122.5g
This solution contain 26,3 g NaOH.
Two parts.Molarity = moles of solute/Liters of solution ( 100 mL = 0.1 Liters )0.5000 M NaOH = X moles/0.1 L= 0.05 moles NaOH--------------------------------now,0.05 moles NaOH (39.998 grams/1 mole NaOH)= 1.99 grams NaOH required====================( could call it 2.00 grams )
Since NaOH and HBr react in a 1:1 ratio, the moles of NaOH needed to titrate HBr can be calculated. Moles of NaOH = moles of HBr. Next, use the concentration and volume of HBr to find the moles present. Finally, use the concentration of NaOH to calculate the volume needed. In this case, approximately 2.41 mL of 0.305M NaOH would be needed.
You could titrate equal volumes of 1M solution of NaOH and 1M solution of HCl to obtain 1M solution of NaCl.
Molarity = moles of solute/Liters of solution. ( 350 ml = 0.350 Liters ) 5.7 M NaOH = moles NaOH/0.350 Liters = 1.995 moles NaOH (39.998 grams/1 mole NaOH) = 78 grams NaOH needed ------------------------------------
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 prepare a 1 normal solution of NaOH, you would need to dissolve 40.00 grams of NaOH in enough water to make 1 liter of solution. This is because the molecular weight of NaOH is 40 g/mol.
This solution contain 26,3 g NaOH.
A 50% NaOH aqueous solution means that the solution contains 50% sodium hydroxide (NaOH) by weight and the rest is water. This concentration indicates that for every 100 grams of the solution, 50 grams is NaOH.
To calculate the grams of NaOH in the solution, you first need to find the moles of NaOH present in 400.0 ml of the solution. The moles of NaOH can be calculated using the formula: moles = molarity x volume (in liters). Once you have the moles of NaOH, you can then calculate the grams using the formula: grams = moles x molecular weight.
Two parts.Molarity = moles of solute/Liters of solution ( 100 mL = 0.1 Liters )0.5000 M NaOH = X moles/0.1 L= 0.05 moles NaOH--------------------------------now,0.05 moles NaOH (39.998 grams/1 mole NaOH)= 1.99 grams NaOH required====================( could call it 2.00 grams )
To calculate the grams of NaOH in the solution, first determine the moles of NaOH using the molarity and volume. Then, convert moles to grams using the molar mass of NaOH. The molar mass of NaOH is 40 g/mol.
Molarity = moles of solute/Liters of solution ( 918 ml = 0.918 liters )rearranged algebraically,moles of solute = Liters of solution * Molaritymoles of NaOH = (0.918 l)(0.4922 M)= 0.45184 moles NaOH=======================so,0.45184 moles NaOH (39.998 grams/1 mole NaOH)= 18.1 grams sodium hydroxide needed============================
Since NaOH and HBr react in a 1:1 ratio, the moles of NaOH needed to titrate HBr can be calculated. Moles of NaOH = moles of HBr. Next, use the concentration and volume of HBr to find the moles present. Finally, use the concentration of NaOH to calculate the volume needed. In this case, approximately 2.41 mL of 0.305M NaOH would be needed.
To calculate the grams of NaOH needed, use the formula: grams = molarity x volume x molar mass. First, convert the volume to liters (4 liters). Next, calculate the grams using 8 M as the molarity and the molar mass of NaOH. This will give you the amount of NaOH required to make 4 liters of 8 M NaOH solution.
To calculate the grams of solute, you need to first find the molar mass of NaOH (40 g/mol). Then, use the formula grams = moles x molar mass. The moles of NaOH in 1.8 L of a 3.5 M solution is 6.3 moles. Therefore, the grams of NaOH in 1.8 L of a 3.5 M solution is 252 grams.
To make a 10% NaOH solution, you would need 100 grams of NaOH per liter of water. So to make 1 liter, you would need 100 grams of NaOH.