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
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P1V1/T1=P2V2/T2 P2=1/2 P1 V1=1 T1=298K
1 P1/298= (1/2) P1V2/373 cross P1
1/298=1/2V2/373
1/298=1/V2 746
v2=746/298
V2=2.5L
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If the formula for density=m/v, you can manipulate this formula to the the mass. After manipulation, you get the equation for mass to be: m=density * volume. With density and volume given, we can find the mass.
mass= (8.9) * 6 = 53.4grams
The percent by mass of 110.6 g of potassium fluoride (NaF) dissolved in 1500.0 g of water is 0.073733 %.
<h3>
What is the mass percentage?</h3>
The mass percent of a solution is defined as the ratio of the mass of solute that is present in a solution, relative to the mass of the solution, as a whole.
Calculate the mass percent of potassium fluoride, using the formula:
Mass percentage = 0.073733 %
Hence, the percent by mass of 110.6 g of potassium fluoride (NaF) dissolved in 1500.0 g of water is 0.073733 %.
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