Final Exam
Concentration by percent (percent by mass and volume) we express percent by mass as
(m/m) mass % = mass of solute / mass of solution x 100 %
Concentration by percent (percent by mass and volume) we express percent by mass / volume as
(m/v) mass/ volume % = mass of solute / volume of solution x 100 %
Concentration by percent (percent by mass and volume) we express percent by volume as
(v/v) volume % volume of solute / volume of solution x 100 %
If 1.39 L of solution contains 0.25 L of solute, what is the percent by volume concentration? A. 0.18% B. 1.6% C. 560% D. 18%
0.25 L / 1.39 L x 100% = 18%
How would you go about preparing the solution? Place the steps in order from first to last. 1) Partially filled the flask with water . 2) Measure out the desired amount of NaCl 3) Add the measured NaCl to the 3.00L volumetric flask 4) Dilute the solution , slowly adding water until the desired volume is reached 5) Mix until NaCl dissolves completely
1) Measure out the desired amount of NaCl. 2) Add the measured NaCl to the 2.00-L volumetric flask. 3) Partially fill the flask with water. 4) Mix until NaCl dissolves completely. 5) Dilute the solution, slowly adding water until the desired volume is reached.
Which of the choices describes a Brønsted-Lowry base? 1. a substance that accepts H+ from another substance 2. a substance that dissociates in water to form OH− 3. a substance that dissociates in water to form H+ 4. a substance that donates H+ to another substance
1. a substance that accepts H+ from another substance
A chemist prepares a solution by adding 1.0 kg of salt to 5.0 kg of water. What is the percent by mass of salt in this solution?
1.0 kg (solute= salt) / 6.0 kg (1.0 solute + 5.0 solvent) x 100 % = 17%
Recall that a percentage is a fraction of ______.
100 %
If the volume of a solution is tripled, how will the concentration of the solution change? 1. The concentration will be INCREASED by a factor of 3. 2. The concentration will DECREASE by a factor of 3. 3. The concentration will be DECREASED by a factor of 9. 4. The concentration will be INCREASED by a factor of 9.
2. The concentration will DECREASE by a factor of 3.
An aqueous NaBr solution has a mass of 185.0 g and contains 25.50 g NaBr. Calculate the mass percent NaBr.
25.50 g NaBr / 185.0 g NaBr = 13.78 (14.0%)
A solution made by adding 17.9 mL of methyl alcohol to enough water to give 473 mL of solution.
3.80 v/v (17.9 ml / 473 ml)
A solution made by adding 67.1 mL of ethylene glycol to enough water to give 213 mL of solution.
31.5 v/v (67.1 ml/ 213 ml)
Which of the choices describes a Brønsted-Lowry acid? 1. a substance that accepts H+ from another substance 2. a substance that dissociates in water to form OH− 3. a substance that dissociates in water to form H+ 4. a substance that donates H+ to another substance
4. a substance that donates H+ to another substance
What is the concentration in PPB of 496 g of solution that contains 8.1 x 10-6 g of solute?
8.1 x 10-6 / 496 g x 10(9) = 16 ppb
If 2.20 kg of solution contains 0.59 kg of solute, what is the percent by mass concentration? A. 27% B. 0.27% C. 370% D. 2.8%
A. 0.59 kg / 2.20 kg = 0.268 x 100 = 26.81 (round up) 27%
How many liters of solute are in 10.2 L of a 69.0% by volume solution? A. 7.04 L B. 704 L C. 14.8 L D. 14.8 L
A. 0.69 x 10.2 L = 7.04 L
Calculate the molar it's, M, of 5.0 liter solution containing 15.0 moles of an unknown compound. A. 3.0 M B. 10 M C. 0.33 M D. 20 M
A. 15.0 moles / 5.0 liter = 3.0 M
Calculate the molarity, M, of a 5.0 liter solution containing 15.0 moles of an unknown compound. A. 3.0 M B. 20 M C. 10 M D. 0.33 M
A. 15.0 moles / 5.0 liter = 3.0 M
Calculate the volume of solution in liters, L, needed to prepare a 0.650 M solution from 180.0 g of magnesium chloride, MgCl2. The molar mass of MgCl2 is 95.21 g/mol. A. 2.91 L B. 1.89 L C. 146 L D. 277 L
A. 180.0 x 1 mole / 95.21 g/mol = 1.891 / 0.650 = 2.908 (round up) 2.91 L
What is he concentration in ppm of 10.5 kg of solution that contains 5.0 x 10-5 kg of solute? A. 4.8 ppm B. 2.1 x 10(11) ppm C. 4,800 ppm D. 4.8 x 10-4 ppm
A. 5.0 x 10-5 kg / 10.5 kg x 10000 = 4.761 (round up) 4.8 ppm
Conjugates of WEAK BASES are?
