• Calculate the freezing point depression (\( \Delta T_f \)):
\( \Delta T_f = 273.15 - 270.65 = 2.5 \, \text{K} \)
• Using the formula for freezing point depression:
\( \Delta T_f = K_f \cdot m = 2.5 = 1.86 \times \frac{n}{0.1} \)
• Solve for moles of methanol (\( n \)):
\( n = 0.1344 \, \text{moles} \)
• Calculate the mass of methanol (\( w \)):
\( w = 0.1344 \times 32 = 4.3 \, \text{g} \)
• Calculate the volume of methanol:
\( \text{Volume} = \frac{4.3}{0.792} = 5.43 \, \text{mL} = 543 \times 10^{-2} \, \text{mL} \)
Answer: \( x = 543 \)
If a substance ‘A’ dissolves in a solution of a mixture of ‘B’ and ‘C’ with their respective number of moles as \(n_a\), \(n_b\), and \(n_c\), the mole fraction of C in the solution is:
A body of mass 1000 kg is moving horizontally with a velocity of 6 m/s. If 200 kg extra mass is added, the final velocity (in m/s) is: