To find the mass \( M \) of potassium hydroxide (KOH) that was dissolved, we start by analyzing the reaction and the information provided in the problem.
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Stoichiometry of the Reaction: The reaction between potassium hydroxide (KOH) and nitric acid (HNO₃) can be represented as:
\[ \text{KOH (aq)} + \text{HNO}_3 \text{ (aq)} \rightarrow \text{KNO}_3 \text{ (aq)} + \text{H}_2\text{O (l)} \]
From the balanced equation, we can see that the molar ratio of KOH to HNO₃ is 1:1.
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Calculating Moles of Nitric Acid: We know that 50 cubic centimeters (cc) of 2.0 M nitric acid is used for neutralization. First, we'll convert 50 cc to liters: \[ 50 , \text{cc} = 0.050 , \text{L} \]
Now we can calculate the moles of nitric acid: \[ \text{Moles of HNO}_3 = \text{Molarity} \times \text{Volume} = 2.0 , \text{M} \times 0.050 , \text{L} = 0.1 , \text{moles} \]
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Calculating Moles of Potassium Hydroxide: Since the stoichiometry is 1:1, the moles of KOH reacted will also be 0.1 moles.
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Concentration of KOH in the Original Solution: The 100 cc (or 0.100 L) solution contains these 0.1 moles of KOH. To find the molarity of KOH in the original solution, we use: \[ \text{Molarity of KOH} = \frac{\text{moles of KOH}}{\text{volume of solution in L}} = \frac{0.1 , \text{moles}}{0.100 , \text{L}} = 1.0 , \text{M} \]
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Calculating Mass of KOH: The molar mass of KOH can be calculated as follows:
- Potassium (K): 39.10 g/mol
- Oxygen (O): 16.00 g/mol
- Hydrogen (H): 1.01 g/mol
\[ \text{Molar mass of KOH} = 39.10 + 16.00 + 1.01 = 56.11 , \text{g/mol} \]
Using the molarity and the volume, we can find the mass of KOH in the solution: \[ \text{Mass of KOH} = \text{moles of KOH} \times \text{molar mass of KOH} = 0.1 , \text{moles} \times 56.11 , \text{g/mol} = 5.611 , \text{g} \]
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Final Result: The mass \( M \) of potassium hydroxide that was dissolved is approximately:
\[ M \approx 5.61 , \text{grams} \]
Thus, the answer is: \[ \boxed{5.61 , \text{g}} \]
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