ncert quest.

Asked by deepitapai | 16th May, 2009, 04:26: PM

Expert Answer:

w= M M' V / 1000

  =0.25  x 32 X 2500 /1000

   =20 g =.02 kg

V = 2.5 L = 2500 ML

.793 kg of methanol present in 1 L of the sample

.02Kg  of methanol present in 1 X.02 / .793   = .02522 L or 25.22mL 

Answered by  | 25th May, 2009, 04:23: PM

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