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0.85 g aqueous solution of NaNO(3) is ap...

0.85 g aqueous solution of `NaNO_(3)` is approximately 90% dessociated at `27^(@)C`. Calculate the osmotic pressure (R=0.0821 L atm `K^(-1) mol^(-1)`)

Text Solution

Verified by Experts

The correct Answer is:
4.68 atm
Calculation of Van't Hoff factor
`NaNO_(3)hArrNa^(+)("aq")+NO_(3)^(-)("aq")`
`alpha=(i-1)/(n-1)or0.9=(i-1)/(2-1)or i=0.9+1=1.9`
Step II. Calculation of osmotic pressure
`"Initial no. of moles of "NaNO_(3)(n_(B))=((0.85g))/((85" g mol"^(-1)))=0.01mol`
`"Osmotic pressure "(pi)=iCRT=(i n_(B)RT)/V`
`=(1.9xx(0.01"mol")xx(0.0821" L atm"^(-1)mol^(-1))xx300K)/((0.1K))=4.68"atm".`
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