An experiment requires 0.7 M NH3(aq). The
stockroom manager estimates that 15 L of
the base is needed. What volume of 15 M
NH3(aq) will be used to prepare this amount
of 0.7 M base?
To determine the volume of 15 M NH3(aq) needed to prepare 0.7 M NH3(aq), we can use the dilution equation:
C1V1 = C2V2
Where:
C1 = Initial concentration of NH3(aq) = 15 M
V1 = Volume of 15 M NH3(aq) we're trying to find
C2 = Final concentration of NH3(aq) = 0.7 M
V2 = Volume of 0.7 M NH3(aq) needed = 15 L
Now, let's plug in the given values and solve for V1:
15 M * V1 = 0.7 M * 15 L
Divide both sides by 15 M:
V1 = (0.7 M * 15 L) / 15 M
Cancel out the units to get the volume:
V1 = 0.7 L
Therefore, to prepare 0.7 M NH3(aq), we will need 0.7 liters of 15 M NH3(aq).
To determine the volume of 15 M NH3(aq) needed to prepare 0.7 M NH3(aq), we can use the concept of dilution.
First, let's define the dilution formula:
(C₁)(V₁) = (C₂)(V₂)
Where:
C₁ = initial concentration
V₁ = initial volume
C₂ = final concentration
V₂ = final volume
In this case:
C₁ = 15 M (concentration of the stock NH3(aq))
V₁ = volume we need to find (unknown)
C₂ = 0.7 M (desired final concentration)
V₂ = 15 L (desired final volume)
Plugging in these values to the dilution formula, we get:
(15 M)(V₁) = (0.7 M)(15 L)
Now we can solve for V₁:
V₁ = (0.7 M)(15 L) / 15 M
Simplifying, we get:
V₁ = 0.7 L
Therefore, to prepare 0.7 M NH3(aq), you will need to use 0.7 liters of 15 M NH3(aq).
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Use c1v1 = c2v2
c = concn
v = volume
15M x v = 0.7M x 15L
v = ? (in L).