A solution containing 10.0 g of an unknown liquid and 100.0 g of water has a freezing point of -3.55 °C. Given [tex]$K_f = 1.86\,°C/m$[/tex] for water, the molar mass of the unknown liquid is:

A. 62.1 g/mol
B. 98.2 g/mol
C. 52.6 g/mol
D. 69.0 g/mol

1. Formula weight of water = 2(1.008 g/mol) + 16.00 g/mol = 18.016 g/mol
2. Moles of water = mass of water/molar mass of water = 100 g/18.016 g/mol = 5.550 mol
Calculate the moles of the unknown liquid, using its formula weight:
1. Moles of the unknown liquid = mass of unknown liquid/molar mass of the unknown liquid
2. Moles of the unknown liquid = 10.0 g/molar mass of the unknown liqui

Next, we can determine the molecular formula of the unknown liquid:

Calculate the molality of the solution:
1. Molality (m) = moles of solute/mass of solvent (in kg)
2. Given that the mass of water is 100.0 g and the density of water at 25°C is approximately 1 g/mL, the mass of water in kg is:
3. Mass of water (kg) = 100.0 g/1000 g/kg = 0.100 kg
4. Molality (m) = (moles of the unknown liquid + moles of water)/0.100 kg = (10.0 g/molar mass of the unknown liquid + 5.550 mol)/0.100 kg
Use the freezing point depression formula to solve for the molar mass of the unknown liquid:
1. ∆T = Kf * m
2. Given that the freezing point depression (∆T) is -3.55°C and the cryoscopic constant (Kf) is 1.86°C/m, we can solve for the molar mass:
3. -3.55°C = 1.86°C/m * [(10.0 g/molar mass of the unknown liquid + 5.550 mol)/0.100 kg]

By solving this equation, we can find the molar mass of the unknown liquid.

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A solution containing 10.0 g of an unknown liquid and 100.0 g of water has a freezing point of -3.55 °C. Given [tex]$K_f = 1.86\,°C/m$[/tex] for water, the molar mass of the unknown liquid is:A. 62.1 g/mol B. 98.2 g/mol C. 52.6 g/mol D. 69.0 g/mol