Respuesta :

joseaaronlara

Answer:

The answer to your question is:  52.92 g of CuNO₃

Explanation:

Data

grams of CuNO₃ = ?

volume = 1 l

concentration = 0.420 M

MW of  CuNO₃ = 64 + 14 + 3(16) = 126 g

Formula

Molarity = moles / volume

moles = Molarity x volume

moles = 0.420 x 1

moles = 0.420 moles

                                    1 mol of CuNO₃  -------------------  126 g of CuNO₃

                                  0.420 moles CuNO₃ -------------    x

                                    x = (0.420 x 126) / 1

                                    x = 52.92 g of CuNO₃

okpalawalter8

We have that the grams of Cuno3 that will be required to make 1.00 l of 0.420 m solution is

[tex]G=52.92g[/tex]

From the question we are told

How many grams of cuno3 will be required to make 1.00 l of 0.420 m solution?

Generally the equation for the Molarity   is mathematically given as

[tex]Molarity =\frac{mass}{volume}\\\\M=\frac{m}{v}\\\\\Therefore\\\\m=M x v\\\\Therefore\\\\m=0.420*1.00\\\\m=0.420\\\\[/tex]

Therefore

Grams of Cuno3 that will be required to make 1.00 l of 0.420 m solution is

[tex]G=\frac{(0.420*126)}{1}[/tex]

[tex]G=52.92g[/tex]

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https://brainly.com/question/19694949

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