Stu Dent weighs out 2.137 g of potassium iodide (KI) salt. He dissolves the salt
ID: 979134 • Letter: S
Question
Stu Dent weighs out 2.137 g of potassium iodide (KI) salt. He dissolves the salt in water and obtains a final solution volume of 100.0 mL using a volumetric flask. Calculate the molarity of KI in the solution. Answer to 4 significant figures with appropriate units.
Kemmi Major weights 153.0 g of ammonium chloride (NH4Cl) salt. She also weighs 500.0 g of DI water. She combines the salt and the water to make her solution. Calculate the molality of the NH4Cl solution. Answer to 4 significant figures with appropriate units.
Doc Inmaking weights 15.42 g of glucose (C6H12O6) and 75.36 g of ethanol (C2H5OH). He dissolves the glucose in the ethanol to make his solution. Calculate the mole fraction of glucose in the solution.
Explanation / Answer
Answer 1) Molarity of KI solution :
166.0028 g KI In 100 ml water = 10 M KI solution ……..(Both side multiplied by 10 as volume and concentration has inverse ratio)
2.137 g KI in 100 ml water = (2.137 /166.0028) X 10…..(Simply 166.0028 = 10 M then 2.137 = How much? Hencecalculation)
= 0.1287 M (appr.)
2.137 g in 100 ml water makes 0.1287 M solution.
Answer 2) Molarity of Ammonium chloride solution(NH4Cl) :
53.5 g NH4Cl in 500 ml water =2 M NH4Cl.
153 g NH4Cl in 500 ml water= (153/53.5) X 2…………. (Simply 53.9 g in 500 ml gives 2 M so how much 153 g in same amount,)
= 5.7196 M solution
Answer 3) Mole fraction of Glucose and Ethanol.
180 g glucose = 1 mol
15.42 g glucose = 15.42/180 mol .= 0.0857
ii) Molecular weight of Ethanol (C2H6O) is 46 g/mol
46 g ethanol = 1 mol
75.36 g ethanol = 75.36/46 = 1.6383
Total no. of mole =0.0857+1.6383=1.7240
Mole fraction of glucose = mole of glucose/total mole = 0.0857/1.7240 = 0.0497.
Mole fraction of ethanol = mole of ethanol / total mole = 1.6386/1.7240 = 0.9503.