Question

How many moles of kf are contained in 244 ml of 0.135 m kf solution? the density of the solution is 1.22 g/ml?

Answer 1

The Molarity of a solution = number of moles / volume.  
Volume = 244ml = 0.244L
 So it follows that number of moles = Molarity * volume 
 Number of moles = 0.135 * 0.244 = 0.03945.
 Hence the number of moles = 0.03945

Answer 2

$\boxed{0.03294\;{\text{mol}}}$ of KF are contained in 244 mL of 0.135 M KF solution.

Further Explanation:

The proportion of any substance in the mixture is termed as its concentration. Various concentration terms are used to find out the concentrations of substances. Some of the concentration terms are listed below:

1. Molality (m)

2. Mole fraction (X)

3. Molarity (M)

4. Parts per million (ppm)

5. Mass percent ((w/w) %)

6. Volume percent ((v/v) %)

Molarity is defined as the amount of solute that is dissolved in one liter of the solution. It is denoted by M and its unit is mol/L.

The expression for molarity of the solution is as follows:

${\text{Molarity of solution}} = \dfrac{{{\text{Amount}}\left( {{\text{mol}}} \right){\text{of solute}}}}{{\;{\text{volume}}\left( {\text{L}} \right)\;{\text{of solution}}}}$     ...... (1)                                

Rearrange equation (1) to calculate the amount of solute.

${\text{Amount of solute}} = \left( {{\text{Molarity of solution}}} \right)\left( {{\text{Volume}}\;{\text{of}}\;{\text{solution}}} \right)$   …… (2)                    

The volume of the solution is first converted from mL to L. The conversion factor for this is,

${\text{1 mL}} = {10^{ - 3}}\;{\text{L}}$

So the volume of the solution can be calculated as follows:

$\begin{aligned}{\text{Volume of solution}}&= \left({{\text{244 mL}}} \right)\left( {\frac{{{{10}^{ - 3}}\;{\text{L}}}}{{{\text{1}}\;{\text{mL}}}}} \right)\\&= 0.244{\text{ L}}\\\end{aligned}$

The molarity of the solution is 0.135 M.

The volume of the solution is 0.244 L.

Substitute these values in equation (2) to calculate the amount of KF.

 $\begin{aligned}{\text{Amount of KF}} &= \left( {0.135{\text{ M}}} \right)\left( {0.244{\text{ L}}} \right)\\&= 0.03294\;{\text{mol}}\\\end{aligned}$

Therefore 0.03294 moles of KF are present in the given solution.

Learn more:

1. Determine the given mass of solute: brainly.com/question/2847466
2. Determine the moles of water produced: brainly.com/question/1405182

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Concentration terms

Keywords: KF, 0.03294 mol, 0.135 M, 244 mL, 0.244 L, molarity, volume, amount of solute, conversion factor, concentration, concentration terms, molality, mole fraction, mol/L.