mohon bantuannya kak

a) C = 40%
H = 6,6%
O = 100-40-6,6 = 53,4%
C : H : O = 40/12 : 6,6/1 : 53,4/16 = 3,33 : 6,6 : 3,33 = 1 : 2 : 1
Mk rumus empiris = (CH2O)n

b) ΔTf = massa/Mr × 1000/massa air × Kf
1,24 = 30/Mr × 1000/500 × 1,86
1,24 = 111,6/Mr
Mr = 111,6/1,24 = 90

c) (CH2O)n = 90
(12+2+16)n = 90
30n = 90
n = 3
Mk rumus molekulnya = C3H6O3