The cook needs to add 21.333 ounces of water to obtain a 3 % concentration.
In matter of units, concentration equals to to the mass of cider vinegar ([tex]m_{c}[/tex]) divided by mass of water ([tex]m_{w}[/tex]), both expressed in ounces. A reduction on concentration implies that mass of cider vinegar remains constant while mass of water is increased.
Now we derive expressions for the initial and final concentrations for cider vinegar:
Initial concentration
[tex]c_{i} = \frac{m_{c}}{m_{w}+m_{c}}[/tex] (1)
Where:
Final concentration
[tex]c_{f} = \frac{m_{c}}{m_{c}+m_{w}'}[/tex] (2)
Where:
Then, if we know that [tex]c_{i} = 0.05[/tex], [tex]c_{f} = 0.03[/tex], [tex]m_{w}+m_{c} = 32\,oz[/tex] and [tex]m_{c}+m_{w}' = 32 + x[/tex], then the addition of water is:
[tex]0.05 = \frac{m_{c}}{32}[/tex] (1b)
[tex]0.03 = \frac{m_{c}}{32+x}[/tex] (2b)
Then we eliminate [tex]m_{c}[/tex] in (1b) and (2b):
[tex]0.05\cdot 32 = 0.03\cdot 32 + 0.03\cdot x[/tex]
[tex]0.02\cdot 32 = 0.03\cdot x[/tex]
[tex]x = 21.333\,oz[/tex]
The cook needs to add 21.333 ounces of water to obtain a 3 % concentration.
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