Multiple Choice
Calculate the concentration (M) of sodium ions in a solution made by diluting 40.0 mL of a 0.474 M solution of sodium sulfide (Na2S) to a total volume of 300 mL.
6. Chemical Quantities & Aqueous Reactions
Dilutions
Struggling with General Chemistry?
Join thousands of students who trust us to help them ace their exams!Watch the first videoMultiple Choice
To 225 mL of a 0.80 M solution of KI, a student adds enough water to make 1.0 L of a more dilute KI solution. What is the molarity of the new solution?
A
0.20 M
B
0.18 M
C
0.36 M
D
0.72 M
Verified step by step guidance1
Start by understanding the concept of dilution, which involves adding solvent to a solution to decrease its concentration. The number of moles of solute remains constant before and after dilution.
Use the dilution formula: \( C_1V_1 = C_2V_2 \), where \( C_1 \) is the initial concentration, \( V_1 \) is the initial volume, \( C_2 \) is the final concentration, and \( V_2 \) is the final volume.
Identify the known values: \( C_1 = 0.80 \text{ M} \), \( V_1 = 225 \text{ mL} \) (convert to liters by dividing by 1000, so \( V_1 = 0.225 \text{ L} \)), and \( V_2 = 1.0 \text{ L} \).
Rearrange the dilution formula to solve for \( C_2 \): \( C_2 = \frac{C_1V_1}{V_2} \).
Substitute the known values into the equation: \( C_2 = \frac{0.80 \times 0.225}{1.0} \). Calculate \( C_2 \) to find the molarity of the new solution.
1:03mWatch next
Master Solution Dilution Process with a bite sized video explanation from Jules
Start learningRelated Videos
Related Practice
