Sodium hydroxide molar mass

Your teacher has asked you to prepare 1.00 liter of a 0. lOO-molar aqueous solution of sodium hydroxide (molar mass 40.0 g). You should weigh out... 

Determine the concentration of the solution when 4.00 grams of sodium hydroxide (molar mass 40.0g mol ) Is dissolved In 200cm of water. 

To a solution of 4.76 g of 1-(4 -methyl-6 -methoxy-2 -pyrimidinyl)-3-methyl-3-pyrazoline-5-one in 200 ml of ether was added an ether solution containing 6 molar equivalents of diazomethane and the reaction mixture was allowed to stand at room temperature for 20 hours. After distilling off the solvent, the residue was dissolved in 160 ml of water, made alkaline (pH 10) with sodium hydroxide solution and extracted three times with 140 ml of benzene. The extract was washed with a small amount of water, dried over sodium sulfate and evaporated to give a crystalline mass. Recrystallization from isopropylether gave 1-(4 -methyl-6 -methoxy-2 -pyrimidinyl)-3-methyl-5-methoxypyrazole (3.96 g, 84%) as colorless prisms, MP 90° to 92°C. 

J.9 You are asked to identify compound X, which was extracted from a plant seized by customs inspectors. You run a number of tests and collect the following data. Compound X is a white, crystalline solid. An aqueous solution of X turns litmus red and conducts electricity poorly, even when X is present at appreciable concentrations. When you add sodium hydroxide to the solution a reaction takes place. A solution of the products of the reaction conducts electricity well. An elemental analysis of X shows that the mass percentage composition of the compound is 26.68% C and 2.239% H, with the remainder being oxygen. A mass spectrum of X yields a molar mass of 90.0 g-moF. (a) Write the empirical formula of X. (b) Write... 

GL 22] [R 1] [P 23] The mass transfer efficiency of the falling film micro reactor as a function of the carbon dioxide volume content was compared quantitatively (Figure 5.30) [5]. The molar ratio of carbon dioxide to sodium hydroxide was constant at 0.4 for all experiments, i.e. the liquid reactant was in slight excess. 

Standard solutions, a solution of nickel in acid with a quoted mass/volume concentration, a solution of sodium hydroxide with a quoted concentration as a molarity and a solution of pesticides with quoted mass/volume concentrations. Matrix RMs - natural materials, river sediment with quoted concentrations of metals, milk powder with a quoted fat content and crab paste with quoted concentrations of trace elements. 

With 0.5M sodium hydroxide as eluent, Sephadex G-50 effects fractionation in the molar mass range 1000-15000 dalton and can be used for a period of 3-4 weeks with a single calibration carried out with proteins and polypeptides of known molar mass, as revealed by Figure 10. Relative retention volumes 0.0 and 1.0 are defined with Blue Dextran and phenol, respectively. 

Due to the dark color of alkali lignins, their molar masses cannot be determined by means of the light-scattering method. However, as shown by Figure 10, elution with sodium hydroxide also brings about a consistent elution pattern of lignin sulfonates and polypeptides. It is assumed that this also applies to the kraft lignins. 

Fractionation on Sephadex G-25 using 0.5M sodium hydroxide as eluent causes the low molar mass lignin components in black liquor to elute in the relative retention volume range 0.3-1.3 with partial separation from each other, as shown in Figure 11. 

A more detailed mass transfer study on the carbon dioxide absorption in sodium hydroxide solution was performed using a falling-film microreactor [319]. Experimental investigations were made at a liquid flow of 50ml/h, with three NaOH concentrations (0.1,1 and 2 M), at a fixed inlet molar ratio C02 NaOH of 0.4, and for a range of C02 concentration of 0.8-100%. A two-dimensional reactor model was developed, and the results are similar to the experimental data at low NaOH concentrations (0.1 and 1 M). The agreement is less pronounced for higher concentrations such as 2 M NaOH, which could be explained by either maldistribution of... 

Sodium hydroxide solution is usually standardized by titrating a pure sample of potassium hydrogen phthalate (KHP), an acid with one acidic hydrogen and a molar mass of 204.22 g. It takes 34.67 mL of a sodium hydroxide solution to titrate a 0.1082-g sample of KHP. What is the molarity of the sodium hydroxide ... 

