Question

What is the freezing point (in degrees Celcius) of 4.62 kg of water if it contains 236.5 g of C a B r 2

Answer 1

The new freezing point of water is -0.476°C.

Step-by-step explanation:

Given that,

Mass of water = 4.62 kg

Mass of CaBr₂ = 236.5 g

We need to calculate the molality of solution

Using formula of molality

`m=(mass\ of\ CaBr_(2))/(molar\ mass\ of\ CaBr_(2)* mass\ of\ water)`

Put the value into the formula

`m=(236.5)/(200*4.62)`

`m=0.256\ mol/kg`

`m=0.256\ m`

We need to calculate the depression in freezing point of water

Using formula of freezing point

`\Delta T=m* k`

Where, k = depression constant

m = molality of solution

Put the value into the formula

`\Delta T=0.256*1.86`

`\Delta T=0.476^(\circ)C`

We know that,

The standard freezing point of water is 0°C

We need to find the new freezing point of water

Using formula for freezing point of water

Freezing point of 4.62 kg of water = freezing point of water-depression in freezing point of water

`\Delta T'=0-\Delta T`

Put the value into the formula

`\Delta T'=0-0.476`

Hence, The new freezing point of water is -0.476°C.