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Spirytus Polish Vodka is 96% alcohol by volume.

How much cranberry juice (no alcohol content) would have to be added to 500mL of this vodka to give the resultant mixture with an alcohol content of 12%?

Guest May 2, 2015

Best Answer 

 #4
avatar+26329 
+5

No amount of vodka at 37.5% by volume alcohol content when added to vodka at 37.5% by volume alcohol content will result in a mixture of 55% by volume alcohol content!

.

Alan  May 2, 2015
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4+0 Answers

 #1
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Volume of alcohol = 0.96*500 mL

 

Total volume = 500 + C   where C is volume of cranberry juice

 

0.96*500/(500 + C) = 0.12

 

0.96*500 = 0.12*(500 + C)

 

0.96*500/0.12 = 500 + C

 

C = 0.96*500/0.12 - 500

 

$${\mathtt{C}} = {\frac{{\mathtt{0.96}}{\mathtt{\,\times\,}}{\mathtt{500}}}{{\mathtt{0.12}}}}{\mathtt{\,-\,}}{\mathtt{500}} \Rightarrow {\mathtt{C}} = {\mathtt{3\,500}}$$

 

So you must add 3500 mL of cranberry juice.

.

Alan  May 2, 2015
 #2
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+5
  • Spirytus Polish Vodka is 96% alcohol by volume.

 

  1. How much cranberry juice would have to be added to 500mL of this vodka  resultant mixture with an alcohol content of 12%

 

Volume of alcohol= 0.96 x 500mL

Total Volume: 500+C, where C is volume of cranberry juice

0.96 x 500/(500 + C) = 0.12

0.96 x 500 = 0.12 x ( 500 + C)

0.96 x 500/0.12 - 500= 3500

 

So you must add  3500 mL of cranberry juice

b. How much Smirnoff Vodka with an alcohol content of 37.5% by volume would have to be added to 500mL of Smirnoff Vodka to get a resultant mixture of 55% alcohol.

Guest May 2, 2015
 #3
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Thanks Alan and anon - I still trip up on these.    

Melody  May 2, 2015
 #4
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+5
Best Answer

No amount of vodka at 37.5% by volume alcohol content when added to vodka at 37.5% by volume alcohol content will result in a mixture of 55% by volume alcohol content!

.

Alan  May 2, 2015

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