1 kg of 0.75 molal aqueous solution of sucrose can be cooled up to -4°C before freezing. The amount

Q: Why does only water freeze and not sucrose?

When an aqueous solution freezes, the solvent molecules (water) arrange themselves into a pure crystalline lattice. The solute molecules (sucrose) are structurally incompatible with this ice lattice and are excluded, remaining in the liquid phase.

Q: Why does the freezing point keep dropping as ice forms?

As pure water turns into ice, the amount of liquid solvent decreases. Since the amount of solute remains constant, the concentration (molality) of the remaining solution increases. A higher molality means a greater depression in freezing point.

Q: How did we get the equation $256.5x = 10^6 - 1000x$?

This comes from cross-multiplying the molality equation: $0.75 = \frac{(1000 - x) \times 1000}{342x}$. Multiplying both sides by $342x$ gives $256.5x = (1000 - x) \times 1000$, which expands to $256.5x = 10^6 - 1000x$.

Q: Can we use molarity instead of molality here?

No. Colligative properties like freezing point depression depend on the ratio of solute particles to solvent molecules. Molality is mass-based and temperature-independent, making it the correct concentration term to use when temperature changes are involved.

Q: What if the solute was NaCl instead of sucrose?

NaCl is a strong electrolyte that dissociates into $\text{Na}^+$ and $\text{Cl}^-$ ions. We would need to multiply the molality by the van't Hoff factor ($i \approx 2$) in the $\Delta T_f$ equation, because colligative properties depend on the total number of dissolved particles.

Question

1 kg of 0.75 molal aqueous solution of sucrose can be cooled up to $-4^{\circ}	ext{C}$ before freezing. The amount of ice (in g) that will be separated out is ......... . (Nearest integer)
[Given, $K_f(	ext{H}_2	ext{O}) = 1.86 	ext{ K kg mol}^{-1}$]

Sigma Percentile · Accepted Answer

### The Chilling Setup

Imagine you have a beaker containing exactly $1 	ext{ kg}$ of an aqueous sucrose solution. The molality of this solution is given as $0.75 	ext{ m}$.

Before we can even think about freezing it, we need to know exactly what's inside this beaker. How much of that $1 	ext{ kg}$ is water, and how much is sucrose?

### Decoding the Initial Composition

Let's assume the mass of the water is $x 	ext{ g}$. Since the total mass is $1000 	ext{ g}$, the mass of the sucrose must be $(1000 - x) 	ext{ g}$.

We know the formula for molality: it's the moles of solute divided by the mass of the solvent in kilograms. The molar mass of sucrose ($	ext{C}_{12}	ext{H}_{22}	ext{O}_{11}$) is $342 	ext{ g mol}^{-1}$.

Plugging our variables into the molality equation, we get:

$0.75 = \frac{\frac{1000 - x}{342}}{\frac{x}{1000}}$

Solving this algebraic equation gives us $x = 795.86 	ext{ g}$. This is our initial mass of water.

Now, we can also find the exact moles of sucrose in our solution:

$n_{	ext{sucrose}} = \frac{1000 - 795.86}{342} = 0.5969 	ext{ mol}$

### The Big Freeze

Now comes the fun part. We take this beaker and cool it down to $-4^{\circ}	ext{C}$.

As the temperature drops below the freezing point of the solution, pure water starts to crystallize out as ice. **Crucially, the sucrose does not freeze with it.** It remains dissolved in the remaining liquid water.

Because water is leaving the liquid phase to become ice, the concentration (molality) of the remaining solution increases. This causes the freezing point to drop even further, until it reaches equilibrium at $-4^{\circ}	ext{C}$.

### The Final Calculation

Let's use the depression in freezing point formula to find out how much liquid water is left at $-4^{\circ}	ext{C}$.

$\Delta T_f = K_f 	imes m_{	ext{new}}$

We know $\Delta T_f = 4 	ext{ K}$, and $K_f = 1.86 	ext{ K kg mol}^{-1}$. Let the mass of the remaining liquid water be $w_f 	ext{ kg}$.

$4 = 1.86 	imes \frac{0.5969}{w_f}$

Solving for $w_f$, we get $w_f = 0.2775 	ext{ kg}$, or $277.5 	ext{ g}$.

So, if we started with $795.86 	ext{ g}$ of water and only $277.5 	ext{ g}$ remains as liquid, the rest must have turned into ice!

