Given unlimited quantities of 95% alcohol and distilled water, prepare 200ml of 40% alcohol.

Question

Question

science chemistry

Answer 1

Answer #1

C1V1 = C2V2

C1 = concentration of stock alcohol = 95%

V1 = volume of stock alcohol = to be calculated

C2 = concentration of desired alcohol = 40%

V2 = volume of desired diluted alcohol = 200mL

So,

V1 = (C2V2) / C1

V1 = (40 % * 200mL) / (95%)

V1 = 84.21 mL

So, volume of 95% alcohol needed =84.21 mL

Amount of water to be added = 200 - 84.21 = 115.79 mL

Answer 2

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