Respuesta :

Willchemistry
Molar mass NaCl = 23 + 35.5 => 58.5 g/mol

Volume = 1.0 L

Molarity = 3.5

mass of solute = molas mass ( NaCl) x volume x molarity

mass of solute = 58.5 x 1.0 x 3.5

= 204.75 g of NaCl

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Answer:

204.75 grams of sodium chloride are required.

Explanation:

[tex]Molarity=\frac{n}{V(Liter)}[/tex]

n = moles of compound

V = volume of the solution in liters.

We have :

Molarity of the solution = 3.5 M

Volume of the solution = 1.0 L

Moles of sodium chloride = n

[tex]3.5 M =\frac{n}{1.0 L}[/tex]

n = 3.5 M × 1.0 L = 3.5 mol

Mass of 3.5 moles of sodium chloride : 3.5 mol × 58.5 g/mol = 204.75 g

204.75 grams of sodium chloride are required.

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