rastvori

Submitted by todorovic on Sat, 01/25/2014 - 22:53
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Pomesaju se 40cm kubnih 20% HNO3 gustine 1,12g/cm kubni i 36 cm kubnih 15% NaOH gustine 1.17g/cm kubni. Izracunati masevni udeo nagradjene soli u rastoru.

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milicaradenkovic - Sun, 01/26/2014 - 08:55

HNO3+NaOH---------------->NaNO3+H2O

mr=V*ro=40*1.12=44.8 g

ms=w*mr=44.8*0.2=8.96 g HNO3

n=ms/Mr=8.96/63=0.142 mol HNO3

mr1=V1*ro1=36*1.17=42.12 g

ms1=w1*mr1=42.12*0.15=6.318 g

n1=ms1/Mr=6.318/40=0.158 mol NaOH u  visku je NaOH pa je kolicina soli

n(NaNO3)=n(HNO3)=0.142 mol

ms2=n(NaNO3)*Mr=0.142*85=12.07 g

mru.=mr+mr1=44.8+42.12=86.92 g

w(NaNO3)=ms2/mru.=12.07/86.92=0.139=13.9 %