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
Given:
We know the molar mass,
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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