Question

1. What is the freezing point of a solution in which 3.5 grams of sodium chloride (NaCl) is added to 230 g of water? Kip water is 1.853°C/m

full explanation please ​

Answer 1

To determine the freezing point of the solution, we need to use the freezing point depression equation:

ΔTf = Kf x molality

where:

ΔTf = change in freezing point

Kf = freezing point depression constant (1.853°C/m for water)

molality = moles of solute per kilogram of solvent

First, we need to find the molality of the solution, which is the number of moles of solute per kilogram of solvent. We'll start by finding the number of moles of NaCl:

m (mass of NaCl) = 3.5 g

M (molar mass of NaCl) = 58.44 g/mol

n (moles of NaCl) = m/M = 3.5 g ÷ 58.44 g/mol = 0.0598 mol

Next, we'll find the mass of the water:

m (mass of water) = 230 g

Now we can use molality to find ΔTf:

molality = n (moles of solute) ÷ mass of solvent (in kg)

molality = 0.0598 mol ÷ 0.23 kg = 0.2598 mol/kg

ΔTf = Kf x molality

ΔTf = 1.853°C/m x 0.2598 mol/kg = 0.481°C

The freezing point of pure water is 0°C, so the freezing point of the solution will be:

Freezing point of solution = Freezing point of pure water - ΔTf

Freezing point of solution = 0°C - 0.481°C

Freezing point of solution = -0.481°C

Therefore, the freezing point of the solution is -0.481°C.