Answer to Question #172528 in General Chemistry for Blank 0822

Question

A 163 grams of solvent with molar mass of 18.0g/mole is present, to create an aqueous solution with the mole fraction of Potassium fluoride (KF) of 0.0855? How many grams of Potassium fluoride with molar mass of 58.1g/mole is needed to achieved the desired result?

Answer 1

Mole fraction of KF= 0.0855

Mass of solvent= 163g

Molar mass of solvent= 18.0g/mol

Mass of KF= ?

Molar mass of KF= 58.1g/mol

Mole of solvent= mass/molar mass= 163/18.0= 9.06mole

Let the mole of KF be x

Mole fraction= mole of KF/mole of KF +mole of solvent

0.0855= x/x+9.06

0.0855x + 0.77463= x

0.77463= x-0.0855x

0.22537x= 0.77463

x= 3.44mol

Now,

Mass= mole x molar mass

= 3.44 x 58.1= 199.7g of KF