What is the conductivity of 1 molar solution of sodium hydroxide at ambient temperature
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.
The increased brightness in HCl and NaOH solutions is due to the higher ion concentration, which enhances the conductivity of the solution and allows more current to flow through the bulb. Vinegar has a lower ion concentration compared to HCl and NaOH, resulting in weaker conductivity and thus a dimmer glow in the light bulb.
NaOH is Sodium Hydroxide.
moles = mass/Mr moles = 100/(23+16+1) moles of NaOH = 2.5mol
The chemical formula for sodium hydroxide is NaOH.
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).
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.
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.
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.
You could titrate equal volumes of 1M solution of NaOH and 1M solution of HCl to obtain 1M solution of NaCl.
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).
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.
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.
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.
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.
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.
To remove any O-benzoylbenzoic acid than may remain . by converted it to salt Dissolved