(a)
Interpretation:
The final concentration of the prepared solution has to be determined.
Concept Introduction:
Dilution equation:
Dilution equation is given by,
(a)
Answer to Problem 8.69EP
The final concentration of the prepared solution is 3.0 M.
Explanation of Solution
Given information:
Initial volume of solution = 30.0 mL
Initial concentration = 5.0 M
Final volume of solution = 50 mL (after 20.0 mL water is added)
Final concentration = ?
Apply dilution equation to calculate the molarity of the concentrated solution,
Therefore, the final concentration of the prepared solution is 3.0 M.
(b)
Interpretation:
The final concentration of the prepared solution has to be determined.
Concept Introduction:
Dilution equation:
Dilution equation is given by,
(b)
Answer to Problem 8.69EP
The final concentration of the prepared solution is 3.0 M.
Explanation of Solution
Given information:
Initial volume of solution = 30.0 mL
Initial concentration = 5.0 M
Final volume of solution = 50 mL (after 20.0 mL water is added)
Final concentration = ?
Apply dilution equation to calculate the molarity of the concentrated solution,
Therefore, the final concentration of the prepared solution is 3.0 M.
(c)
Interpretation:
The final concentration of the prepared solution has to be determined.
Concept Introduction:
Dilution equation:
Dilution equation is given by,
(c)
Answer to Problem 8.69EP
The final concentration of the prepared solution is 4.5 M.
Explanation of Solution
Given information:
Initial volume of solution = 30.0 mL
Initial concentration = 7.5 M
Final volume of solution = 50 mL (after 20.0 mL water is added)
Final concentration = ?
Apply dilution equation to calculate the molarity of the concentrated solution,
Therefore, the final concentration of the prepared solution is 4.5 M.
(d)
Interpretation:
The final concentration of the prepared solution has to be determined.
Concept Introduction:
Dilution equation:
Dilution equation is given by,
(d)
Answer to Problem 8.69EP
The final concentration of the prepared solution is 1.5 M.
Explanation of Solution
Given information:
Initial volume of solution = 60.0 mL
Initial concentration = 2.0 M
Final volume of solution = 80 mL (after 20.0 mL water is added)
Final concentration = ?
Apply dilution equation to calculate the molarity of the concentrated solution,
Therefore, the final concentration of the prepared solution is 1.5 M.
Want to see more full solutions like this?
Chapter 8 Solutions
Bundle: General, Organic, And Biological Chemistry, Loose-leaf Version, 7th + Lms Integrated For Owlv2 With Mindtap Reader, 1 Term (6 Months) Printed Access Card
- Given the equation of the line, y=150x-8.1221 (cells per mL is in 10^8), what is the CFU per mL of the original suspension if the thousand-fold dilution of that suspension has an absorbance of 0.3 at 600nm? * A. 3.7 x 10^12 B. 3.7 x 10^9 C. 3.7 x 10^1 D. None of these is correctarrow_forwardA volumetric flask made of Pyrex glass is calibrated at 20 °C.It is filled to the 100 mL mark with 35 °C acetone. After the flask is filled the acetone cools and the flask warms so that the combination of acetone and flask reaches a uniform temperature of 32 °C. The combination is then cooled to 20 °C. a) What is the volume of the acetone when it is at 20 °C? b)At the temperature of 32 °C, does the level of acetone lie above or below the 100 mL mark?arrow_forwardAn aqueous solution is 3.50 % by mass hydrochloric acid, HCl, and has a density of 1.02 g/mL. The mole fraction of hydrochloric acid in the solution is ____.arrow_forward
- You have a 6 M solution of stock A. You do the following serial dilutions: Add 1 ml of stock A to tube B. Add 9 ml H2O. Mix.Then take 1 ml from tube B and put it into tube C. Add 19 ml H2O. Mix. Then put 5 ml from tube C into tube D. Add 5 ml H2O. A. what is the final (cumulative) dilution of stock A (dilution factor)? B. What is the final concentration (include units) after all these dilutions?arrow_forwardYou have a 1000X stock solution of sodium chloride (NaCl). The concentration of NaCl in the stock solution is 5M. a)How would you make 300mL of a 1X solution? b)What is the concentration of NaCl in your 1X solution?arrow_forwardConsider a buffer solution that contains 0.55 M NH2CH2CO2H and 0.35 M NH2CH2CO2Na. pKa(NH2CH2CO2H)=9.88. a. Calculate its pH. b. Calculate the change in pH if 0.155 g of solid NaOH is added to 250 mL of this solution. c. If the acceptable buffer range of the solution is ±0.10 pH units, calculate how many moles of H3O+ can be neutralized by 250 mL of the initial buffer.arrow_forward
- The pH probe/meter uses following equations: Ecell = L + 0.0592 log a1 = L - 0.0592 pH Where L = L1 + EAg/AgCI + Easy= constants L1 = - 0.0592 log a2 a1 = activity of analyte solution a2 = activity of internal solution How will measured pH value be affected vs “real” pH if HCl in pH electrode, became 0.15M instead of 0.1M ? a.impact can not be determined b.measured pH is higher than "real" pH. c.measured pH is lower than "real" pH. d.measured pH is same as "real" pH.arrow_forward20 mL solution is 2 M NaOH. If 0.5 M HCl is to be added to this solution, calculate the pH of the resulting solution: A. when 10 mL of the 0.5 M HCl is added to the original solution of the base (initial pH). B. When 50 mL H20 is added to the original solution of the base.arrow_forwardCalculate the molarity of dilute Ca(OH)2 solution if the titration of 30.00 mL of 0.05231 M HCl required 24.76 mL of the acid. An 0.2500 g hydroxide tablet sample submitted in the lab used 36.75 mL during titration. How much is the hydroxide in mg present?arrow_forward
- What volume of 0.10NH2 SO3 will be required to neutralize a solution containing 10.0grams of Ca(OH)2?a) 0.27Lb) 27Lc) 2.7Ld) 270L What volume of 1.5N NaOH is needed to react with 25ml 4.0N HCl?a) 66.67mlb) 6.67mlc) 50mld) 70mlarrow_forwardThe pH probe/meter uses following equations: Ecell = L + 0.0592 log a1 = L - 0.0592 pH Where L = L1 + EAg/AgCI + Easy= constants L1 = - 0.0592 log a2 a1 = activity of analyte solution a2 = activity of internal solution How will measured pH value be affected vs “real” pH if the temperature of the sample is 30C when pH was measured ? a.measured pH is lower than real pH b.impact can not be determined c.measured pH is higher than real pH d.measured pH is same as real pHarrow_forwardA chemist is performing a precipitation titration on a 1.00 L sample containing an unknown concentration of the cadmium ion, Cd2+. If it takes 24.52 mL of a 0.250 M solution of Na2CO3 to reach the end point, calculate the concentration of the cadmium ion?arrow_forward
- Basic Clinical Lab Competencies for Respiratory C...NursingISBN:9781285244662Author:WhitePublisher:Cengage