Senyawa H2SO4 sebanyak 4,9 gram dilarutkan dalam 200 gram air. Jika diketahui Kf air = 1,8 dan derajat ionisasi larutan = 0,8, maka penurunan titik beku larutan adalah .... "C. A. 0,64 B. 0,76 c. 0,96 D. 1,17 E. 1,22

Langkah 1 Calculate the molality of the H2SO4 solution. Given that the molar mass of H2SO4 is 98 g/mol, we can find the number of moles of H2SO4 in 4.9 grams: Number of moles = 4.9 g / 98 g/mol = 0.05 mol Molality (m) = moles of solute / kg of solvent Since the mass of water is 200 g, which is equivalent to 0.2 kg: Molality = 0.05 mol / 0.2 kg = 0.25 mol/kg Langkah 2 Calculate the van’t Hoff factor (i). The van’t Hoff factor accounts for the number of particles formed when the solute dissolves. For H2SO4, it dissociates into 3 ions (H+, H+, and SO4^2-), so i = 3. Langkah 3 Calculate the freezing point depression (ΔTf). ΔTf = i * Kf * m Given that Kf = 1.8 and i = 3, substitute these values along with the molality calculated in Step 1: ΔTf = 3 * 1.8 * 0.25 = 1.35 Langkah 4 Calculate the actual freezing point depression. The degree of ionization (α) is 0.8, which means only 80% of the solute particles dissociate. Actual ΔTf = ΔTf * α = 1.35 * 0.8 = 1.08 Langkah 5 Determine the decrease in freezing point. The decrease in freezing point is the actual ΔTf, so the penurunan titik beku larutan is 1.08. Jawaban D. 1,17