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What volume of 90% alcohol by weight (d ...

What volume of 90% alcohol by weight `(d = 0.8 g mL^(-1))` must be used to prepared `80 mL` of 10% alcohol by weight `(d = 0.9 g mL^(-1))`

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The correct Answer is:
A
Use `M = (% "by weight" xx 10 xx d)/(Mw_(2))`
`M_(1) V_(1) = M_(2) V_(2)`
`(90 xx cancel(10) xx 0.8)/(cancel(46)) xx V = (10 xx cancel(10) xx 0.9)/(46) xx 80`
`V = (10 xx 0.9 xx 80)/(90 xx 0.8) = 10 mL`
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