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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))` |
Answer» Correct Answer - 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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