Answer:
We should add 330.2 grams of NaCl
Explanation:
Step 1: Data given
Volume of water = 1.46 L = 1460 mL
Density of water = 1.00 g/ mL
The solution freezes at -14.4 °C
Kf = 1.86 °C / m
Step 2: Calculate molality
ΔT = i*Kf*m
⇒ with ΔT = The change in temperature = -14.4 °C
⇒ with i = The van't Hoff factor for NaCl = 2
⇒ with Kf = 1.86°C/m
⇒ with m = the molality = TO BE DETERMINED
m = ΔT / (i*Kf)
m = 14.4 / (2*1.86)
m =3.87 molal
Step 3: Calculate moles of NaCl
Molality = moles Nacl / mass Water
3.87 molal = moles NaCl / 1.46 kg
Moles NaCl = 3.87 * 1.46
Moles NaCl = 5.65 moles
Step4: Calculate mass of NaCl
Mass NaCl = Moles NaCl * molar mass NaCl
Mass Nacl =5.65 moles * 58.44 g/mol
Mass NaCl =330.2 grams
We should add 330.2 grams of NaCl
