Sebanyak 100 ml asam sulfat 49% (massa jenis = 1,4 kg/liter) dilarutkan dalam 100 ml air (massa jenis 1 kg/liter), hitunglah kemolalan dan fraksi mol H2SO4 dalam larutan itu. (Mr H2SO4 = 98)
0r4n94l4y
Massa Larutan = 100 ml × 1,4 gr/ml = 140 gram massa H2SO4 = (49/100) × 140 gram = 68,6 gram massa H2O = 140-68,6 = 71,4 gram ditambah 100 gram air jadi massa air = 100+71,4= 171,4 gram m H2SO4 = (68,6/98) × (1000/171,4) = (7/10) × (5000/857) = 4,084014002 m X H2SO4 = {(68,6/98) / [(171,4/18) + (68,6/98)]} = [0,7/(9,52 + 0,7)] = (0,7/10,2) = 0,0685
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0r4n94l4y
Massa Larutan = 100 ml × 1,4 gr/ml = 140 grammassa H2SO4 = (49/100) × 140 gram = 68, 69 grammassa H2O = 140-68,69 = 71,4 gramditambah 100 gram air jadi massa air = 100+71,4= 171,4 gramm = (68,6/98)×(1000/171,4)= 4,08 m
X = = {(68,6/98) / [(174/18) + (68,6/98)]} = 0,7/ 9,52 + 0,7 = 0,0685
0r4n94l4y
baru diedit karena kesalahan itu awalnya T_T
massa H2SO4 = (49/100) × 140 gram = 68,6 gram
massa H2O = 140-68,6 = 71,4 gram
ditambah 100 gram air jadi massa air = 100+71,4= 171,4 gram
m H2SO4 = (68,6/98) × (1000/171,4) = (7/10) × (5000/857) = 4,084014002 m
X H2SO4 = {(68,6/98) / [(171,4/18) + (68,6/98)]} = [0,7/(9,52 + 0,7)] = (0,7/10,2) = 0,0685