Volume = 25 mL = 25*10-3 L.
Calculate the molarity of NaOH as follows:
Molarity = mole of solute/volume of solutions (L)
Molartity = 4.55*10-3/25.00*10-3
Molarity = 0.182 mol/L
Therefore, option A is the correct answer.
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Question 3 (1 point) What is the molarity of a NaOH solution with 4.55 x 10-3...
Question 1 (1 point) How many moles are in 35.66 ml of 4.55 M NaOH solution? Report your answer in moles of NaOH but do not include units in your response. Your Answer: Answer Question 2 (1 point) For the balanced reaction: 3 NaOH + H2PO4 3 moles of NaOH will react with → Naz PO, + 3 H2O mole H3PO4
Exp 9 titration of vinegar procedure b
molarity of NaOH = .0859
ROCEDURE B. DETERMINATION OF CONCENTRATION OF ACETIC ACID IN UNKNOWN SAMPLES Titration of vinegar solutions Trial 1 Trial 2 Trial 3 1. Volume of vinegar solution being titrated in milliliters (mL) 25.00 mL 25.00 mL 25.00 ml 2. Volume of vinegar solution being titrated in Liters (L) 0.0250 L 0.0250 L 0.0250 L 3. Final buret reading in ml 16. 20m 16.316L 16.25ml 4. Initial buret reading in...
Molarity for NaOH = 0.08732. In a second titration with the same solution of NaOH as used in Question #1, the student weighs out a sample of KHP of 0.359 g. Calculate the volume of the NaOH solution needed to neutralize this sample of KHP. 3. A monoprotic weak acid with the general formula of HA will react with a base, such as NaOH. Write the neutralization equation which describes the reaction. 4. If K, for the weak acid, HA is 1.8...
Molarity(m)=Moles (mol) of solute/Liters (L) of solution Molarity of standard NaOH from bottle = 0.1005 M Molarity of NaOH added in titration (mL) = 10.47 mL 1. Moles of NaOH added in titration ______
Exact molarity of 1M NaOH used is 1.083M, the volume of NaOH
used is 10.0mL, the final volume of stock NaOH is 100.00mL, and the
molarity of stock NaOH solution is 0.1083M.
KHP had a molar mass of 204.2 g/mole.
acid base titration, part A
B. Titration of KHP 1. 2. 3. Mass of KHP Initial buret reading Final buret reading Trial 1 Trial 2 0.30T usly & 0.00ml D.DD_ml 15.3 L 142 ml t is 1/0 ml _mol mol...
309 Week One Data and Calculations Standardization of NaOH Solution Sample 1 Sample 2 Sample 3 Mass of KHP 10.5749 10.8939 1.26g. Initial Volume NaOH, V. (mL) Read to nearest 0.01 ml. No.25mL 8.86 m L 10.45 mL Final Volume NaOH, V:(mL) 0.31 ml 25.73 m2 Read to nearest 0.01 mL 24.47 ml Total Volume NaOH used mL) 110.62 mL 16.92 mL 24.03 mL Moles KHP 2.81 × 10-3 4.37x10-3 6.16 x 10 3 Moles NaOH 12.81x 10-3 4.37 ×...
Find:
Moles of NaOH (mol)
Volume of NaOH (L)
Molarity of NaOH mol/L
Average Molarity
Deviation from mean
Average deviation
Percent Relative Average Deviation
Weighing Results for Part A Trial 1 Trial 2 Trial 3 In al 4 Mass of conical flask + KHp/g 93.228 g Mass of conical flask/g Mass of KHp/g 92. 185 g 0.443 g ^ 86.554 85.911 1 0.363 g g g | 93.242 g 92.785 g 0.457 86.345 g 85.991g 10.354g Titration Results for Part...
Question 3 (1 point) Solution Composition: Molarity What is the molarity of a solution that contains (7.75x10^-1) mol of NH4Cl in (4.134x10^2) mL of solution? (Example answer: 1.23e-2 or 1.23) Note: Your answer is assumed to be reduced to the highest power possible. Your Answer: X10 Answer
10 Question (1 point) A 100.0 mL solution of NaOH reaches the equivalence point when 37.51 mL of a 0.0600 M solution of HCl is added from the burette. 1st attempt Feedback hi See Periodic Table See Hint How many moles of NaOH were originally in the solution? mol VIEW SOLUTION C TRY AG 1 OF 13 QUESTIONS COMPLETED < 10/13 > a ORI е со
moles of oxalix acid
moles/molarity of NaOH
average molarity
volume/molarity of hcl used
average molarity
NAME SECTION DATA SHEET LOCKER A CTOR Write a balanced equation for the reaction between H.CO. and NaOH: 90.33/mol 0.878 0836 TRIAL NUMBER a. Mass of oxalic acid used (g) b. Final Buret reading (ml.] c. Initial Buret reading (ml] d. Volume of NaOH used (b-c) Moles of oxalic acid used f. Moles of NaOH used (2 xe) g. Molarity of NaOH (f/d/1000) Average Molarity...