Math  /  Data & Statistics

QuestionWhen 30.0 g of glycine (C2H5NO2)\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NO}_{2}\right) are dissolved in 1000 g of a certain mystery liquid XX, the freezing point of the solution is 6.80C6.80{ }^{\circ} \mathrm{C} less than the freezing point of pure XX. Calculate the mass of potassium bromide that must be dissolved in the same mass of XX to produce the same depression in freezing point. The van't Hoff factor i=1.78i=1.78 for potassium bromide in XX. Be sure your answer has a unit symbol, if necessary, and is rounded to the correct number of significant digits. \square ×10\square \times 10 \square - \square

Studdy Solution

STEP 1

What is this asking? We need to find out how much potassium bromide we need to dissolve in liquid *X* to lower its freezing point by the same amount as when we dissolved glycine in it. Watch out! Don't forget that potassium bromide breaks into ions, while glycine doesn't, so we need to account for that using the van't Hoff factor!

STEP 2

1. Calculate the molality of the glycine solution.
2. Calculate the molal freezing-point depression constant of *X*.
3. Calculate the molality of the KBr solution.
4. Calculate the mass of KBr.

STEP 3

First, let's **calculate the molar mass of glycine** C2H5NO2\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NO}_{2}.
We've got 2 carbons, 5 hydrogens, 1 nitrogen, and 2 oxygens.
So that's (212.01)+(51.01)+14.01+(216.00)=75.07 g/mol(2 \cdot 12.01) + (5 \cdot 1.01) + 14.01 + (2 \cdot 16.00) = 75.07 \text{ g/mol}.
Awesome!

STEP 4

Now, we can **find the number of moles of glycine** we have.
We've got 30.0 g\text{30.0 g} of glycine, and we just figured out that one mole weighs 75.07 g\text{75.07 g}.
So, we have 30.0 g÷75.07 g/mol=0.3996 mol30.0 \text{ g} \div 75.07 \text{ g/mol} = 0.3996 \text{ mol} of glycine.
Sweet!

STEP 5

We know we dissolved the glycine in 1000 g\text{1000 g} of liquid *X*, which is the same as 1.000 kg\text{1.000 kg}. **Molality** is moles of solute per kilogram of solvent.
So, the molality of our glycine solution is 0.3996 mol÷1.000 kg=0.3996 m0.3996 \text{ mol} \div 1.000 \text{ kg} = \mathbf{0.3996} \ \mathbf{m}.
Boom!

STEP 6

The **freezing point depression** is given by the formula ΔTf=iKfm\Delta T_f = i \cdot K_f \cdot m, where ΔTf\Delta T_f is the change in freezing point, ii is the van't Hoff factor, KfK_f is the molal freezing-point depression constant, and mm is the molality.
For glycine, the van't Hoff factor is just 1\mathbf{1} because it doesn't break into ions in solution.

STEP 7

We know the freezing point was lowered by 6.80C\text{6.80}^\circ \text{C}, and we just calculated the molality.
So, we can **solve for** KfK_f!
We have 6.80C=1Kf0.3996 m6.80^\circ \text{C} = 1 \cdot K_f \cdot 0.3996 \text{ m}.
Dividing both sides by 0.3996 m0.3996 \text{ m} gives us Kf=17.0C/mK_f = 17.0^\circ \text{C/m}.
Excellent!

STEP 8

Now, we want to find the molality of the KBr solution that gives the same freezing point depression.
We know ΔTf=6.80C\Delta T_f = 6.80^\circ \text{C}, i=1.78i = 1.78 for KBr, and we just found Kf=17.0C/mK_f = 17.0^\circ \text{C/m}.
Let's **plug those values** back into our freezing point depression formula: 6.80C=1.7817.0C/mm6.80^\circ \text{C} = 1.78 \cdot 17.0^\circ \text{C/m} \cdot m.

STEP 9

Multiplying 1.781.78 by 17.0C/m17.0^\circ \text{C/m} gives us 30.3C/m30.3^\circ \text{C/m}.
So, we have 6.80C=30.3C/mm6.80^\circ \text{C} = 30.3^\circ \text{C/m} \cdot m.
Dividing both sides by 30.3C/m30.3^\circ \text{C/m} gives us m=0.225 mm = \mathbf{0.225} \ \mathbf{m} for the KBr solution.
Perfect!

STEP 10

We know the molality of the KBr solution and the mass of the solvent (1.000 kg\text{1.000 kg}).
We can use this to **find the number of moles of KBr**: 0.225 m=moles of KBr÷1.000 kg0.225 \text{ m} = \text{moles of KBr} \div 1.000 \text{ kg}.
Multiplying both sides by 1.000 kg\text{1.000 kg} gives us 0.225 mol\text{0.225 mol} of KBr.
Right on!

STEP 11

The **molar mass of KBr** is 39.10+79.90=119.00 g/mol39.10 + 79.90 = 119.00 \text{ g/mol}.
To find the mass of KBr, we multiply the number of moles by the molar mass: 0.225 mol119.00 g/mol=26.8 g0.225 \text{ mol} \cdot 119.00 \text{ g/mol} = \mathbf{26.8} \ \mathbf{g}.
Fantastic!

STEP 12

We need 26.8 g\text{26.8 g} of potassium bromide to achieve the same freezing point depression.

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