Question

38. What is the expected boiling point of a solution prepared by dissolving 6.37 g of sodium iodide (Nal) in 48.6 g of water (H 0)? K'b (ater)-0.512"Cm a) 0.90°C d) 103.26"C b) 100.45°C e) None of the above c) 100.90°C

Answer 1

Answer: c) 100.90°C

Step-by-step explanation:

Elevation in boiling point is given by:

`\Delta T_b=i* K_b* m`

`\Delta T_b=T_b-T_b^0` = elevation in boiling point

i= vant hoff factor = 2 (for NaI[/tex]

`NaI\rightarrow Na^++I^-`

`K_b` = boiling point constant =
`0.512^0C/m`

m= molality

`\Delta T_b=i* K_b* \frac{\text{mass of solute}}{\text{molar mass of solute}* \text{weight of solvent in kg}}`

Weight of solvent i kg =48.6 g= 0.0486kg

`\Delta T_b=0.9^0C`

`(T_b-1000)^0C=0.9^0C`

`T_b=100.9^0C`

Thus the boiling point of the solution will be
`100.90^0C`.