Question

question 16 of 30
what is the predicted change in the boiling point of water when 4.00 g of
barium chloride (bacl₂) is dissolved in 2.00 kg of water?

k_b of water = 0.51°c/mol
molar mass bacl₂ = 208.23 g/mol
i value of bacl₂ = 3

a. 0.0016°c
b. -0.0049°c
c. 0.015°c
d. -1.0°c

Explanation:

Step1: Calculate moles of \( BaCl_2 \)

Moles = mass / molar mass. Mass of \( BaCl_2 = 4.00 \, g \), molar mass = \( 208.23 \, g/mol \).
\( n = \frac{4.00 \, g}{208.23 \, g/mol} \approx 0.01921 \, mol \)

Step2: Calculate molality (m)

Molality = moles of solute / kg of solvent. Solvent (water) mass = \( 2.00 \, kg \).
\( m = \frac{0.01921 \, mol}{2.00 \, kg} \approx 0.009605 \, mol/kg \)

Step3: Use boiling point elevation formula \( \Delta T_b = i \cdot K_b \cdot m \)

\( i = 3 \), \( K_b = 0.51^\circ C/mol \), \( m = 0.009605 \, mol/kg \).
\( \Delta T_b = 3 \times 0.51^\circ C/mol \times 0.009605 \, mol/kg \approx 0.015^\circ C \)

Answer:

C. \( 0.015^\circ C \)