Solutions are homogeneous mixtures containing one or more solutes in a solvent. The solvent that makes up most of the solution, whereas a solute is the substance that is dissolved inside the solvent. Concentrations are often expressed in terms of relative unites (e.g. percentages) with three different types of percentage concentrations commonly used: \[\text{Mass/Volume Percent}= \dfrac{\text{Mass of Solute (g)}}{\text{Volume of Solution (mL)}} \times 100\% \label{3}\] For example, In the United States, alcohol content in spirits is defined as twice the percentage of alcohol by volume (v/v) called proof. What is the concentration of alcohol in Bacardi 151 spirits that is sold at 151 proof (hence the name)? It will have an alcohol content of 75.5% (w/w) as per definition of "proof". When calculating these percentages, that the units of the solute and solution should be equivalent units (and weight/volume percent (w/v %) is defined in terms of grams and mililiters).
Sometimes when solutions are too dilute, their percentage concentrations are too low. So, instead of using really low percentage concentrations such as 0.00001% or 0.000000001%, we choose another way to express the concentrations. This next way of expressing concentrations is similar to cooking recipes. For example, a recipe may tell you to use 1 part sugar, 10 parts water. As you know, this allows you to use amounts such as 1 cup sugar + 10 cups water in your equation. However, instead of using the recipe's "1 part per ten" amount, chemists often use parts per million, parts per billion or parts per trillion to describe dilute concentrations.
Here is a table with the volume percent of different gases found in air. Volume percent means that for 100 L of air, there are 78.084 L Nitrogen, 20.946 L Oxygen, 0.934 L Argon and so on; Volume percent mass is different from the composition by mass or composition by amount of moles.
The molarity and molality equations differ only from their denominators. However, this is a huge difference. As you may remember, volume varies with different temperatures. At higher temperatures, volumes of liquids increase, while at lower temperatures, volumes of liquids decrease. Therefore, molarities of solutions also vary at different temperatures. This creates an advantage for using molality over molarity. Using molalities rather than molarities for lab experiments would best keep the results within a closer range. Because volume is not a part of its equation, it makes molality independent of temperature.
In a solution, there is 111.0 mL (110.605 g) solvent and 5.24 mL (6.0508 g) solute present in a solution. Find the mass percent, volume percent and mass/volume percent of the solute.
Mass Percent =(Mass of Solute) / (Mass of Solution) x 100%| =(6.0508g) / (110.605g + 6.0508g) x 100% =(0.0518688312) x 100% =5.186883121% Mass Percent= 5.186% Volume Percent =(Volume of Solute) / (Volume of Solution) x 100% =(5.24mL) / (111.0mL + 5.24mL) x 100% =(0.0450791466) x 100% =4.507914659% Volume Percent= 4.51% Mass/Volume Percent =(Mass of Solute) / (Volume of Solution) x 100% =(6.0508g) / (111.0mL + 5.24mL) x 100% =(0.0520) x 100% =5.205% Mass/Volume Percent= 5.2054%
With the solution shown in the picture below, find the mole percent of substance C.
Moles of C= (5 C molecules) x (1mol C / 6.022x1023 C molecules) = 8.30288941x10-24mol C Total Moles= (24 molecules) x (1mol / 6.022x1023 molecules)= 3.98538691x10-23mol total XC= (8.30288941x10-24mol C) / (3.98538691x10-23mol) = .2083333333 Mole Percent of C = XC x 100% =(o.2083333333) x 100% =20.83333333 Mole Percent of C = 20%
A 1.5L solution is composed of 0.25g NaCl dissolved in water. Find its molarity.
Moles of NaCl= (0.25g) / (22.99g + 35.45g) = 0.004277 mol NaCl Molarity =(Moles of Solute) / (Liters of Solution) =(0.004277mol NaCl) / (1.5L) =0.002851 M Molarity= 0.0029M
0.88g NaCl is dissolved in 2.0L water. Find its molality.
Moles of NaCl= (0.88g) / (22.99g + 35.45g) = 0.01506 mol NaCl Mass of water= (2.0L) x (1000mL / 1L) x (1g / 1mL) x (1kg / 1000g) = 2.0kg water Molality =(Moles of Solute) / (kg of Solvent) =(0.01506 mol NaCl) / (2.0kg) =0.0075290897m Molality= 0.0075m References
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