Question

Potassium fluoride (kf), a salt, has a molecular weight of 58.10 grams. how many grams would be needed to mix 1.0 liter of 0.10 m salt solution? 0.17 g 580 g 17 g 5.8 g

Answer 1

n(KF)=m(KF)/M(KF)

n(KF)=vc

m(KF)/M(KF)=vc

m(KF)=M(KF)vc

m(KF)=58.10*1.0*0.10=5.81 g≈ 5.8 g

Answer 2

Answer: 5.8 g

Explanation: Molarity : It is defined as the number of moles of solute present in one liter of solution.

$Molarity=\frac{\text{given mass of compound}}{\text{molar mass of compound}\times \text{volume of solution in L}}$

Given: Molar mass of KF= 58.10 g

Molarity of KF= 0.10

Volume of solution= 1.0 L

$0.10=\frac{\text{given mass of KF}}{58.10}\times 1.0$

${\text{given mass of KF}}=5.8g$