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get what mass of kcl in grams must be added to 500 ml of a 0.15 m kcl solution to produce a 0.40 m solution from EN Bilgi.
How many grams of KCl are needed to make 500 mL of 2.45 M KCl?
91.4 grams C = (mol)/(volume) 2.45M = (mol)/(0.5L) 2.45 M * 0.5 L = mol mol = 1.225 Convert no. of moles to grams using the atomic mass of K + Cl 1.225 mol * ((39.1 + 35.5)g)/(mol) 1.225cancel(mol) * (74.6g)/cancel(mol) =1.225*74.6g = 91.4g
How many grams of
K C l
are needed to make 500 mL of 2.45 M
K C l ?
Chemistry Solutions Molarity
Madi May 6, 2016 91.4 grams
C = m o l v o l u m e 2.45 M = m o l 0.5 L 2.45 M ⋅ 0.5 L = m o l m o l = 1.225
Convert no. of moles to grams using the atomic mass of K + Cl
1.225 m o l ⋅ ( 39.1 + 35.5 ) g m o l 1.225 m o l ⋅ 74.6 g m o l = 1.225 ⋅ 74.6 g = 91.4 g Answer link
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How many grams of potassium chloride (KCl) must be added to make 500 mL of 1.00M KCl solution?
Click here👆to get an answer to your question ✍️ How many grams of potassium chloride (KCl) must be added to make 500 mL of 1.00M KCl solution?
How many grams of potassium chloride (KCl) must be added to make 500 mL of 1.00M KCl solution?A
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Correct option is C)1000 ml 1 M KCL contains 74.5 grams of KCl
1 ml 1 M KCL contains
1000 74.5 grams of KCl
Thus, 500 ml 1 M KCL contains
1000 74.5 × 500 grams of KCl
Its value comes out to be almost 37.3 g
Thus option C is the correct answer.
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2.5: Preparing Solutions
Preparing a solution of known concentration is perhaps the most common activity in any analytical lab. The method for measuring out the solute and the solvent depend on the desired concentration and …
2.5: Preparing Solutions
Last updated Jun 15, 2020
2.4: Basic Equipment
2.6: Spreadsheets and Computational Software
David Harvey DePauw University
Preparing a solution of known concentration is perhaps the most common activity in any analytical lab. The method for measuring out the solute and the solvent depend on the desired concentration and how exact the solution’s concentration needs to be known. Pipets and volumetric flasks are used when we need to know a solution’s exact concentration; graduated cylinders, beakers, and/or reagent bottles suffice when a concentrations need only be approximate. Two methods for preparing solutions are described in this section.
Preparing Stock Solutions
A stock solution is prepared by weighing out an appropriate portion of a pure solid or by measuring out an appropriate volume of a pure liquid, placing it in a suitable flask, and diluting to a known volume. Exactly how one measure’s the reagent depends on the desired concentration unit. For example, to prepare a solution with a known molarity you weigh out an appropriate mass of the reagent, dissolve it in a portion of solvent, and bring it to the desired volume. To prepare a solution where the solute’s concentration is a volume percent, you measure out an appropriate volume of solute and add sufficient solvent to obtain the desired total volume.
Example 2.5.1 2.5.1
Describe how to prepare the following three solutions: (a) 500 mL of approximately 0.20 M NaOH using solid NaOH; (b) 1 L of 150.0 ppm Cu2+ using Cu metal; and (c) 2 L of 4% v/v acetic acid using concentrated glacial acetic acid (99.8% w/w acetic acid).
(a) Because the desired concentration is known to two significant figures, we do not need to measure precisely the mass of NaOH or the volume of solution. The desired mass of NaOH is
0.20 mol NaOH L × 40.0 g NaOH mol NaOH ×0.50 L=4.0 g NaOH
0.20 mol NaOHL×40.0 g NaOHmol NaOH×0.50 L=4.0 g NaOH
To prepare the solution, place 4.0 grams of NaOH, weighed to the nearest tenth of a gram, in a bottle or beaker and add approximately 500 mL of water.
(b) Since the desired concentration of Cu2+ is given to four significant figures, we must measure precisely the mass of Cu metal and the final solution volume. The desired mass of Cu metal is
150.0 mg Cu L ×1.000 M × 1 g 1000 mg =0.1500 g Cu
150.0 mg CuL×1.000 M ×1 g1000 mg=0.1500 g Cu
To prepare the solution, measure out exactly 0.1500 g of Cu into a small beaker and dissolve it using a small portion of concentrated HNO3. To ensure a complete transfer of Cu2+ from the beaker to the volumetric flask—what we call a quantitative transfer—rinse the beaker several times with small portions of water, adding each rinse to the volumetric flask. Finally, add additional water to the volumetric flask’s calibration mark.
(c) The concentration of this solution is only approximate so it is not necessary to measure exactly the volumes, nor is it necessary to account for the fact that glacial acetic acid is slightly less than 100% w/w acetic acid (it is approximately 99.8% w/w). The necessary volume of glacial acetic acid is
4 mL CH 3 COOH 100 mL ×2000 mL=80 mL CH 3 COOH
4 mL CH3COOH100 mL×2000 mL=80 mL CH3COOH
To prepare the solution, use a graduated cylinder to transfer 80 mL of glacial acetic acid to a container that holds approximately 2 L and add sufficient water to bring the solution to the desired volume.
Exercise 2.5.1 2.5.1
Provide instructions for preparing 500 mL of 0.1250 M KBrO3.Answer
Preparing Solutions by Dilution
Solutions are often prepared by diluting a more concentrated stock solution. A known volume of the stock solution is transferred to a new container and brought to a new volume. Since the total amount of solute is the same before and after dilution, we know that
C o × V o = C d × V d (2.5.1) (2.5.1)Co×Vo=Cd×Vd where C o Co
is the stock solution’s concentration,
V o Vo
is the volume of stock solution being diluted,
C d Cd
is the dilute solution’s concentration, and
V d Vd
is the volume of the dilute solution. Again, the type of glassware used to measure
V o Vo and V d Vd
depends on how precisely we need to know the solution’s concentration.
Note that Equation 2.5.1 2.5.1
applies only to those concentration units that are expressed in terms of the solution’s volume, including molarity, formality, normality, volume percent, and weight-to-volume percent. It also applies to weight percent, parts per million, and parts per billion if the solution’s density is 1.00 g/mL. We cannot use Equation
if we express concentration in terms of molality as this is based on the mass of solvent, not the volume of solution. See Rodríquez-López, M.; Carrasquillo, A. J. Chem. Educ. 2005, 82, 1327-1328 for further discussion.
Example 2.5.2 2.5.2
A laboratory procedure calls for 250 mL of an approximately 0.10 M solution of NH3. Describe how you would prepare this solution using a stock solution of concentrated NH3 (14.8 M).