molarna masa

marank's picture

 Rastvor 6.35 grama nekog neelektrolita u 500 ml vode mrzne na -0,456 stepeni C.Izracunati molekulsku masu jedinjenja.(k=1,86Kkg/mol).Hvala.

milicaradenkovic's picture

deltaT=Kb*b*i

i=1 jer je u pitanju neelektrolit

0.456=1.86*b

b=0.456/1.86=0.245 mol/kg

m(vode)=ro*V=500*1=500 g

n=b*m(vode)

n=0.245*0.5=0.1225 mol

Mr=m/n=6.35/0.1225=52 g/mol