How many grams of #NH_4OH# must be dissolved in 75 g of water to lower the freezing point -3.0°C?

Following Michael's wise counsel, the solution is provided below, but if your professor's real concern was that paper chemistry is rarely useful, then this is the real issue.

Remember,

wherein

Thus, we require,

of that substance would, in theory, reduce the freezing point by the amount mentioned.

You need to dissolve 45 grams of NH₄OH in 75 grams of water to lower the freezing point by -3.0°C.

To calculate the grams of NH₄OH required to lower the freezing point of 75 g of water by -3.0°C, we can use the equation:

ΔTf = i * Kf * m

Where: ΔTf = change in freezing point (in °C) i = van't Hoff factor (for NH₄OH, i = 2) Kf = freezing point depression constant for water (1.86 °C kg/mol) m = molality of the solution (mol solute/kg solvent)

First, we calculate the change in freezing point: ΔTf = -3.0°C

Now, we rearrange the equation to solve for m: m = ΔTf / (i * Kf)

Substituting the known values: m = (-3.0°C) / (2 * 1.86 °C kg/mol)

m ≈ -0.806 mol/kg

Now, we need to find the number of moles of NH₄OH required to achieve this molality in 75 g of water: moles of NH₄OH = m * mass of water (in kg)

mass of water = 75 g = 0.075 kg

moles of NH₄OH = -0.806 mol/kg * 0.075 kg

moles of NH₄OH ≈ -0.060 mol

Since we cannot have a negative number of moles, it suggests that NH₄OH must be added to the water to reach the desired freezing point depression.

The molar mass of NH₄OH is approximately 35.05 g/mol.

Finally, we find the grams of NH₄OH required: grams of NH₄OH = moles of NH₄OH * molar mass of NH₄OH

grams of NH₄OH ≈ -0.060 mol * 35.05 g/mol

grams of NH₄OH ≈ -2.103 g

So, approximately 2.103 grams of NH₄OH must be dissolved in 75 g of water to lower the freezing point by -3.0°C.