Acidic
How many grams of solute are in 2.47 kg of a 19% by mass solution? A. 13 kg B. 0.47 kg C. 0.13 kg D. 47 kg
B. 0.19 x 2.47 kg = 0.4693 (round up) 0.47 kg
How many moles of potassium nitrate, KNO3, are in 0.90 liters of a 3.0 M KNO3 solution? A. 2.1 mol B. 2.7 mol C. 0.30 mol D. 3.3 mol
B. 3.0 M KNO3 x 0.90 L = 2.7 mol KNO3
What volume in liters, L, of a 0.850 M solution can Jorge prepare from 0.350 L of a 3.00 M solution? A. 0.0992 L B. 1.24 L C. 1.05 L D. 2.55 L
B. 3.00 M x 0.350 L = 1.05 / 0.850 = 1.235 (round up) 1.24 L
Conjugates of WEAK ACIDS are?
Basic
For VERY SMALL concentration we calcite the parts per million by using
Concentration (PPB) = mass of solute / mass of solution x 10 to the 9th power
For SMALL concentration we calcite the parts per million by using
Concentration (PPM) = mass of solute / mass of solution x 10 to the 6th power
Linnea diluted 1.50 L of a 0.90 M solution into a final volume of 4.0 L. What is the concentration of the solution Linnea prepared? A. 1.4 M B. 2.4 M C. 6.7 M D. 0.34 M
D. 0.90 M x 1.50 L = 1.35 / 4.0 L = 0.3375 (round up) 0.34 M
Calculate the molarity, M, of a 3.50 liter solution containing 5.00 moles of sodium chloride, NaCl. A. 1.50 M B. 17.5 M C. 0.700 M D. 1.43 M
D. 5.00 moles / 3.50 liter = 1.428 (round up) 1.43 M
What are the Seven (7) STRONG ACIDS that completely ionize into water?
HCl- Hydrochloric Acid, HBr- Hydrobromic acid, HI- Hydroiodic acid, HNO3- Nitric acid, H2SO4- Sulfuric acid, HClO4- Perchloric acid, HCIO3- Chloric acid
what is a CONJUGATE base?
Its identical to the acid, but WITHOUT the H+
What is the formula for molarity?
Molarity = moles of solute/liters of solution
A very dilute concentration is expressed in _________ or _________.
Parts per million (PPM) or parts per billion (PPB)
What does [NaOH] mean?
The concentration of NaOH (the square brackets are used to represent the concentration of a solute)
Percent by volume =
Volume of solute / volume of solution x 100%
A Homogeneous mixture is called?
a solution
A 1045 mL sample of drinking water was found to contain 13.0 mg of lead. Calculate the concentration of lead in milligrams per liter.
concentration: 12.44 mg/L (convert 1045 mL into liters by dividing by 1000) 13.0 mg / 1.045 = 12.44
An aqueous potassium carbonate solution is made by dissolving 7.27 moles of K2CO3 in sufficient water so that the final volume of the solution is 3.30 L. Calculate the molarity of the K2CO3 solution.
concentration: 2.20 M (7.27 moles / 3.30 L)
A 2355 g sample of ground water is found to contain 0.0557 g of arsenic. Calculate the concentration of arsenic in parts per million, ppm.
concentration: 23.65 ppm Example, 0.0557 / 2355.0557 = 0.000023651 x 1,000,000 = 23.65
If a solution contains a small amount of solute, we say it is ______. If it contains a large amount of solute, we say it is ____________. A solution that contains the maximum amount of dissolved solute is _________.
dilute/ concentrated/ saturated
The amount of solute that is present in a solution is called?
it's concentration
When the solute is a solid, we express concentration by
mass / volume percent
How many grams of sodium hydroxide are present in 252.0 mL of a 0.700 M NaOH solution?
mass of sodium hydroxide = 7.054 g (convert 252.0 to liters by moving the decimal to the left 3x 0.2520 x 0.700 M = 0.1764 x (Na- 22.99, O- 15.99, H- 1.008) 39.99 = 7.054)
Calculate the percent mass per volume, % (m/v), of a dextrose solution containing 7.00 g of dextrose in 275 mL of solution. Note that mass is not technically the same as weight, but the abbreviation % (w/v) is often used interchangeably with % (m/v).
mass/volume %: 2.54% (m- 7.00 g/v- 275 mL)
While working in a pharmaceutical laboratory, you need to prepare 3.00 L of a 1.75-M NaCl solution. What mass of NaCl would be required to prepare this solution?
mass: 306.81 g (1.75 x 3.00)
The first solution contains 0.200 mol of NaOH in 2.65 L of solution.
molarity: 0.0754 M (0.200 mol / 2.65 L)
The second solution contains 13.5 g of NaCl in 795 mL of solution.
molarity: 0.290 M (13.5 g x 1 mole / 58.443 g = 0.230 moles / 795 mL)
If 6.09 mL of 10.0 M NaOH is used for a reaction, calculate the number of moles of NaOH that were used.
number of moles of NaOH= 0.0609 mol (6.09 x 1 liter / 1000 x 10.0)
The three (3) common ways of describing concentration by
percent, by fraction, or by molarity.
The substance that dissolves is called?
the solute
The liquid in which the solute dissolves in is?
the solvent