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algol [13]
4 years ago
13

Determine the concentration of the following dye standard made by a student who pipettes out 2.50 mL of a 0.250 M stock solution

and transfers it into a volumetric flask and dilutes the dye to a final volume of 100.0 mL with DI water.
Chemistry
1 answer:
Thepotemich [5.8K]4 years ago
7 0

Answer:

Concentration of dye in diluted solution is 0.00625 M

Explanation:

The given problem can be solved by using laws of dilution

According to laws of dilution-   C_{1}V_{1}=C_{2}V_{2}

where C_{1} and C_{2} are initial and final concentration respectively

          V_{1} and V_{2} are initial and final volume respectively

Here, C_{1}=0.250 M, V_{1}=2.50mL and V_{2}=100.0mL

So, C_{2}=\frac{C_{1}V_{1}}{V_{2}}

or, C_{2}=\frac{(0.250M)\times (2.50mL)}{100.0mL}

or, C_{2}=0.00625M

So, concentration of dye in diluted solution is 0.00625 M

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Sati [7]
<span>Answer: <u>Molarity of solution is 0.751 M. </u>

Reason:
Given: weight of solute (LiF) = 17.8 g, volume of solution = 915 ml = 0.915 l

We know that,
Molarity = </span>\frac{\text{weight of solute (g)}}{\text{Molecular weight X Volume of solution(l)}}<span>

<em>Molecular Weight of LiF = 25.9 g/mol. </em>

</span>∴<span>, Molarity = </span>\frac{17.8}{25.9X0.915}
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3 years ago
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What are the molality and mole fraction of solute in a 35.9 percent by mass aqueous solution of formic acid (HCOOH)?
sdas [7]

ِAnswer:

1- The molarity of HCOOH = 9.515 M.

2- The mole fraction of HCOOH = 0.18.

Explanation:

<em>1- The molarity of HCOOH:</em>

  • We can calculate the molarity of HCOOH using the relation:

M = (10pd)/molar mass.

p is the percent by mass of HCOOH = 35.9 %.

d is the specific gravity of HCOOH = 1.22 g/cm³.

Molar mass of HCOOH = 46.03 g/mol.

∴ M = (10pd)/molar mass = (10)(35.9 %)(1.22 gcm³) / (46.03 g/mol) = 9.515 M.

<em>2- The mole fraction of HCOOH:</em>

  • We can suppose that we have a 100 g solution, that contains 35.9 g of HCOOH and 64.1 g of water.

<em>The mole fraction of HCOOH = (no. of moles of HCOOH) / (no. of moles of HCOOH + no, of moles of water).</em>

no. of moles of HCOOH = mass / molar mass = (35.9 g)/(46.03 g/mol) = 0.78 mol.

no. of moles of water = mass / molar mass = (64.1 g)/(18.0 g/mol) = 3.56 mol.

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7 0
3 years ago
Please and thank you
Virty [35]

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Therefore the free energy, ΔG is

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