\,\text{L} = 0.165\,\text{mol}$

• Calculate mass: $0.165\,\text{mol} \times 163.94\,\text{g/mol} = 27.05\,\text{g}$

Practice Problems

Application of Molarity in Medical and Laboratory Contexts

Practice problems help reinforce the use of molarity in real-world and laboratory scenarios.

• Hypernatremia: A medical condition where a patient has high levels of sodium in their blood due to low water content. If a patient has a sodium level of 165 mM and 30.0 mL of blood is drawn, calculate the mass (in mg) of sodium present.

• Solution Preparation: Calculate the molecular weight of an unknown compound if 2.64 g is dissolved in water to yield 150 mL of solution at 0.075 M concentration.

• Volume Calculation: What volume (in μL) of 0.125 M HBr contains 0.170 moles HBr?

• Molarity of Benzene Solution: Calculate the molarity of benzene in a solution prepared by dissolving 30.00 mL of benzene (C6H6, density = 0.877 g/mL) in dichloromethane to make a final volume of 750.0 mL.

Sample Table: Molarity Calculations and Conversions

The following table summarizes key conversion steps in molarity calculations:

Step	Operation	Example Value
Convert mass to moles	Divide by molar mass	$\frac{2.37\,\text{g}}{39.998\,\text{g/mol}} = 0.05925\,\text{mol}$
Convert volume to liters	Divide by 1000	$\frac{300.0\,\text{mL}}{1000} = 0.300\,\text{L}$
Calculate molarity	Divide moles by liters	$\frac{0.05925\,\text{mol}}{2.50\,\text{L}} = 0.0237\,\text{M}$
Calculate mass from moles	Multiply by molar mass	$0.165\,\text{mol} \times 163.94\,\text{g/mol} = 27.05\,\text{g}$

Summary

Molarity is a central concept in solution chemistry, allowing for the calculation of concentrations, masses, and volumes in chemical and medical contexts. Mastery of molarity calculations is essential for success in GOB Chemistry and related fields.