Intermolecular forces as a whole are also known as
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This is a straightforward dilution calculation that can be done using the equation
where <em>M</em>₁ and <em>M</em>₂ are the initial and final (or undiluted and diluted) molar concentrations of the solution, respectively, and <em>V</em>₁ and <em>V</em>₂ are the initial and final (or undiluted and diluted) volumes of the solution, respectively.
Here, we have the initial concentration (<em>M</em>₁) and the initial (<em>V</em>₁) and final (<em>V</em>₂) volumes, and we want to find the final concentration (<em>M</em>₂), or the concentration of the solution after dilution. So, we can rearrange our equation to solve for <em>M</em>₂:
Substituting in our values, we get
So the concentration of the diluted solution is 0.05875 M. You can round that value if necessary according to the appropriate number of sig figs. Note that we don't have to convert our volumes from mL to L since their conversion factors would cancel out anyway; what's important is the ratio of the volumes, which would be the same whether they're presented in milliliters or liters.
When 0.34 of HNO₃ is titrated to equivalence using 0.14 l of 0.1 m NaOH then the concentration of HNO₃ is 0.041 M
The reaction of neutralization of HNO₃ with NaOH is
HNO₃ + NaOH → H₂O + NaNo₃
When 1 mole of HNO₃ react with 1 mole of NaOH, based on chemical rection the moles of NaOH at equivalence point are equal to moles of HNO₃ present in solution: -
With the mole and volumes, we can find molarity as follows:
Moles of NaOH = moles HNO₃
⁼ 0.14 L X (0.1 mol NaOH/L) = 0.014 mole NaOH
=0.014 mol HNO₃
Molarity: -
= 0.041 M
Thus, from above solution we concluded that the concentration of HNO₃ solution is 0.041 M.
Learn more about molarity: brainly.com/question/8732513
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