First, calculate the mass of sodium in g with the help of molar mass and number of moles.
Number of moles = 
(1)
Molar mass of sodium =
Substitute the given value of number of moles and molar mass of sodium in formula (1)
(1)
(1)
mass of sodium in g = 
Now, according to conversion factor, 1 g = 1000 mg
So, of sodium = 
= 
of sodium
Thus, mass of sodium in mg =
Answer:
yes solution is always a mixture but not all mixtures are solution
Explanation:
A solution.is a homogeneous mixture of substance that have uniform composition throughout
And a mixture hVe twoo or more substances that are not chemically.combine
The volume of the 0.15 M LiOH solution required to react with 50 mL of 0.4 M HCOOH to the equivalence point is 133.3 mL
<h3>Balanced equation </h3>
HCOOH + LiOH —> HCOOLi + H₂O
From the balanced equation above,
The mole ratio of the acid, HCOOH (nA) = 1
The mole ratio of the base, LiOH (nB) = 1
<h3>How to determine the volume of LiOH </h3>
- Molarity of acid, HCOOH (Ma) = 0.4 M
- Volume of acid, HCOOH (Va) = 50 mL
- Molarity of base, LiOH (Mb) = 0.15 M
- Volume of base, LiOH (Vb) =?
MaVa / MbVb = nA / nB
(0.4 × 50) / (0.15 × Vb) = 1
20 / (0.15 × Vb) = 1
Cross multiply
0.15 × Vb = 20
Divide both side by 0.15
Vb = 20 / 0.15
Vb = 133.3 mL
Thus, the volume of the LiOH solution needed is 133.3 mL
Learn more about titration:
brainly.com/question/14356286
