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ankoles [38]
3 years ago
10

You need 270 ml of a 65% alcohol solution. on hand, you have a 90% alcohol mixture. how much of the 90% alcohol mixture and pure

water will you need to obtain the desired solution?
Chemistry
1 answer:
olganol [36]3 years ago
8 0

You must add 75 mL water to 195 mL 90 % alcohol to make 270 mL of 65 % alcohol.

<em>Step 1.</em> Calculate the volume of 90 % alcohol needed

You can use the dilution formula

<em>V</em>1×<em>C</em>1 = <em>V</em>2×<em>C</em>2

where

<em>V</em>1 and<em> V</em>2 are the volumes of the two solutions

<em>C</em>1 and <em>C</em>2 are the concentrations

You can solve the above formula to get

<em>V</em>2 = <em>V</em>1 × <em>C</em>1/<em>C</em>2

<em>V</em>1 = 270 mL; <em>C</em>1 = 65 %

V2 = ?; _____<em>C</em>2 = 90 %

∴<em>V</em>2 = 270 mL × (65 %/90 %) = 195 mL

You need 195 mL of 90 % alcohol to make 270 mL of 65 % RA

<em>Step 2</em>. Calculate the amount of water to add.

Volume of water = 270 mL – 195 mL = 75 mL

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Data Given:

                  Initial Volume = V₁ =  36.7 L

                  Initial Pressure = P₁ =  145 kPa

                  Initial Temperature = T₁ =  65 °C + 273 = 338 K

                  Final Volume = V₂ = ?

                  Final Pressure = P₂ = 101.325 kPa           (Standard Pressure)

                  Final Temperature = T₂ = 273 K                (Standard Temperature)

Formula used:
                      As number of moles are constant, so Ideal Gas equation in following form is used,
                                        P₁ V₁ / T₁  =  P₂ V₂ / T₂
Solving for V₂,
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Putting Values,

                  V₂ = (145 kPa × 36.7 L × 338 K) ÷ (273 K × 101.325 kPa)

                  V₂ = 1798667 ÷ 27661.25

                  V₂ = 65.02 L
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