Deducing the unit missing from the solution to a basic quantitative problem, a student sets up and solves the following equation to solve a problem in solution stoichiometry. Fill in the missing part of the student's equation.
[tex]\frac{(1.05 mol)\left(\frac{1 mL}{10^{-3} L}\right)}{(4.7 \square)}=2.2 \times 10^2 mL[/tex]
Answer (2)
Rearrange the equation to isolate the missing unit.
Simplify the equation by performing the division and multiplication.
Calculate the value of the missing unit.
Identify the missing unit as molarity ( L m o l ).
The missing unit is L m o l .
Explanation
Problem Setup and Goal The student is trying to solve a stoichiometry problem and has set up the equation: ( 4.7 □ ) ( 1.05 mol ) ( 1 0 − 3 L 1 mL ) = 2.2 × 1 0 2 mL We need to find the missing unit represented by the square.
Isolating the Missing Unit To find the missing unit, we will isolate it by rearranging the equation. First, we can rewrite 2.2 × 1 0 2 mL as 220 mL . The equation then becomes: ( 4.7 □ ) ( 1.05 mol ) ( 1 0 − 3 L 1 mL ) = 220 mL Now, we multiply both sides by ( 4.7 □ ) :
( 1.05 mol ) ( 1 0 − 3 L 1 mL ) = 220 mL × ( 4.7 □ ) ( 1.05 mol ) ( 1 0 − 3 L 1 mL ) = 1034 mL × □
Calculating the Missing Unit Next, we divide both sides by 1034 mL to isolate the missing unit: □ = 1034 mL ( 1.05 mol ) ( 1 0 − 3 L 1 mL ) □ = 1 0 − 3 L × 1034 1.05 mol □ = 1.034 L 1.05 mol □ ≈ 1.015 L mol
Determining the Missing Unit The missing unit is approximately 1.015 L mol . Since L mol is the unit for molarity (M), the missing unit is M or L mol . We can approximate this to 1 L mol .
Final Answer Therefore, the missing unit is L mol , which represents molarity (M).
Concise Answer The missing unit is L m o l .
Examples
In chemistry, when preparing solutions, it's crucial to know the concentration of the solute in the solvent. This problem demonstrates how to ensure the units in your calculations are consistent to arrive at the correct concentration, typically measured in molarity (mol/L). For instance, if you're titrating an acid with a base, knowing the exact molarity of your solutions is essential for accurate results. Similarly, in pharmacology, precise concentration calculations are vital for drug dosages to ensure patient safety and treatment effectiveness. Accurately determining the units and their relationships is fundamental to quantitative problem-solving in various scientific and practical contexts.
The missing unit in the student's equation is L m o l , indicating molarity. This value is essential for understanding the concentration of the solution involved in stoichiometric calculations. The need for consistent units is a fundamental aspect of solving chemistry problems effectively.
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