molarity question
What is the mass of oxalic acid, that is required to make 500ml of a 0.0500M solution?
(M.WT = 90g/mol)

molarity question What is the mass of oxalic acid, that is required to make 500ml of...
Find the molarity of the following: A. Dissolve 5.00g of NaCl in enough water to make 500ml of solution. (M.WT of Nacl = 58,45g/mol) B. If 0.435g of a drug if dissolved in enough water to give 750ml of solution, what is the molarity of the drug in solution? (M.WT = 158.0g/mol)
Sample Data Sheet: TITRATION AND MOLARITY Part I: Preparing the Oxalic Acid solution 1. Mass of oxalic acid + weighing paper. _____1.5765_____________ g 2. Mass of weighing paper. _____n/a_____________ g 3. Volume of oxalic acid solution. 250 mL 4. Concentration of oxalic acid=___________ (show calculation above) Part III: Completing the Neutralization Trial #1 Trial #2 Trial #3 Volume of Oxalic Acid 15ml 15ml 15ml Final Buret Reading of NaOH (mL) 15.87 15.74 19.43 Initial...
Calculate the mass of oxalic acid needed to make a 250.0 mL standard solution that is 0.0500 mol L-1 . You will use oxalic acid dihydrate, (COOH)2.2H2O. M((COOH)2.2H2O) = 126.07 g mol-1 )
Oxalic acid, H2C204, is used in the restoration of old wood. What is the molarity of oxalic acid if 29.78 mL of 0.538 M NaOH is required to titrate a 14.75 mL of an oxalic acid solution given the following reaction? (3 decimal places, fill in the number ONLY!) H2C2O4 (aq) + 2 NaOH (aq) → Na2C204 (aq) + 2 H20 (1)
Data Table 1 Mass of flask and oxalic acid (g) 117.43 Mass of empty flask (g) 116.93 Mass of oxalic acid (g) 0.5 Moles of oxalic acid (mol) Final volume of NaOH (mL) 17 Initial volume of NaOH (mL) 5 Volume of NaOH used (mL) 12 Moles of NaOH (mol) Molarity of NaOH (M) Data Table 2 Mass of flask and vinegar (g) 126.61 Mass of empty flask (g) 121.63 Mass of vinegar (g) 4.98 Final volume of NaOH (mL)...
You titrated a 22.00 mL solution of 0.0300 M oxalic acid with
freshly prepared solution of KMnO4. If it took 48.99 mL of this
solution, what is the molarity of the KMnO4?
0.022 L of 0.300 M oxalic acid
(0.022)(0.03) = 0.00066 mol oxalic acid
0.0006 mol oxalic acid x (2 mol KMnO4)/(5 mol oxalic acid)=
0.000264 mol KMnO4
0.000264 mol/0.04899 L = 0.05388855 M KMnO4
= 0.005388 M KMnO4
where did I go wrong?
Question 7 0.1 / 1...
What mass of oxalic acid, H2C2O4, is required to prepare 250 mL of a solution that has a concentration of (4.53x10^-1) M H2C2O4?
4. Calculate the mass of Oxalic acid (H2C2O4*2H2O) required to neutralize 20.0 ml of 0.10 M NaOH solution using the balanced chemical equation for this reaction. 5. A 0.120 g sample of pure oxalic acid (H2C2O2*2H2O) was dissolved in water and neutralized with 21.0 ml of NaOH. Calculate the molarity of NaOH. Do not use scientific notation, but do use the proper number of significant digits and units.
3.75 moles of KCI dissolved in sufficient water to make 500mL of solution. What is the molarity of this solution? (dw=1.00 g/mol)
A sample of oxalic acid dihydrate (126.07g/mL) with mass of 0.1473g was titrated by the addition of 35.87mL potassium hydroxide That solution of potassium hydroxide required 22.48mL to neutralize 10.00mL of nitric acid, determine molarity of the nitric acid.