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A sugar syrup weighing \(214.2 \mathrm{~g}\), contains \(34.2 \mathrm{~g}\) of water. The molal concentration is (a) \(0.55\) (b) \(5.5\) (c) 55 (d) \(0.1\)

Short Answer

Expert verified
Re-evaluate the exercise setup, as calculations show a molality of \(29.24 \)mol/kg, not matching given options.

Step by step solution

01

Calculate Moles of Solute

First, we determine the solute, which in this case is the sugar. The entire syrup weighs 214.2 g and contains 34.2 g of water, so the weight of the sugar is \(214.2 \, \mathrm{g} - 34.2 \, \mathrm{g} = 180 \, \mathrm{g}\). To find the number of moles of sugar, we need its molar mass. Assuming we are using glucose (a common sugar) with a molar mass of \(180.16 \, \mathrm{g/mol}\), the number of moles of glucose is \(\frac{180 \, \mathrm{g}}{180.16 \, \mathrm{g/mol}} \approx 1 \, \mathrm{mol}\).
02

Calculate Mass of Solvent in Kilograms

Now, we need the mass of the solvent (water) in kilograms. The problem states there are 34.2 grams of water: \(\frac{34.2 \, \mathrm{g}}{1000} = 0.0342 \, \mathrm{kg}\).
03

Use Molality Formula

Molality is calculated using the formula \(m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}}\). Using the values from Steps 1 and 2: \(m = \frac{1 \, \mathrm{mol}}{0.0342 \, \mathrm{kg}} \approx 29.24 \, \mathrm{mol/kg}\).
04

Interpret Result and Find Correct Answer

The calculated molality of \(29.24 \, \mathrm{mol/kg}\) does not match any of the given options. Upon reviewing the steps and assuming a typographical or computational error in the options or setup, the correct calculations as shown here are based on assumed inputs. Check the course material for accurate result interpretation.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Moles of Solute
In chemistry, the term "moles" refers to the measure of quantity for a substance. Specifically, it relates to the number of molecules or atoms of that substance. When we talk about moles of solute, we are discussing the number of moles of the substance being dissolved. In our exercise, this substance is sugar, likely glucose.
Glucose, a simple sugar, often serves as a model example for exercises like this. To find the moles of solute, it's crucial to know the weight of the solute and its molar mass.
  • The weight of the sugar given: 180 g (after subtracting water weight).
  • The assumed molar mass of glucose: 180.16 g/mol.
By dividing the weight of the glucose by its molar mass, we calculate the moles of the solute:\(\text{Moles of glucose} = \frac{180 \ \mathrm{g}}{180.16 \ \mathrm{g/mol}} \approx 1 \ \mathrm{mol}. \)
What is Mass of Solvent?
The mass of solvent refers to the weight of the substance that dissolves the solute. In many chemical problems, water is used as the solvent because it's a "universal solvent" known for dissolving a diverse range of substances. For our exercise, water, weighing 34.2 grams, is the solvent.
It is necessary to convert the mass of the solvent to kilograms to comply with the molality calculation requirements. Kilogram is the standard unit in these calculations.
  • Given mass of water: 34.2 g
  • Conversion to kilograms: \( \frac{34.2 \ \mathrm{g}}{1000} = 0.0342 \ \mathrm{kg} \)
Knowing the mass of the solvent helps us calculate the molality, a measure of concentration expressed in terms of moles of solute per kilogram of solvent.
Exploring Glucose Molar Mass
The molar mass is a critical property in chemistry and essential to understand for anyone studying chemical reactions and solutions. Molar mass signifies the mass of one mole of a substance. For glucose — a common sugar found in numerous biochemical contexts — the molar mass is approximately 180.16 grams per mole.
This property arises from the sum of the atomic masses of all the atoms within a single glucose molecule, which includes carbon, hydrogen, and oxygen atoms.
  • Each glucose molecule contains: 6 carbon atoms, 12 hydrogen atoms, and 6 oxygen atoms.
  • Calculations involve adding their individual atomic masses based on the periodic table.
  • Atomic masses: Carbon (C): ~12 g/mol, Hydrogen (H): ~1 g/mol, Oxygen (O): ~16 g/mol.
The molar mass is a fundamental aspect in calculating the moles of a substance when working with solutions, like the sugar in our syrup exercise, making it indispensable for molality determinations.

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Most popular questions from this chapter

\(K_{t}\) for water is \(1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1} .\) If your automobile radiator holds \(1.0 \mathrm{~kg}\) of water, how many grams of ethylene glycol \(\left(\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}_{2}\right)\) must you add to get the freezing point of the solution lowered to \(-2.8^{\circ} \mathrm{C}\) ? (a) \(39 \mathrm{~g}\) (b) \(93 \mathrm{~g}\) (c) \(72 \mathrm{~g}\) (d) \(27 \mathrm{~g}\)

An aqueous solution of sucrose \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\), containing \(34.2 \mathrm{~g} / \mathrm{L}\), has an osmotic pressure of \(2.38\) atmospheres at \(17^{\circ} \mathrm{C}\). For an aqueous solution of glucose \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) to be isotonic with this solution, it would have (a) \(18.0 \mathrm{~g} / \mathrm{L}\) (b) \(16.2 \mathrm{~g} / \mathrm{L}\) (c) \(36.6 \mathrm{~g} / \mathrm{L}\) of glucose (d) \(14.0 \mathrm{~g} / \mathrm{L}\)

When mercuric iodide is added to the aqueous solution of potassium iodide, the (a) freezing point is raised (b) freezing point does not change (c) freezing point is lowered (d) boiling point does not change

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A \(0.001\) molal solution of \(\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{4}\right]\) in water had a freezing point depression, \(0.0056^{\circ} \mathrm{C} . \mathrm{K}_{\mathrm{f}}\) for water is \(1.86^{\circ} \mathrm{cm}^{-1} .\) The number of moles of \(\mathrm{AgNO}_{3}\) required to react with one mole of the complex in aqueous solution is

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