Question

A non-volatile solute has a molecular weight of 180 g mol\(^{-1}\). Assume that the solute does not associate or dissociate in water, and the boiling-point constant (ebullioscopic constant) of water is 0.51 K kg mol\(^{-1}\).
The amount (in g) of solute added to 500 g of water to elevate the boiling point by 0.153 K is ............... (answer in integer)

Hint:

When calculating the amount of solute to elevate the boiling point, use the equation \( \Delta T_b = K_b \times m \), where molality \( m \) is moles of solute per kilogram of solvent.

Answer 1

The elevation in boiling point (\( \Delta T_b \)) is related to the amount of solute by the equation: \[ \Delta T_b = K_b \times m \] where \( K_b \) is the ebullioscopic constant, and \( m \) is the molality of the solution. First, calculate the molality: \[ m = \frac{\Delta T_b}{K_b} = \frac{0.153}{0.51} = 0.3 \, {mol/kg} \] Next, calculate the number of moles of solute needed for 500 g of water (0.5 kg): \[ {moles of solute} = 0.3 \times 0.5 = 0.15 \, {mol} \] Now, calculate the mass of the solute using its molar mass: \[ {mass of solute} = 0.15 \times 180 = 27 \, {g} \] Thus, the amount of solute needed is 27 grams.