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Semmy [17]
3 years ago
14

A stock solution of sodium sulfate NaSO4 has a concentrate of 1.00 m. The volume of this solution is 50 ml. What volume of 0.25

M solution could be made from the stock solution
Chemistry
1 answer:
mylen [45]3 years ago
5 0
<h3>Answer:</h3>

200 mL

<h3>Explanation:</h3>

Concept tested: Dilution formula

We are given;

  • Concentration of stock solution as 1.00 M
  • Volume of the stock solution as 50 mL
  • Molarity of the dilute solution as 0.25 M

We are required to calculate the volume of diluted solution;

  • The stock solution is the original solution before dilution while diluted solution is the solution after dilution.
  • Using the dilution formula we can determine the volume of diluted solution;

M1V1 = M2V2

Rearranging the formula;

V2 = M1V1 ÷ M2

     = (1.00 M × 50 mL) ÷ 0.25 M

     = 200 mL

Therefore, a volume of 200mL of 0.25 M solution could be made from the stock solution.

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7 0
3 years ago
a heliox tank contains 32% helium and 68% oxygen. the total pressure in the tank is 395 kPa. What is the partial pressure of oxy
trapecia [35]

Answer:

Pp O2 = 82.944 KPa

Explanation:

heliox tank:

∴ %wt He = 32%

∴ %wt O2 = 68%

∴ Pt = 395 KPa

⇒ Pp O2 = ?

assuming a mix of ideal gases at the temperature and volumen of the mix:

∴ Pi = RTni/V

∴ Pt = RTnt/V

⇒ Pi/Pt = ni/nt = Xi

⇒ Pi = (Xi)*(Pt)

∴ Xi: molar fraction (ni/nt)

⇒ 0.68 = mass O2/mass mix

assuming mass mix = 100 g

⇒ mass O2 = 68 g

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⇒ moles O2 = (68 g)(mol/32 g) = 2.125 mol O2

⇒ mass He = 32 g

∴ molar mass He = 4.0026 g/mol

⇒ moles He = (32 g)(mol/4.0026 g) = 7.995 mol He

⇒ nt = nO2 + nHe = 2.125 mol + 7.995 mol = 10.12 moles

molar fraction O2:

⇒ X O2 = nO2/nt = (2.125 mol/10.12 mol) = 0.2099

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⇒ Pp O2 = (0.2099)(395 KPa)

⇒ Pp O2 = 82.944 KPa

6 0
3 years ago
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