A quantity of \(10 \mathrm{~g}\) of acetic acid is dissolved in \(100 \mathrm{~g}\) of each of the following solvents. In which solvent, the molality of solution is maximum? Assume no any dissociation or association of acetic acid in the solvent. (a) Water (b) Ethanol (c) Benzene (d) Same in all solvents

Short Answer

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a) Water b) Ethanol c) Benzene d) Same in all solvents Answer: d) Same in all solvents

Step by step solution

01

Calculate the moles of acetic acid

First, we need to find the moles of acetic acid. The molecular weight of acetic acid (CH3COOH) is: MW(CH3COOH) = 12(2) + 1(4) + 16(2) + 1(1) = 60 g/mol So, moles of acetic acid = mass / molecular weight = \(\frac{10}{60}\) = \(\frac{1}{6}\) mol
02

Convert the mass of each solvent to kg

We are given that 100 g of each solvent is used. To determine molality, we need the mass of the solvent in kg. Water: 100 g = 0.1 kg Ethanol: 100 g = 0.1 kg Benzene: 100 g = 0.1 kg
03

Calculate the molality of the solution with each solvent

Now, we can calculate the molality of the solution with each solvent: Molality (m) = moles of solute / mass of solvent (in kg) (a) Water Molality (m_water) = \(\frac{1/6}{0.1}\) = 1.67 mol/kg (b) Ethanol Molality (m_ethanol) = \(\frac{1/6}{0.1}\) = 1.67 mol/kg (c) Benzene Molality (m_benzene) = \(\frac{1/6}{0.1}\) = 1.67 mol/kg In each case, the molality of the solution is 1.67 mol/kg.
04

Compare the molalities and choose the maximum molality solvent

Since the molalities of all three solutions are the same, the maximum molality is the same in all solvents. Hence, the solution is (d) Same in all solvents.

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