0.0165 (L) * 0.750 (mol/L) = 0.0124 mol NaOH in 16.5 mL
mass divided by molar mass gives moles: 20 g NaOH / 40 g.mol-1 = 0.5 mol NaOH
The solution has a molar concentration of 0,33.
I believe the molarity is 1. molarity = number of moles / liters of solution molarity = 3 / 3 = 1
Need total molar mass sodium hydroxide which you can easily get from your periodic table. 2.2 moles NaOH (39.998 grams/1 mole NaOH) = 88 grams in mass ==============
The molar weight of NaOH is approx. 40 g/molSo 12.0 g of NaOH would be approx 12/40 = 0.3 mol NaOH
mass divided by molar mass gives moles: 20 g NaOH / 40 g.mol-1 = 0.5 mol NaOH
I assume you mean 32.0 grams of NaOH and 450 milliliters of NaOH. Molarity = moles of solute/Liters of solution ( 450 ml = 0.450 liters ) get moles of NaOH 32.0 grams NaOH (1 mole NaOH/39.998 grams) = 0.800 moles NaOH Molarity = 0.800 moles NaOH/0.450 liters = 1.78 Molar NaOH
The solution has a molar concentration of 0,33.
The number of moles is 0,0038.
I believe the molarity is 1. molarity = number of moles / liters of solution molarity = 3 / 3 = 1
Need total molar mass sodium hydroxide which you can easily get from your periodic table. 2.2 moles NaOH (39.998 grams/1 mole NaOH) = 88 grams in mass ==============
The molar weight of NaOH is approx. 40 g/molSo 12.0 g of NaOH would be approx 12/40 = 0.3 mol NaOH
Molar mass of NaOH is 39.998 grams. So, 2.75 mol NaOH (39.998g NaOH/1 mol NaOH) = 109.9945 grams NaOH ( you do the sigi figi )
The number of moles is 1,8.
The same number of moles for each.
The minimum molarity of HCl can be calculated by using the equation M1V1 = M2V2. Given that V1 = 18.6 ml, V2 = 18.6 ml, and M2 = 0.1 M (concentration of NaOH), you can solve for M1 to find the minimum molarity of HCl.
There is 0.5 moles of NaOH per litre To calculate 0.5 molar NaOH first know the molecular weight of NaOH i.e 40 now multiply the number of moles of NaOH you have (0.5) found as above. so to find the number of grams of NaOH we needed to start with (0.5) * (40) = 20g So dissolve 20g of NaOH in one litre of the solution to prepare 0.5 molar solution