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laiz [17]
3 years ago
8

The concentration of a diluted "wine" sample was found to be 0.24 %(v/v) ethanol. Assuming the wine was diluted by 75x before an

alysis, what was the original concentration of ethanol in the "wine"
Chemistry
1 answer:
Softa [21]3 years ago
3 0

Answer:

The original concentration of ethanol was 18 % (v/v)

Explanation:

For the analysis of a sample, mostly dilutions are made of the original concentrated sample. The analysis results obtained from the diluted sample are then calculated for the concentrated samples.

According to the given data, the wine sample was diluted 75 times. This means that the initial concentration of ethanol was 75 times more than the final concentration i.e. 0.24 % (v/v). So mathematically:

original concentration (v/v)= final concentration (v/v) x times diluted

original concentration (v/v)= 0.24 % x 75

original concentration (v/v) = 18 %

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Read 2 more answers
A chef plans to mix 100% vinegar with italian dressing. the italian dressing contains 13% vinegar. the chef wants to make 150 mi
charle [14.2K]
To determine the volume of both concentration of vinegar, we need to set up two equations since we have two unknowns. 

For the first equation, we do a mass balance:

mass of 100% vinegar + mass of 13% vinegar = mass of 42% vinegar

Assuming they have the same densities, then we can write this equation in terms of volume.

V(100%) + V(13%) = V(42%)
   we let x = V(100%)
             y = V(13%)

x + y = 150

For the second equation, we do a component balance:

1.00x + .13y = 150(.42)
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The two equations are
x + y = 150
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Solving for x and y,
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Therefore, you need to mix 50 mL of the 100% vinegar and 100 mL of the 13% vinegar.

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