Question

A. The freezing point of water H₂O is 0.00°C at 1 atmosphere.

How many grams of iron(II) nitrate (179.9 g/mol), must be dissolved in 228.0 grams of water to reduce the freezing point by 0.300°C ?

_____ g iron(II) nitrate.

B. The freezing point of water is 0.00°C at 1 atmosphere.

If 14.42 grams of ammonium acetate, (77.10 g/mol), are dissolved in 165.8 grams of water ...

The molality of the solution is _______ m.

The freezing point of the solution is _____°C.

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Answer 1

Freezing point is the temperature at which solid and liquid phases of a substance are in equilibrium with each other.

Freezing point of pure solvent decreases by the addition of non volatile solute.

The difference between the freezing points of pure solvent and solution is called depression in freezing point and is denoted by ∆T_f

Depression in freezing point is related to the Molality of the solution.

∆T_f = i * K_{​​​​​f} * m

_{​​​​​​}where i is the Van't Hoff factor

Kf is the Molal elevation constant.

m is the Molality of the solution

A)

Given that freezing point of pure water= 0°C

Freezing point of solution = -0.3°C

Weight of solvent= 228 g

Van't Hoff factor for Iron (II) Nitrate = 2

0°C - (-0.3°C) =

2 * (1.86 °C - kg/mol) * w * 1000/ ( 180 g/mol * 228)

Weight of ferrous nitrate = 0.3 °C * 180 g/mol * 228 /( 2 * 1.86 *1000)

= 3.31 g

B)

Molality of Ammonium acetate = weight of solute*1000/( Molar mass of solute*Volume of the solution)

= 14.42 g*1000/(77.1 g/mol * 165.8 )

= 1.13 m

Freezing point of solution is given by

Van't Hoff factor for ammonium acetate = 2 as it gives 2 ions on dissociation

0°C - T = 2 *( 1.86 °C-Kg/mol ) * 1.13 m

T = 0°C - 4.2°C = -4.2 °C