How would you prepare 100.0 ml of.400 m CaCl2 from a stock solution of 2.00 M CaCl2?

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Dejanras Dejanras · Answer 1 · 2018-12-07T00:04:40+01:00

Answer is: add 80.00 mL of water to 20.00 mL of stock solution.

V₁(CaCl₂) = ? mL; initial volume of calcium chloride solution.

c₁(CaCl₂) = 2.00 M; initial concentration.

V₂(CaCl₂) = 100.0 mL; final volume.

c₂(CaCl₂) = 0.400 M; final concentration.

Use c₁V₁ = c₂V₂.

2.00 M · V₁(CaCl₂) = 0.400 M · 100.0 mL.

V₁(CaCl₂) = (0.400 M · 100.0 mL) ÷ 2.00 M.

V₁(CaCl₂) = 20 ml.

V(H₂O) = 100 mL - 20 mL.

V(H₂O) = 80 mL.

okpalawalter8 okpalawalter8 · Answer 2 · 2022-04-25T16:30:43+02:00

prepare 100.0 ml of.400 m CaCl2 we have to add V(H2O) = 80 mL of water to V(CaCl2) = 20 ml of stock solution .

Question Parameter(s):

How would you prepare 100.0 ml of.400 m CaCl2

a stock solution of 2.00 M CaCl2

Generally, the equation for the concentration is mathematically given as

c₁V₁ = c₂V₂.

Therefore

2.00 M · V₁(CaCl₂) = 0.400 M · 100.0 mL.

V(CaCl2) = (0.400 M · 100.0 mL)/ 2.00 M.

V(CaCl2) = 20 ml.

In conclusion

V(H2O) = 100 mL - 20 mL.

V(H2O) = 80 mL.