Answer :
Final answer:
In 38.1 mL of a 0.300 M NaOH solution, there are 0.00963 moles of NaOH. To find this, multiply the molarity (0.300 moles per liter) by the volume in liters (0.0381 L).
Explanation:
In 38.1 mL of a 0.300 M (moles per liter) NaOH (sodium hydroxide) solution, the number of moles of NaOH can be calculated by first converting the volume to liters.
To do this, divide 38.1 mL by 1000, which equals 0.0381 liters.
Then, you can use the formula Moles = Molarity (M) x Volume (L), where the molarity is 0.300 M and the volume in liters is 0.0381 L.
This calculation results in 0.00963 moles of NaOH in the solution.
Therefore, there are 0.00963 moles of NaOH in the provided 38.1 mL of the 0.300 M NaOH solution, indicating the amount of the solute present in the given volume of the solution.
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Final answer:
There are 0.01143 moles of NaOH in a 38.1 mL solution of 0.300 M NaOH. This was calculated by multiplying the given solution molarity (0.300 moles/L) by the volume in liters (0.0381 L).
Explanation:
To calculate the number of moles in this solution, we'll need to use the formula for molarity, which is moles of solute divided by liters of solution. Given that the molarity (M) is 0.300 and the volume is 38.1 mL (or 0.0381 L as we need the volume in liters for the calculation), we can simply multiply these values.
So, the calculation would be:
Moles of NaOH = Molarity x Volume (in L) = 0.300 mol/L * 0.0381 L = 0.01143 mol.
Therefore, 0.01143 moles of NaOH are present in 38.1 mL of a 0.300 M NaOH solution.
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