The correct answer for the question that is being presented above is this one:
Given that:
delta Tb = Kbm Kb H2O = 0.52 degrees C/m
<span>delta Tf = Kfm Kf H2O = 1.86 degrees C/m
</span>
We need to know the formula for Molality.
molality = mol solute / kg solvent
<span>We are given the amount of solute in grams
Since amount of solute is given in moles, we have to convert 25 g NaCl to moles. Divide by molar mass. </span>
<span>25 g NaCl / 58.5 g/mol = 0.427 mol </span>
<span>Then, use the formula for molality. </span>
<span>molality = mol solute / kg solvent </span>
<span>= 0.427 / 1 </span>
<span>= 0.427 m </span>
<span>Use now the formula to get the boiling point.</span>
<span>delta Tb = Kbm </span>
<span>= (0.52)(0.427) </span>
<span>= 0.22C </span>
Answer: It is approximately 24.985837
Answer:
The answer to your question is V = 0.108 L or 108 ml
Explanation:
Data
Volume = ?
mass = 0.405 g
Temperature = 273°K
Pressure = 1 atm
Process
1.- Convert mass of Kr to moles
83.8 g of Kr -------------------- 1 mol
0.405 g ------------------- x
x = (0.405 x 1) / 83.8
x = 0.0048 moles
2.- Use the Ideal gas law to solve this problem
PV = nRT
- Solve for V
V = nRT / P
- Substitution
V = (0.0048)(0.082)(273) / 1
- Simplification
V = 0.108 / 1
- Result
V = 0.108 L
