A stream of dry air was passed through a...

A stream of dry air was passed through a bulb containing a solution of 7.5 g of a compound dissolved in 75 g of water and then through another bulb containing pure water. The loss of mass in the first bulb was 2.81 g and in the second bulb was 0.054 g. Calculate the molecular mass of the compound.

Text Solution

Verified by Experts

` M_(2) = (P^(0)(W_(2)M_(1)))/((P^(0) - P)W_(1))`
OR
`M_(2) = (W_(2) M_(1))/(W_(1)) xx ((P^(0))/(P^(0)-P))`
`("Total loss in mass of solution bulb and the solvent bulb")/("Loss in mass of solvent bulb")`
= `(7.5 xx 18)/(75) xx (2.864)/(0.054)`
`M_(2) = 95.47`.

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312.5

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None

Dry air was successively passed through a solution of 5 g solute in 80 g water and then through pure water. The loss in weight of solution was 2.5 and that of pure water was 0.04 g. What is the mol . Mass of the solute ?

160 amu

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200 amu

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The mass of glucose to be dissolved in 50 g of water to get 0.3 Molar solution is

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0.27 g

2.7 g

5.4 g.