Assume that the electrical conductivity of a solution depends on the total concentration of dissolved ions and that you measure the conductivity of three different solutions while carrying out titration procedures: (a) Begin with 1.00 L of 0.100 M KCl, and titrate by adding 0.100 M AgNO3. (b) Begin with 1.00 L of 0.100 M HF, and titrate by adding 0.100 M KOH. (c) Begin with 1.00 L of 0.100 M BaCl2, and titrate by adding 0.100 M Na2SO4. Which of the following graphs corresponds to which titration?
Ch.4 - Reactions in Aqueous Solution
Chapter 4, Problem 46a
How many moles of solute are present in each of the following solutions? (a) 35.0 mL of 1.200 M HNO3

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Determine the formula for calculating moles from molarity and volume: \( \text{moles} = \text{molarity} \times \text{volume in liters} \).
Convert the volume from milliliters to liters: \( 35.0 \text{ mL} = 0.0350 \text{ L} \).
Substitute the given values into the formula: \( \text{moles} = 1.200 \text{ M} \times 0.0350 \text{ L} \).
Perform the multiplication to find the number of moles.
Ensure the units are consistent and the calculation is correct.

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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Molarity (M)
Molarity is a measure of concentration defined as the number of moles of solute per liter of solution. It is expressed in moles per liter (mol/L). Understanding molarity is crucial for calculating the amount of solute in a given volume of solution, as it directly relates the volume of the solution to the number of moles present.
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Volume Conversion
In chemistry, it is often necessary to convert volumes from milliliters (mL) to liters (L) when working with molarity. Since 1 L equals 1000 mL, converting mL to L involves dividing the volume in mL by 1000. This conversion is essential for accurately applying the molarity formula to determine the number of moles of solute.
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Calculating Moles from Molarity
To find the number of moles of solute in a solution, the formula used is: moles = molarity × volume (in liters). This relationship allows for the straightforward calculation of moles when both the molarity and the volume of the solution are known. It is a fundamental concept in stoichiometry and solution chemistry.
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Molar Mass Calculation Example
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