Multiple Choice
How many moles of K⁺ are present in 343 mL of a 1.27 M solution of K₃PO₄?
6. Chemical Quantities & Aqueous Reactions
Molarity
Struggling with General Chemistry?
Join thousands of students who trust us to help them ace their exams!Watch the first videoMultiple Choice
The molarity of an aqueous solution containing 75.3 g of glucose (C₆H₁₂O₆) in 35.5 mL of solution is ________.
A
2.1 M
B
5.3 M
C
11.8 M
D
8.4 M
Verified step by step guidance1
First, calculate the molar mass of glucose (C₆H₁₂O₆). The molar mass is the sum of the atomic masses of all atoms in the molecule. Glucose contains 6 carbon atoms, 12 hydrogen atoms, and 6 oxygen atoms. Use the atomic masses: Carbon (C) = 12.01 g/mol, Hydrogen (H) = 1.01 g/mol, Oxygen (O) = 16.00 g/mol.
Next, determine the number of moles of glucose in the solution. Use the formula: \( \text{moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \). Substitute the mass of glucose (75.3 g) and the molar mass calculated in the previous step.
Convert the volume of the solution from milliliters to liters, as molarity is expressed in moles per liter. Use the conversion: \( 1 \text{ mL} = 0.001 \text{ L} \). Therefore, 35.5 mL is equivalent to 0.0355 L.
Calculate the molarity of the solution using the formula: \( \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \). Substitute the number of moles of glucose and the volume of the solution in liters.
Review the calculated molarity and compare it with the given options to determine the correct answer.
Related Videos
Related Practice
