Question

Calculate the freezing point depression for 1 kilogram of sodium chloride (NaCl) in 5 kilograms of

water. The Ky of water is -1.86 °C/m.

Answer 1

`\\ \rm\dashrightarrow \Delta T_f=ik_fm`

• m is molality

Moles of NaCl

• 1000/23
• 43.4mol

Molality

• 43.4/5
• 8.68

`\\ \rm\dashrightarrow \Delta T_f=i(-1.86)(8.68)`

`\\ \rm\dashrightarrow \Delta T_f=-16.14i`

i is vant Hoff factor

For NaCl vant Hoff factor is 2( Actually 1.9)

`\\ \rm\dashrightarrow \Delta T_f=-16.14(2)=-32.28°C`

Answer 2

Moles of NaCl

`\\ \tt{:}\dashrightarrow (1000g)/(23g/mol)`

`\\ \tt{:}\dashrightarrow 43.4mol`

Find Molality

`\boxed{\sf Molality=(Moles\:of\:solute)/(Volume \:of\: solution\:in\:kg)}`

• Volume of solution=1kg+5kg=6kg

`\\ \tt{:}\dashrightarrow m=(43.4)/(6)=7.2m`

Vant Hoff factor

`\\ \tt{:}\dashrightarrow NaCl\longrightarrow Na^++Cl^-`

• i=2

Freezing point depression

`\\ \tt{:}\dashrightarrow \Delta T_f=iK_fm`

`\\ \tt{:}\dashrightarrow \Delta T_f=2(-1.86)(7.2)`

`\\ \tt{:}\dashrightarrow \Delta T_f=-26.7°C`