What is the molality of a solution of 30.1 g of propanol (CH₃CH₂CH₂OH)in 152 mL water,if the density of water is 1.00 g/mL? A)3.29 m B)0.00330 m C)0.303 m D)0.501 m E)5.01 m

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Molality (m) is defined as the number of moles of solute per kilogram of solvent. First, we need to calculate the moles of propanol (C3H8O). The molar mass of propanol is approximately 60.1 g/mol (3*12 + 8*1 + 16 = 60). Thus, 30.1 g of propanol is 30.1 g / 60.1 g/mol = 0.501 moles. The mass of water (the solvent) is 152 mL * 1.00 g/mL = 152 g, which is 0.152 kg. Molality = moles of solute / kg of solvent = 0.501 moles / 0.152 kg = 3.29 m.

What Is the Molality of a Solution of 30