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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))`

Text Solution

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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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Knowledge Check

  • What volume of 75% alcohol by weight (d-0.80g//cm^(3)) must be used to prepare 150 cm^(3) of 30 % alcohal by mass (d=0.90g//cm^(3)) ?

    A
    67.5 mL
    B
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    C
    44.44 mL
    D
    None of these

  • What volume of 75% alcohol by mass (d=0.8 g/ cm^(3) ) must be used to prepare 150cc. Of 30% alcohol by mass (d=0.9g//cm^(3)) ?

    A
    44.44mL
    B
    56.25mL
    C
    67.5mL
    D
    33.56mL

  • X mL of a 60% w/w alcohol by weight (d = 0.6gm//mL) must be used to prepare 200 mL of 12% alcohol by weight (d=0.6gm//mL) . Then the value of X will be :

    A
    20
    B
    40
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    60
    D
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