Larutan 18% berat glokosa ( mr = 180) dalam air akan mendidih pada suhu ... c (kb air =0,41).... Question from @aisyahnarf1

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Larutan 18% berat glokosa ( mr = 180) dalam air akan mendidih pada suhu ... c (kb air =0,41)

andriprasetio57 Anggap air 100ml
rho glukosa=1

mGLUKOSA=
18/100×100=18

tb=molalitasxKb
m/mr × 1000/pelarut × kb
18/180×1000/100x0,41
=0,41

100+0,41=100,41

2 votes Thanks 1

fatihahalhusna $\Delta T_b =K_b *\frac{\% *1000}{Mr*(100-\%)}=0,41 *\frac{18 *1000}{180*(100-18)}=0,41 *\frac{100}{82}=0,5^0C\\\\Titik~didih~T_b=\boxed{100,5^0C}$

3 votes Thanks 6