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What is the molarity of the solution produced when 145 g NaCl is dissolved in sufficient water to prepare 2.75 L of solution? *

a. 0.0190
b. 52.7
c. 399
d. 0.902

Respuesta :

olatunbosunadeoke

Answer: Option D) 0.902M

Explanation:

Sodium chloride has a chemical formula of NaCl

Given that,

Amount of moles of NaCl (n) = ?

Mass of NaCl in grams = 145g

For molar mass of NaCl, use the molar masses:

Sodium, Na = 23g;

Chlorine,Cl = 35.5g

NaOH = (23g + 35.5g)

= 58.5g/mol

Since, amount of moles = mass in grams / molar mass

n = 145g / 58.5g/mol

n = 2.48 mole

Amount of moles of NaCl (n) = 2.48 mole

Volume of NaCl solution (v) = 2.75L

Concentration of NaCl solution (c) = ?

Since concentration (c) is obtained by dividing the amount of solute dissolved by the volume of solvent, hence

c = n / v

c = 2.48 mole / 2.75L

c = 0.9018M

Approximately, c = 0.902M

[c is the concentration in moles per litres which is also known as molarity]

Thus, the molarity of the solution is 0.902M

Cricetus

The molarity of the solution will be "0.902 M"

Given:

  • Mass of NaCl = 145 g
  • Volume of NaCl = 2.75 L

We know the molar mass,

  • Sodium, Na = 23 g
  • Chlorine, Cl = 35.5 g

Now,

The molar mass of NaCl will be:

= [tex]23+35.5[/tex]

= [tex]58.5 \ g/mol[/tex]

The amount of moles will be:

= [tex]\frac{Mass}{Molar \ mass}[/tex]

= [tex]\frac{145}{58.5 }[/tex]

= [tex]2.48 \ mole[/tex]

hence,

The molarity will be:

→ [tex]c = \frac{n}{v}[/tex]

By substituting the values,

     [tex]= \frac{2.48}{2.75}[/tex]

     [tex]= 0.902 \ M[/tex]

Thus the above approach i.e., "option d" is correct.

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