Answer: B, D
Kb=1000×300008.314×3502×100=3.39 K kg mol−1, so the last statement (about 3.4) holds. Kb∝1/ΔvapH, so doubling ΔvapH halves it. Kb∝M1, so doubling the solvent molar mass doubles Kb and the first statement is false. For 0.50 mol kg−1: ΔTb=0.50×3.39=1.70 K, giving 351.7 K, not 351.4 K.