Solution and Concentration in Chemistry
A solution is a homogeneous mixture, made up of a solute dissolved in a solvent. Ex. In a water (aqueous) solution of sodium chloride, the sodium chloride is the solute and the water is the solvent. Solute – the component that is dissolved or is the least abundant component in the solution.
SolidGasSolidHydrogen in Palladium SolidLiquidSolidMercury in silver SolidSolidSolidSilver in gold Kinds of Solution 1. Saturated Solution – one that contains as much of the solute as can be dissolved at the given temperature. 2. Unsaturated solution – there is less solute than can be dissolved at a given temperature 3. Supersaturated solution – unstable condition in which there is more solute in solution than can normally exist at a given time.
Weight percentwt. %100 x grams of solute / grams of solvent + grams of solute Volume percentvol. %100 x liters of solute / liter of solution Parts per millionPpmmilligrams of solute / kg. of solution or milligrams / liter Parts per billionPpbmicrograms / liter grams per volume—grams of solute / liters of solution Concentration Units 1. Percentage Composition a. mass or weight percent of solute – expresses the concentration of the solution as percent of solute in a given mass of solution wt. % = wt. of solute x 100 = wt. of solute x 100 wt. f solute + solvent wt. of solution 1. a. 1.
What is the weight percent of sodium hydroxide in a solution that is made by dissolving 8. 00 g of sodium hydroxide in 50. 0 g of water? 1. a. 2. What weight of potassium chloride and water are needed to make 250 g of 5. 00% solution? b. weight-volume percent – expresses the concentration as grams of solute per 100 mL of solution. wt-vol % = wt. of solute, gram x 100 vol. of solution, mL 1. b. 1. A solution was prepared by dissolving 1. 30 grams silver nitrate in water and diluting to 50. 0 mL.
Calculate the mass-volume percent silver nitrate. 1. b. 2. A 3. 0% H2O2 solution is commonly used as a topical antiseptic to prevent infection. What volume of this solution will contain 10. g of H2O2? c. volume percent (with respect to the solute). This can be used to express the concentration of a solution made by mixing two liquids. vol. % = volume of solute x 100 volume of solution Volume percent is used to express the concentration of alcohol in beverages. Wines generally contain 12% alcohol by volume. This translates into 12 mL of alcohol in each 100 mL of wine.
The beverage industry also uses the concentration unit of proof (twice the volume percent). Pure alcohol is 100% and therefore 200 proof. Scotch whiskey is 86 proof, or 43% alcohol. 2. Parts per million (ppm) and parts per billion (ppb) This concentration is used for very dilute solutions such as water analysis or in biological preparations.
A solution whose solute concentration is 1ppm contains 1 g of solute for each million (106) grams of solution or, equivalently, 1 mg of solute per kilogram of solution. Also, 1 ppm also corresponds to 1 mg of solute per liter of solution. pm = mass of solute x 106 ; ppm = wt in mg of solute mass of solution vol in L of solution For solution that are even more dilute, parts per billion (ppb) is used. A concentration of 1ppb represents 1 gram of solute per billion (109) grams of solution, or microgram (? g) of solute per liter of solution ppb = mass of solute x 109 ; ppb = wt in ? g of solute mass of solution vol in L of solution 3. mole fraction (X) – the ratio of the no. of moles of component to the total no. of moles (solute and solvent) present in the solution.
Xsolute = no. of moles of one component total no. of moles of all components XA + XB + …….. + XN N = total no. of components Note: Mole fraction have no units because the units in the numerator and the denominator cancel. The sum of the mole fraction of all components of a solution must equal to 1. 3. 1. What is the mole fraction of the solute in a 100g sample of a 10% sulfuric acid solution? 3. 2. A solution of hydrochloric acid contains 36% HCl by mass. Calculate the mole fraction of HCl in the solution 4. Molality (m) – the number of moles of solute per kilogram of solvent olality (m) = no. of moles solute no. of kilograms solvent 4. 1. A solution of hydrochloric acid contains 36% HCl by mass. Calculate the molality of HCl in the solution.
Find the molality of a solution of 20. 0% (by mass) ethanol C2H5OH, in water. 5. Molarity (M) – the number of moles of solute per liter of solution. The units of molarity are moles per liter. Molarity (M) = no. of moles of solute Liters of solution Since no. of moles solute = weight of the solute (g) molar mass of the solute (g/mol) Molarity can also be expressed as
M = weight of the solute (g) molar mass of the solute (g/mol) x Liters of solution 5. 1. What will be the molarity of the resulting solution if 3. 50 grams of sodium hydroxide are dissolved in water to make 125 mL? 5. 2. How many grams of potassium hydroxide are required to prepare 600mL of 0. 450M KOH solutions? 6. Normality (N) – expresses as the no. of equivalents of solute per liter of solution. Normality is based on an alternative chemical unit of mass called the equivalent weights. Normality (N) = no. of equivalents of solute L of solution where,no. of equiv. f solute = wt. of solute (g) equiv. wt. of solute then, N = wt. of solute (g) equiv. wt of solute x L of solution 6. 1.
What is the normality of a potassium dichromate, K2Cr2O7, solution made by dissolving 5. 00 grams of potassium dichromate crystals in 250 mL of water if it is to be used for the reaction Cr2O7 -2 + 6 I – + 14 H + 2 Cr +3 + 3 I2 + 7 H2O Anuang KYLLINGA MONOCEPHALA commonly known as ANUANG Kyllinga triceps Blanco Kyllinga mindorensis Steud. Kyllinga monocephala Rottb. Var. mindorensis Boeck. Kyllinga gracilis Kuntz
Local names: Anuang (Tag. ); barubotones (Bis. ); bolobotones (Bis. ); borobotones (Bis. ); botoncillo (Sp. , Bis. ); bosobotones (Bis. ); boesa-ngadadakkel (Ilk. ); baki-baki (S. L. Bis. ); bosikad (C. Bis. ); boskad (Bis. ); busikad (P. Bis. ); katutu (Mag. ); kurukamoting-orig (Bik. ); malabotones (Bis. ); mutha (Tag. ); mustra (Tag. ); pungos (S. L. Bis. ); sangsangitan (Bon. ); sud-sud (Bis. ); uli-uli (Bag. ); busikad (Bis. ). Anuang is found throughout the Philippines in waste places, open grasslands, etc. , at low and medium altitudes. It is pantropic in distribution.
The plant is more or less glabrous, arising from creeping rootstocks. The stems are usually solitary, 10 to 40 centimeters high. The leaves are up to 15 centimeters in length or longer, 3 to 4 millimeters wide; with the bracts similar. The spikes are ovoid, simple, white, 8 to 13 millimeters long. The spikelets are very numerous, 3 to 3. 5 millimeters long, the flowering glume distinctly winged along the keel. The nut is up to 1. 5 millimeters long. According to Kirtikar and Basu the oil distilled from the roots is dark yellowish-green in color, of pleasant odor and pungent taste.
The fresh root in decoction is used as a sudorific in malaria with chills. According to Guerrero the rhizome yields a decoction employed as a diuretic, and, mixed with oil, it is externally employed to combat certain forms of dermatosis. Nadkarni states that the root in decoction is refrigerant, demulcent and tonic; it is given to relieve thirst in fevers and in diabetes. The oil distilled from the root is used to relieve pruritus of the skin. Internally the oil is given in torpor of the liver. Other properties are similar to those of Cyperus rotundus. http://en. wikipilipinas. org/index. php? title=Anuang
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