Respuesta :
Answer:
The new freezing point of water is -0.476°C.
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
[tex]m=\dfrac{mass\ of\ CaBr_{2}}{molar\ mass\ of\ CaBr_{2}\times mass\ of\ water}[/tex]
Put the value into the formula
[tex]m=\dfrac{236.5}{200\times4.62}[/tex]
[tex]m=0.256\ mol/kg[/tex]
[tex]m=0.256\ m[/tex]
We need to calculate the depression in freezing point of water
Using formula of freezing point
[tex]\Delta T=m\times k[/tex]
Where, k = depression constant
m = molality of solution
Put the value into the formula
[tex]\Delta T=0.256\times1.86[/tex]
[tex]\Delta T=0.476^{\circ}C[/tex]
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
[tex]\Delta T'=0-\Delta T[/tex]
Put the value into the formula
[tex]\Delta T'=0-0.476[/tex]
Hence, The new freezing point of water is -0.476°C.
