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What is the conductivity of 1 molar solution of sodium hydroxide at ambient temperature

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16y ago

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How do you prepare 0.02m Naoh from 1M Naoh solution?

To prepare 0.02M NaOH from 1M NaOH solution, you will need to dilute the 1M solution. Use the formula: C1V1 = C2V2, where C1 is the concentration of the stock solution (1M), V1 is the volume of the stock solution you will use, C2 is the desired concentration (0.02M), and V2 is the final volume of the diluted solution. Calculate the volume of 1M NaOH solution (V1) needed to make the desired 0.02M concentration and dilute it with water to reach the desired volume (V2).


Is the conductivity of NaOH different to NH3?

Yes, the conductivity of NaOH is different than NH3. NaOH is a strong electrolyte, meaning it fully dissociates into ions in solution and conducts electricity well. NH3 is a weak electrolyte, so it partially dissociates in solution and has lower conductivity.


How do you prepare 10 ml of 1M naoh?

To prepare 10 ml of 1M NaOH solution, you would need to dissolve 0.4 grams of NaOH (molar mass of NaOH is 40 g/mol) in enough distilled water to make a total volume of 10 ml. First, measure out 0.4 grams of NaOH using a balance. Then, add this solid NaOH to a small beaker and add distilled water while stirring until the total volume reaches 10 ml. Make sure to handle NaOH with care as it is a caustic substance.


How many mL of 1m NaOH would it take to neutralize 100 mL of an unknown HCL concentration?

You would need to know the concentration of the unknown HCl solution in order to determine the volume of 1M NaOH needed to neutralize it using the equation: M1V1 = M2V2. Without the concentration of HCl, it is not possible to calculate the volume of NaOH required for neutralization.


How can we prepare 1M solution of NaCl?

You could titrate equal volumes of 1M solution of NaOH and 1M solution of HCl to obtain 1M solution of NaCl.


Will mixing equal quantities of 1m solutions of the following give an acidic basic or neutral solution?

Mixing equal quantities of 1M HCl and 1M NaOH solutions will give a neutral solution because they will react to form water and a salt (NaCl).


How many ml of 1m naoh do you need to add to a 250ml flask to make a 0.12 m naoh solution?

1/.12 = 8.3333333333 This is the dilution factor you need to achieve. 250ml / 8.333333 = 30ml. This is a very important concept for chemistry so it may be important to try this method. The logical way works for me.


What is the molarity of Na OH solution?

The molarity of a NaOH solution is determined by the concentration of NaOH in moles per liter of solution. It is calculated by dividing the moles of NaOH by the volume of solution in liters. For example, a 0.1 M NaOH solution would contain 0.1 moles of NaOH per liter of solution.


What is the make up of 0.1 sodium hydroxide?

0.1 M sodium hydroxide solution contains 0.1 moles of sodium hydroxide per liter of solution. This corresponds to 0.1 moles of NaOH per 40 g (1 mole) of NaOH, resulting in 4 g of NaOH in 1 liter of 0.1 M NaOH solution.


What is the resulting molarity of a solution made by mixing equal volumes of solutions of HCl and NaOH each of molarity 1M?

As a thought experiment - suppose you mixed one liter of 1M HCl with 1 liter of 1M NaOH. The resulting solution (neglecting any density changes associated with mixing) would contain 2 liters with 1 mole of Cl- and 1 mole of Na+ with the balance being water (the H+ from the HCl and the OH- from the NaOH would just become part of the water). This would give you a 0.5 M solution of NaCl.


What amount of NaOH is required to prepare 1N NaOH solution and why only that amount is used?

To prepare a 1N NaOH solution, you would need to dissolve 40 grams of NaOH in water to make 1 liter of solution. This amount is used because 1N solution means 1 mole of NaOH per liter of solution, and the molar mass of NaOH is 40 g/mol, so 40 grams of NaOH is needed to have 1 mole in 1 liter of solution.


Why do you give washings with 1M NaOH in synthesis of anthraquinone?

To remove any O-benzoylbenzoic acid than may remain . by converted it to salt Dissolved