"\u0394T = k_bm \\\\\nk_b = +6.26 \\;\u00b0C\/m \\\\\nM(biphenyl) = 12 \\times 12 + 10 =154 \\;g\/mol \\\\\nn(biphenyl) = \\frac{12.5}{154}=0.0811 \\;mol \\\\\nMolality \\;m= \\frac{0.0811}{0.1} = 0.811 \\;mol\/kg \\\\\n\u0394T = 6.26 \\times 0.811 = 5.08 \\;\u00b0C \\\\\nT = 156 + 5.08 = 161.08 \\;\u00b0C"
Comments
Leave a comment