How many moles of KF are contained in 347 g of water in a 0.175 mol kg^-1 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

To solve this problem, we need to use the formula:

moles of solute = concentration × volume

where concentration is given in units of mol kg^-1, and volume is the mass of water divided by its density (which is about 1000 g/L).

First, we need to convert the concentration from mol kg^-1 to mol L^-1:

0.175 mol kg^-1 = 0.175 mol / (1000 g / L) = 0.175 mol / 1 L = 0.175 mol L^-1

Next, we need to calculate the volume of the solution using the mass of water and its density:

volume = mass / density = 347 g / (1 g/mL) = 347 mL = 0.347 L

Now we can use the formula to calculate the moles of KF:

moles of KF = concentration × volume = 0.175 mol L^-1 × 0.347 L = 0.060725 mol

Finally, we need to round the answer to the appropriate number of significant figures, which is three:

moles of KF = 0.0607 mol

Therefore, the correct answer is C) 6.07 × 10^-2 mol KF, which is equivalent to 6.07 × 10^-2 / 0.0585 ≈ 1.036 moles of KF.

How Many Moles of KF Are Contained in 347 G