If a 100.00 g sucrose solution has a boiling point of 101.5 degrees celsius, how

Question

If a 100.00 g sucrose solution has a boiling point of 101.5 degrees celsius, how many moles of sucrose are in the solution? How many grams?

What is the percent mass concentration of NaCl in a solution that is prepared by adding 15.0 g of NaCl to 90.00 g water?

What happens to each of the following properties of a solution when the concentration of solute is decreased?

a. vapor pressure

b. boiling point

c. freezing point

What is the concentration of OH- in each of the following solutions?

a. 0.0010 M KOH

b. 0.0010 M HCl

c. 0.0050 M NaOH

Explanation / Answer

we know that ?T = Kb*molality ....................(1) ...... ................Here, ?T = 101.5 - 100 = 1.5 degree C (Boiling point of pure solvent, water = 100 deg C) b or the molal elevation constant for the solvent, water= 0.52 deg C/molal Let the mass of sucrose taken in grams be x. Then, mass of solvent = Mass of solution - mass of sucrose = (100 - x) g No. of moles of sucrose = mass in grams/gram molecular mass =x/342.97 mol Molality = (x/342.97)/(100-x) mol/g = 1000x/{342.97*(100-x)} mol/kg = 1000x/{342.97*(100-x)} molal [using formula (2)] Putting the values of known parameters in the formula (1): 1.5 = 0.52 * 1000x/{342.97*(100-x)} =>1.5 * {342.97*(100-x)} = 514x and calculate res part...........................................................................................................................................................................................................................................................................................................................................................................................-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- Boiling point and freezing point>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Addition of solute to form a solution stabilizes the solvent in the liquid phase, and lowers the solvent chemical potential so that solvent molecules have less tendency to move to the gas or solid phases. As a result, liquid solutions slightly above the solvent boiling point at a given pressure become stable, which means that the boiling point increases. Similarly, liquid solutions slightly below the solvent freezing point become stable meaning that the freezing point decreases. Both the boiling point elevation and the freezing point depression are proportional to the lowering of vapor pressure in a dilute solution. These properties are colligative in systems where the solute is essentially confined to the liquid phase. Boiling point elevation (like vapor pressure lowering) is colligative for non-volatile solutes where the solute presence in the gas phase is negligible. Freezing point depression is colligative for most solutes since very few solutes dissolve appreciably in solid solvents. Boiling point elevation (ebullioscopy).......................... ........>>>>>>>>>>>>>>>>>>Vapor pressure lowering

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