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Hypothetically, if you weighed out 0.8036 g of NaOH, what would the exact NaOH solution concentration be

Respuesta :

Complete question

The complete question is shown on the first uploaded image

Answer:

The value is [tex]C = 0.02009 \ M [/tex]

Explanation:

From the question we are told that

The mass of NaOH is [tex]m= 0.8036 g[/tex]

In the question DI (stands for de-ionized water )

So the number of moles of NaOH present in the given mass is

[tex]n = \frac{m}{Z}[/tex]

Here Z is the molar mass of NaOH with value [tex]Z = 40 \ g/ mol[/tex]

So

[tex]n = \frac{0.8036}{40}[/tex]

[tex]n =0.02009 \ mols [/tex]

Given that the volume of the flask is V = 1 L then we can evaluated the concentration as

[tex]C = \frac{n}{V}[/tex]

=>    [tex]C  =  \frac{0.02009}{1}[/tex]

=>    [tex]C  =  0.02009 \ M [/tex]  

Ver imagen okpalawalter8