Molarity:
1.0 mol (NaOH) /L(solution) = 1.0 M NaOH
The conversions of molality, b, to and from the molarity , c,
for one-solute solutions are:
c = ρ.b / [1 + b.M]
and
b = c / [ρ -c.M]
where ρ is the mass density of the solution, b is the molality, and M is the molar mass of the solute.
This depends on the mass of NaOH dissolved in 1 L water.
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
To find the molarity of NaOH, first calculate the moles of HCl: ( \text{moles of HCl} = \text{volume (L)} \times \text{molarity} = 0.020, \text{L} \times 5.0, \text{M} = 0.10, \text{moles} ). In a neutralization reaction, HCl and NaOH react in a 1:1 ratio, so 0.10 moles of NaOH are needed. The molarity of NaOH can be calculated using the formula ( \text{Molarity} = \frac{\text{moles}}{\text{volume (L)}} = \frac{0.10, \text{moles}}{0.100, \text{L}} = 1.0, \text{M} ). Thus, the molarity of the NaOH is 1.0 M.
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.
The answer is 0,625 moles.
This depends on the mass of NaOH dissolved in 1 L water.
To calculate the molarity, you first need to convert the grams of NaOH to moles using the molar mass of NaOH (40 g/mol). Then, you divide the moles of NaOH by the volume of solution in liters (450 ml = 0.45 L) to get the molarity. Molarity = moles of NaOH / volume of solution in liters Moles of NaOH = 95 g / 40 g/mol = 2.375 mol Molarity = 2.375 mol / 0.45 L = 5.28 M
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 molarity of the solution can be calculated by dividing the moles of solute by the volume of solution in liters. In this case, 2 moles of NaOH in 1620 mL (1.62 L) of water gives a molarity of approximately 1.23 M.
The reaction between HNO3 and NaOH is a 1:1 molar ratio. This means that the moles of HNO3 required to neutralize the NaOH is the same as the moles of NaOH. Given that 20.0 ml of HNO3 is needed to neutralize 10.0 ml of a 1.67 M NaOH solution, the molarity of the HNO3 solution is twice the molarity of the NaOH solution, which is 3.34 M.
To find the molarity, you need to know the amount in moles of NaOH and the volume in liters. First, convert 10 mL to liters by dividing by 1000 (10 mL = 0.01 L). Then, calculate the number of moles of NaOH using the molarity formula, Molarity = moles/volume. Given that you have 0.05 moles of NaOH and a volume of 0.01 L, the molarity would be 5 M.
Molarity = moles solute/Liters solution get moles NaOH 0.240 grams NaOH (1 mole NaOH/39.998 grams) = 0.0060 moles NaOH ----------------------------------as one to one OH- has this many moles also Molarity = 0.0060 moles OH-/0.225 Liters = 0.0267 M OH- ----------------------- -log(0.0267 M OH-) = 14 - 1.573 = 12.4 pH -------------
To calculate the normality of NaOH for a 1:1 molar ratio, you can use the formula: Normality = Molarity x Number of equivalents. Since NaOH is a monoprotic base, it provides one equivalent per mole. Therefore, for a solution of NaOH with a molarity of 1 M, the normality would be 1 N.
The balanced equation for the reaction is 1 mole of NaOH to 1 mole of HNO3. Using the titration data, you can calculate the moles of HNO3 used. From there, you can determine the moles of NaOH present in the 4.37 ml solution. Finally, dividing the moles of NaOH by the volume of the NaOH solution in liters will give you the molarity.
To find the molarity, we first need to calculate the number of moles of NaOH. The molar mass of NaOH is 40 g/mol (sodium=23g/mol, oxygen=16g/mol, hydrogen=1g/mol). Thus, 80g NaOH is 2 moles (80g / 40g/mol). Given 1L of solution, the molarity is 2 moles / 1 L = 2 M.
I believe the molarity is 1. molarity = number of moles / liters of solution molarity = 3 / 3 = 1
To find the molarity of NaOH, first calculate the moles of HCl: ( \text{moles of HCl} = \text{volume (L)} \times \text{molarity} = 0.020, \text{L} \times 5.0, \text{M} = 0.10, \text{moles} ). In a neutralization reaction, HCl and NaOH react in a 1:1 ratio, so 0.10 moles of NaOH are needed. The molarity of NaOH can be calculated using the formula ( \text{Molarity} = \frac{\text{moles}}{\text{volume (L)}} = \frac{0.10, \text{moles}}{0.100, \text{L}} = 1.0, \text{M} ). Thus, the molarity of the NaOH is 1.0 M.