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0.2 g of a substance displaced 65.2 mL o...

`0.2 g` of a substance displaced `65.2 mL` of air at 300 K and 749 mm pressure. Calculate the molecular mass of the substance. (Aqueous Tension at 300 K is `26.7mm`)

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Given: Mass of substance=0.2g, Volume of air (V)=65.2 mL, Temperature=300K Pressure (P) = 749 -26.7=722.3.
Molar mass, `M = (w xx RT xx 760 xx 1000)/( ( P - p ) xx V) = ( 760 xx 1000 xx 0. 2 xx 0. 0 821 xx 300)/( 722 . 3 xx 65.2) = 79.5 g mol ^(-1)`
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