Question

please help me answer question 1 to 3,

Chapter 9, Solutions, Review & Extra Credit. 2 Points for each XC correct answer; 6 points total, max. Partial Credit given. Suggestions...work in teams [it will improve the learning experience] 1. Solutions-A solution forms when a solute (the smaller quantity) dissolves in a solvent (the larger quantity). In a solution, the particles of solute are evenly distributed in the solvent. A solute may be a gas, liquid or solid. A polar solute dissolves in a polar solvent (while nonpolar solutes dissolve in nonpolar solvents). Ionic solutes dissolve in water, one of the most common solvents in nature, as the polar water molecules attracts and bonds with the ions. The polar 0-H bond leads to hydrogen bonding between the partially positive hydrogen atoms (in the water molecule) are attracted to the strongly electronegative atoms, like, 0, N, F, Cl in other molecules. Hydration is the name of the process where the dissolving are surrounded by water molccules, reducing thcir attraction to other ions, and helping to kcep them in solution. ions 2 Electrolytes and Nonelectrolytes -A solute that is able to conduct electricity consists of ions when dissolved, and is called an clectrolyte. Solutes that do not conduct electricity (and hence do not consist of ions when dissolved) are called nonelectrolytes. When an ionic solute dissolves in a polar solvent, the process of separating into the component ions is called dissociation. Strong clectrolytes are completely dissociated, while weak electrolytes exist primarily as the molecules, with only a few ions produced. Nonelectrolytes do not dissociate at all, and hence exist in solution only as the molecules. Electrolytes are important components of bodily fluids. An equivalent is the amount of an electrolyte that carries one mole of (positive or negative) charge. One mole of Na is one cquivalent. One mole of Mg:' has two cquivalents. In fluid replacement solutions, the concentration of clcctrolytes is expressed as mEq/L of solution, where ImEq is 10-J Eq. Q. How many g Na" must he dissolved to produce a 100 mEq/L of Na solution? 100 mEqlL of Na' contains 1/10 of a mole of Na per liter of volume, which is 0.10 x 2.0Amole 2.3g Na per liter. 7.3 Solubility -The term solubility describes the maximum amount ofsolute that dissolves in a given amount of solvent (100 grams), at a given temperature. When a solution contains all that solute that can dissolve (at that temperature) it is saturated An unsaturated solution contains less than the maximum amount of the solute. [In a saturated solution the rates of dissolution and recrystallization are equal, so the additional amount added will simply remain undissolved on the bottom of the container. The solubility of most solids increases as the temperature increases, while the solubility of most gases decreases as the temperature increases. Henry's Law states that the solubility of a gas in a liquid is directly related to the pressure of the gas above the liquid Q2-Since the solubility of CaCl, is 59.5g at 0°C. and 159g at 100°C, if we add 75g of CaCl2 to 100g ofwater at 0°C, is the solution saturated or unsaturated?{saturated] And if we heat that solution to 100°C, is the solution saturated or unsaturated funsaturated] How many more g of CaCl, must he added to create a saturated solution? [159g -75g = 84g] XCHI-See Solubility Table on page 295.(a) If we add 45.0g of NaN0 to 50 g of water at 20°C, is the solution saturated or unsaturated? (b) If saturated, how much has dropped to the bottom of the beaker? If unsaturated, how much more can be added to the solution? (c) Now raise the temp to 50 c. Now is the solution saturated or unsaturated? If saturated, how much has dropped to the bottom of the beaker? If unsaturated, how much more can be added to the solution? 7.4 Percent Concentration -The amount of solute dissolved in a fixed amount of solvent is called the concentration of the solution. Concentrations are typically expressed as Mass o/.) (% m/m) which is the mass of the solute (g) divided by the mass of the solution [solute + solvent] (g) x 1000/0; or as Volume o.) (% v/v) which is the volume of the solute in a fixed volume of solution (100mL of solution) x 1000/0; or less commonly in the laboratory, but common in hospitals and pharmacies, as a Mass/Volume (-) (m/v%) which is the mass in g of the solute in 100mL of the solution. Q3 What is the mass % (m/m %) of 1.0 mole of NaCl dissolved in 141.5g ofwater? The molar mass of NaCI is 23.0 35.5 58.5g Therefore the total solution is the 58.5g of NaCI plus the 141.5g of H:,O 200.0 g of solution; and the mass %= 58.5 g NaCl/ 200 g solution x 100= 0.293.x 100 29.3%

thank you

Answer 1

Answers for Q1 , Q2, Q3 and XC#1 are given in below attached images , please go through it .

83 Answes mole e Nacl water desscved n 14S Mars OL ate = 141S Aele Mars G Nacl Mole q Nacl x Moeulas weight aNacl Mole. culoor weight e Noacl = Mo.utq Na +Molutod 230+ 35 Sa/mol . 9/mol Mas Nal - 1.0 r6l x 58Sg/rac S8.S Mas 0 Nacl Mars Nacd (solut =S8.Si Mas &lvent oade Mag Sluent Solute Mars o Mars Solution = S859+ 1415 9 Solutan = 2009 Mas Solute Mars = Mars Mars a solution Mars SB.q メ100 Loo 29.25 Mas y 29.3 %. Mars

83 Answes mole e Nacl water desscved n 14S Mars OL ate = 141S Aele Mars G Nacl Mole q Nacl x Moeulas weight aNacl Mole. culoor weight e Noacl = Mo.utq Na +Molutod 230+ 35 Sa/mol . 9/mol Mas Nal - 1.0 r6l x 58Sg/rac S8.S Mas 0 Nacl Mars Nacd (solut =S8.Si Mas &lvent oade Mag Sluent Solute Mars o Mars Solution = S859+ 1415 9 Solutan = 2009 Mas Solute Mars = Mars Mars a solution Mars SB.q メ100 Loo 29.25 Mas y 29.3 %. Mars

xc#1 aAnswer 329 /i00g oater 9/oog owater Solubility o NaNo at 20'c Solubtbty NaNog at 5oc to, at 20c olubibhy NaNog at soc solubibtyj o NaNog so wates iss7 watea at add us.o9 Na No3 o so 20c the solution beomes satrated and 19 we 4S. O NaNo dsolues at the botbm we ase the temperature soc, the solution unsatorated beomes At socS79 a NaNo dsolues tn Sog ate, but have ony Us S3-45912 So, exta 12a - N>NO should be added to the Solution to saborated solubon become

& How many a Na must be crsdved to pioduce mEq/L e Na* solution Answe 1 mole e Na equivalent mEq oo m moles e Na* mmoles OO = o.l mole . Mole weight Gram molecula wweg ht Grarn molecular weiqht e Na 2a 23 weight Mole Gram moleular weght eaght Na O rao x 23 eight Na -* 239. e Na must be diusoued to produie 2.3 MElL Na solution

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