How many moles of KF are contained in 347 g of water in a 0.175 m KF solution? A)1.65 × 10^-2 mol KF B)5.04 × 10^-2 mol KF C)6.07 × 10^-2 mol KF D)3.22 × 10^-2 mol KF E)1.98 × 10^-2 mol KF

Question

Quizplus · Accepted Answer

First, we need to find the number of moles of KF in the solution:

mass of water = 347 g 
molarity of KF solution = 0.175 M

We can use the formula:

moles of solute = molarity x volume (in L)

Since we have the mass (in g) of water instead of its volume (in L), we need to convert:

mass of water = volume of water x density of water

The density of water at room temperature is about 1 g/mL, so:

volume of water = mass of water / density of water 
volume of water = 347 g / 1 g/mL 
volume of water = 347 mL = 0.347 L

Now we can calculate the number of moles of KF:

moles of KF = 0.175 M x 0.347 L 
moles of KF = 0.0607 mol

Therefore, the answer is C: 6.07 × 10^-2 mol KF.

How Many Moles of KF Are Contained in 347 G