secer

Submitted by goranspax on Fri, 04/17/2015 - 18:03
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Na kojoj temperaturi ce vodeni rastvor 1.00g secera C12H22O11, 100cm3 rastvor imati osmotski pritisak 66.4 kPa

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milicaradenkovic - Fri, 04/17/2015 - 18:51

p=c*R*T

p*V=n*R*T

n=m/Mr=1/342=0.002924 mol C12H22O11

T=p*V/n*R

T=66.4*10(3)*100*10(-6)/8.314*0.002924

T=273.137 C

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goranspax - Fri, 04/17/2015 - 20:11

A sta mi je ovo 10(3)*100*10(-6)? Zar tu ne bi trebala da mude zapremina u dm3 ili cm3?

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goranspax - Fri, 04/17/2015 - 20:11

A sta mi je ovo 10(3)*100*10(-6)? Zar tu ne bi trebala da mude zapremina u dm3 ili cm3?

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milicaradenkovic - Sat, 04/18/2015 - 08:02

Pretvaranje kPa u Pa i cm3 u m3 .