$	ext{Mass of ice} = 795.86 - 277.5 = 518.36 	ext{ g}$

Rounding to the nearest integer, we get our final answer: **$518 	ext{ g}$**.

Animated Solution for Chemistry - Solutions: 1 kg of 0.75 molal aqueous solution of sucrose can be cooled up to $- 4^{\circ} C$ before freezing. The amount of ice (in g) that will be separated out is ......... . (Nearest integer) [Given, $K_{f} (H_{2} O) = 1.86 K kg mol^{- 1}$ ]

Enter Numerical Value:

Visualized Solution

$I ni t ia l S t a t e o f S o l u t i o n$

$D e f inin g t h e C o m p os i t i o n$

$A ppl y in g M o l a l i t y F or m u l a$

$C a l c u l a t in g I ni t ia l M a ss o f W a t er$

$C a l c u l a t in g M o l es o f S u cr ose$

$C oo l in g t h e S o l u t i o n$

$D e p r ess i o n in F r eez in g P o in t$

$S e tt in g u p t h e F ina l E q u a t i o n$

$C a l c u l a t in g R e mainin g L i q u i d W a t er$

$C a l c u l a t in g M a ss o f I ce S e p a r a t e d$

$T h e W a y F or w a r d$

The Sigma Insight: Colligative Properties

The Chilling Setup

Decoding the Initial Composition

The Big Freeze

The Final Calculation

Similar Questions

A solution contain $62 g$ of ethylene glycol in $250 g$ of water is cooled upto $- 1 0^{\circ} C$ . If $K_{f}$ for water is $1.86 K kg mol^{- 1}$ , then amount of water (in g) separated as ice is

40 g of glucose (Molar mass = 180) is mixed with 200 mL of water. The freezing point of solution is ............ K. (Nearest integer) [Given, $K_{f} = 1.86 K kg mol^{- 1}$ , density of water $= 1.00 g cm^{- 3}$ , freezing point of water $= 273.15 K$ ]

How much amount of NaCl should be added to 600 g of water ( $ρ = 1.00 g/mL$ ) to decrease the freezing point of water to $- 0. 2^{\circ} C$ ? ...... (The freezing point depression constant for water $= 2 K kg mol^{- 1}$ )

83 g of ethylene glycol dissolved in 625 g of water. The freezing point of the solution is ………K. (Nearest integer) [Use, molal freezing point depression constant of water $= 1.86 K kg mol^{- 1}$ , Freezing point of water $= 273 K$ and Atomic masses : $C = 12.0 u, O = 16.0 u, H = 1.0 u$ ]

$K_{f}$ for water is $1.86 K kg mol^{- 1}$ . If your automobile radiator holds $1.0 kg$ of water, then how many grams of ethylene glycol ( $C_{2} H_{6} O_{2}$ ) must you add to get the freezing point of the solution lowered to $- 2. 8^{\circ} C$ ?

Ethylene glycol is used as an antifreeze in a cold climate. Mass of ethylene glycol which should be added to $4 kg$ of water to prevent it from freezing at $- 6^{\circ} C$ will be ( $K_{f}$ for water $= 1.86 K kg mol^{- 1}$ and molar mass of ethylene glycol $= 62 g mol^{- 1}$ )

If sodium sulphate is considered to be completely dissociated into cations and anions in aqueous solution, the change in freezing point of water ( $Δ T_{f}$ ), when $0.01$ mole of sodium sulphate is dissolved in $1 kg$ of water, is ( $K_{f} = 1.86 K kg mol^{- 1}$ )

Molal depression constant for a solvent is $4.0 K kg mol^{- 1}$ . The depression in the freezing point of the solvent for $0.03 mol kg^{- 1}$ solution of $K_{2} SO_{4}$ is (Assume complete dissociation of the electrolyte)

$C_{6} H_{6}$ freezes at $5. 5^{\circ} C$ . The temperature at which a solution $10 g$ of $C_{4} H_{10}$ in $200 g$ of $C_{6} H_{6}$ freeze is ......... $^{\circ} C$ . (The molal freezing point depression constant of $C_{6} H_{6}$ is $5.1 2^{\circ} C / m$ .)

Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Initial State of Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Defining the Composition

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Applying Molality Formula

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Initial Mass of Water

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Moles of Sucrose

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Cooling the Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Depression in Freezing Point

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Setting up the Final Equation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Remaining Liquid Water

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Mass of Ice Separated

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward