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Chapter 1: Problem 176

The density of a solution prepared by dissolving \(120 \mathrm{~g}\) of urea (mol. Mass \(=60 \mathrm{u}\) ) in \(1000 \mathrm{~g}\) of water is \(1.15 \mathrm{~g} / \mathrm{mL}\). The molarity of this solutions is: [2012] (a) \(1.02 \mathrm{M}\) (b) \(0.50 \mathrm{M}\) (c) \(2.05 \mathrm{M}\) (d) \(1.78 \mathrm{M}\)

Short Answer

Expert verified
The molarity of the solution is approximately 2.05 M, which matches option (c).

Step by step solution

01

Find the mass of the solution

To find the mass of the solution, add the mass of the solute (urea) to the mass of the solvent (water). Here, it is given as:\[\text{Mass of solution} = 120 \text{ g (urea)} + 1000 \text{ g (water)} = 1120 \text{ g}\]
02

Calculate the volume of the solution

Use the density formula \( \text{Density} = \frac{\text{Mass}}{\text{Volume}} \) to calculate the volume:\[\text{Volume of solution} = \frac{1120 \text{ g}}{1.15 \text{ g/mL}} = 973.91 \text{ mL}\]
03

Convert volume from mL to L

Since molarity requires the volume in liters, convert the obtained volume in mL to liters:\[\text{Volume of solution in L} = \frac{973.91}{1000} \text{ L} = 0.97391 \text{ L}\]
04

Calculate the moles of urea

Determine the number of moles of urea, using the formula \( \text{moles} = \frac{\text{mass}}{\text{molar mass}} \):\[\text{Moles of urea} = \frac{120 \text{ g}}{60 \text{ g/mol}} = 2 \text{ moles}\]
05

Compute the molarity of the solution

Molarity is calculated using the formula \( \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{volume of solution in L}} \):\[\text{Molarity} = \frac{2 \text{ moles}}{0.97391 \text{ L}} \approx 2.05 \text{ M}\]
06

Choose the correct answer

With a calculated molarity of approximately 2.05 M, the correct multiple choice answer is:\[(c) \ 2.05 \text{ M}\]

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Key Concepts

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

Density in Solutions
Density refers to how much mass is contained in a given volume. It is an essential concept, especially in solutions, as it describes how compact the molecules are packed. Density is calculated using the formula:
  • \( \text{Density} = \frac{\text{Mass}}{\text{Volume}} \)
For example, if you dissolve a substance like urea in water, the density of the resulting solution might change compared to the pure solvent. Knowing the density helps you find the volume of the solution when the mass is known, which is necessary for calculating molarity.
In practical terms, measuring density can inform us about the concentration and characteristics of the solution, providing insights into its composition and behavior during reactions.
Understanding Moles
The concept of moles is central to chemistry. A mole counts entities, much like a dozen counts 12 items. One mole represents Avogadro's number, which is roughly \( 6.022 \times 10^{23} \) entities. In chemistry, moles are used to count atoms, ions, or molecules in a given sample.
To calculate moles of a substance, use the formula:
  • \( \text{Moles} = \frac{\text{mass}}{\text{molar mass}} \)
This means if you have a substance with a known mass and molar mass, you can determine how many moles are present. For example, with urea (molar mass = 60 g/mol), if you have 120 g, then you have 2 moles of urea. This calculation is vital in determining concentrations and reacting proportions in chemical equations.
Molar Mass and Its Importance
Molar mass is the mass of one mole of a substance, typically expressed in g/mol. It connects the macroscopic mass of a material to the atomic scale number of particles involved.
Molar mass is crucial because it enables the conversion between mass and moles, which is a frequent necessity in chemistry calculations. It simplifies the comparison of substances and their behavior during chemical reactions.
A molar mass of a compound is determined by summing the atomic masses of the elements in its chemical formula. For example, urea is CO(NH2)2:
  • Carbon (C) = 12 u
  • Oxygen (O) = 16 u
  • Nitrogen (N) = 14 u (each)
  • Hydrogen (H) = 1 u (each)
Thus, the molar mass of urea is 60 g/mol. This value is then utilized to calculate the number of moles from a given mass.
Essentials of Solutions Chemistry
Solutions chemistry revolves around understanding how different substances mix and react in a solvent to form a homogeneous mixture. Here, key terms such as solute, solvent, and concentration are fundamental.
  • **Solute**: The substance dissolved in the solvent.
  • **Solvent**: The liquid in which the solute is dissolved. In most chemical situations, it's water.
  • **Concentration**: Indicates how much solute is present in a given volume of solution, typically in molarity (M).
Molarity, defined as moles of solute per liter of solution, helps quantify the concentration:
  • \( \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{volume of solution in L}} \)
If the solution density is known, as in our exercise, the entire solution volume can be found using mass and density. Then, dividing moles by the solution's volumetric measure gives molarity, offering insight into how concentrated the solution is.

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Most popular questions from this chapter

Ratio of \(\mathrm{C}_{\mathrm{p}}\) and \(\mathrm{C}_{\mathrm{v}}\) of a gas ' \(\mathrm{X}\) ' is \(1.4\). The number of atoms of the gas 'X' present in \(11.2\) litres of it at NTP will be (a) \(6.02 \times 10^{21}\) (b) \(60.2 \times 10^{23}\) (c) \(6.02 \times 10^{23}\) (d) \(1.02 \times 10^{23}\)

What is the volume (in litres) of oxygen at STP required for complete combustion of \(32 \mathrm{~g}\) of \(\mathrm{CH}_{4}\) ? (mol. wt of \(\mathrm{CH}_{4}=16\) ) (a) \(89.6\) (b) \(189.6\) (c) \(98.4\) (d) \(169.5\)

\(4 \mathrm{~g}\) caustic soda is dissolved in \(100 \mathrm{cc}\) of solution. The normality of solution is (a) 1 (b) \(0.8\) (c) \(0.6\) (d) \(0.10\)

One mole of fluorine is reacted with two mole of hot and concentrated KOH. The products formed are \(\mathrm{KF}, \mathrm{H}_{2} \mathrm{O}\) and \(\mathrm{O}_{2} .\) The molar ratio of \(\mathrm{KF}, \mathrm{H}_{2} \mathrm{O}\) and \(\mathrm{O}_{2}\) respectively is (a) \(1: 2: 1\) (b) \(1: 2: 2\) (c) \(0.5: 1: 2\) (d) \(2: 1: 0.5\)

The amount of Zinc (atomic weight \(=65\) ) necessary to produce \(224 \mathrm{~mL}\) of \(\mathrm{H}_{2}\) by the reaction with an acid will be (a) \(0.65 \mathrm{~g}\) (b) \(7.6 \mathrm{~g}\) (c) \(6.5 \mathrm{~g}\) (d) \(8.5 \mathrm{~g}\)
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