Question

The normal boiling point of Ethyl acetate is \(77^{\circ}\text{C}\). A solution of non-volatile non-electrolyte solute in ethyl acetate boils at \(78^{\circ}\text{C}\). \(K_b\) for ethyl acetate is \(2.77 ^{\circ}\text{C kg mol}^{-1}\). What is the molality of the solution ?

• A. \(0.361 \text{m}\)
• B. \(0.052 \text{m}\)
• C. \(0.075 \text{m}\)
• D. \(0.25 \text{m}\)

Hint:

Use delta Tb = Kb times m.

Answer 1

The Correct Option is A

Step 1: Find elevation
Boiling point of the solution is \(78^{\circ}C\) and that of pure solvent is \(77^{\circ}C\). So \(\Delta T_b = 78 - 77 = 1\ \text{K}\).

Step 2: Apply the formula
For a non-volatile non-electrolyte, \(\Delta T_b = K_b m\).
\[ m = \frac{\Delta T_b}{K_b} = \frac{1}{2.77} = 0.361\ \text{mol/kg} \]

Step 3: Check the options
\(0.052\), \(0.075\) and \(0.25\) m would give elevations of \(0.14\), \(0.21\) and \(0.69\) K, none equal to 1 K.

Final Answer:
The molality is 0.361 m. \[ \boxed{\text{(A)}\ 0.361\ \text{m}} \]