maseni udeo rastvora cijanidne kiseline

Submitted by mika on Fri, 02/14/2014 - 08:35
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2.U kojoj masi vode treba rastvoriti 100dm3 cijanovodonika (HCN) na temperaturi 20C , kada je pritisak 120kPa,da se dobije 40%rastvor cijanidne kiseline?

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milicaradenkovic - Fri, 02/14/2014 - 16:59

p*V=n*R*T

n=p*V/R*T=100*10(-3)*120*10(3)/8.314*293.15=4.92 mol

m=n*Mr=4.92*27=132.84 g HCN

w(HCN)=m(HCN)/mr

mr=m(HCN)/w(HCN)=132.84/0.4=332.1 g

m(vode)=mr-m(HCN)=332.1-132.84=199.26 g

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mika - Fri, 02/14/2014 - 17:59

hvala puno.