A student titrates an unknown amount of potassium hydrogen phthalate (KHQH4O4, often abbreviated KHP) with 20.46 mL of a 0.1000 M NaOH solution. KHP (molar mass = 204.22 g/mol) has one acidic hydrogen. How many grams of KHP were titrated (reacted completely) by the sodium hydroxide solution ... 

The active ingredient in aspirin is acetylsalicylic acid. A 2.51-g sample of acetylsalicylic acid required 27.36 mL of 0.5106 MNaOH for complete reaction. Addition of 15.44 mL of 0.4524 M HCI to the flask containing the aspirin and the sodium hydroxide produced a mixture with pH = 3.48. Find the molar mass of acetylsalicylic acid and its Ka value. Acetylsalicylic acid is a monoprotic acid. 

Three molar equivalents of bromine is added with cooling to a mixture of 1 molar equivalent of quinoline and 3 gram-atom equivalents of sulfur, and the resulting mixture is heated to 180-200" in an oil bath for 2 hours. The cooled reaction mass is extracted several times with hot concentrated hydrochloric acid, and the combined extracts are made alkaline with sodium hydroxide. Ether extraction of the alkaline mixture followed by distillation of the ethereal solution gives a 50% yield of 3-bromoquinoline boiling at 158—162°/24 mm. 

The units of molarity are mole/liter (of solution), but they are commonly replaced with a capital M, which symbolizes molarity. Yet, there will be times when you will need to replace M with mole/liter when analyzing units and solving problems. If a sodium hydroxide solution is labeled 2 M (read as two-molar), it means that 2 moles of NaOH are dissolved in 1 L of solution, 2 moles/liter. If you need to brush up on mass-mole conversions, review the pertinent material in Chapter 5. In all the problems dealing with molar solutions, molarity will be written as a conversion factor to emphasize the canceling and retention of units, just as was done with the percent concentrations. The molarity term for a solution that is 0.55 M in NaOH could be written in four ways to make the required conversion factor ... 

Problem 1 What is the molarity of a solution that contains 50.0 g of sodium hydroxide, NaOH, in 850 mL of solution The molar mass of NaOH is 40.0 g. 

Acetylsalicylic acid is the active ingredient in aspirin. It took 35.17 mL of 0.5065 M sodium hydroxide to react completely with 3.210 g of acetylsalicylicacid. Acetylsalicylic acid has one acidic hydrogen. What is the molar mass of acetylsalicylic acid ... 

Propan-2-ol (molar mass 60 g mol-1 and boiling temperature 82 °C) can be prepared by the reaction between 6.15 g of 2-bromopropane (molar mass 123 g mol and boiling temperature 59 °C) and excess aqueous 2.0 mol dm" sodium hydroxide. This reaction is slow at room temperature. 

What is the molar concetitraiion of a solution of 80g of sodium hydroxide jn 1 dm of solution The amount in moles of NaOH in 80g of sodium hydroxide can be calculated from its molar mass. 

On hydrolywby aqueous sodium hydroxide, A gives an oily liquid, B. B has the percent compoatkm mass C. 77.1% N, 15.1% H, 7.5%, and a molar mass of 93 gmoI . B is basic and gives a precipitate with bromine water. 

Inspection of this equation shows that both volume and molarity of the base are inversely proportional to the molar mass, M. Conditions that would produce a calculated molar mass that is too low include having either the volume or the molarity of the sodium hydroxide solution too large. Factor I will mean a larger volume of NaOH must be used to reach the endpoint as some of the NaOH has reacted with the CO2. Factor II will also mean a larger volume of NaOH must be used to reach the end point for the NaOH is not as concentrated as anticipated. Choice (C) correctly identifies both factors. This eliminates choice (A), which only identifies the first factor. The error in Factor III has the opposite effect, eliminating choices (B) and (D). If not all of the acid has been titrated because some is left in the flask, the volume of NaOH used will be too low, resulting in a molar mass that is higher than the true value. ... 

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