To answer this question, we must first calculate the number of moles of NaCl needed to make 250 mL of a 3. 00 M solution. To do this, we must use the following equation: Molarity = moles of solute / volume of solution.
In other words, if we want to make 250 mL of a 3. 00 M solution, we need to calculate the number of moles of NaCl needed to achieve this concentration level.
Using the equation, we can calculate the number of moles needed as follows:
Moles = Molarity x volume
Moles = 3.00 x 0.250
Moles = 0.75
Therefore, 0.75 moles of NaCl is required to make 250 mL of a 3.00 M solution.
How many moles are in 250mL of NaCl?
There are 0. 15 moles of NaCl in 250 mL of the solution. This is because NaCl is a 1:1 electrolyte, meaning that there is an equal ratio of sodium and chlorine. Since the molecular weight of sodium chloride is 58.
44 g/mol, one mole of sodium chloride would weigh 58. 44 grams. Therefore, 250 mL of NaCl would weigh 58. 44/1000 x 250 = 14. 61 grams. To calculate the amount of moles, divide the weight of the solution (14.
61 g) by the molecular weight (58. 44 g/mol). This gives 0. 15 moles of NaCl in 250 mL of the solution.
What is the number of moles in 3 l of 3 m NaCl?
The number of moles in 3 liters of 3 molar NaCl can be calculated using the following equation: moles = (concentration (M) x volume (L))/1000. In this case, moles = (3M x 3L)/1000 = 0.009 moles.
How many moles of sodium hydroxide are there in 250 mL of a 0.1 molar solution of sodium hydroxide?
There are 0.025 moles of sodium hydroxide in 250 mL of a 0.1 molar solution of sodium hydroxide. This can be determined using the following equation:
Moles = (Molarity)(Volume in Litres)
Molarity is expressed in moles per litre (mol/L), so we must first convert the volume of 250 ml into litres before plugging in into the equation:
Moles = (0.1 mol/L)(0.250 L)
Moles = 0.025 moles sodium hydroxide
How much of a 2.0 M NaCl stock solution would you need to make 250 mL of 0.15 M NaCl solution?
In order to make 250 mL of 0. 15M NaCl solution from a 2. 0M NaCl stock solution, you would need to use 37. 5 mL of stock solution. This calculation can be done using a simple dilution formula, which states that the volume is inversely proportional to concentration: V1/C1 = V2/C2.
This formula can be rearranged to solve for V1, the volume of stock solution needed: V1 = (C2 x V2)/C1. Plugging in the given values gives V1 = (0. 15 M x 250 mL)/2. 0 M, which is equivalent to 37. 5 mL.
What volume of 12.0 m HCl be used to prepare 300 mL of a 3.00 m solution?
In order to prepare 300 mL of a 3. 00 m solution of 12. 0 m HCl, the volumenecessary would be 8. 73 mL. This can be calculated through simple dilute concentration calculations. First, calculate the total moles of the solution.
This is done by taking the desired molarity of the solution (3. 00 m) and multiplying it with the volume (300 mL). This would give 0. 90 mol. Next, we need to calculate the moles of the stock solution (12.
0 m HCl). This is done by taking the molarity (12. 0 m) and multiplying it with the volume. The volume of the stock solution necessary is 0. 083 L or 83 mL. Since 1 mL is equal to 1 cm^3, 83 mL of stock solution would be equal to 83 cm^3, which is equal to 8.
73 mL. Therefore, 8. 73 mL of the 12. 0 m HCl should be used to prepare 300 mL of a 3. 00 m solution.
How much will be NaOH required to make 0.1 N 250 mL NaOH solution?
To make a 0. 1 N 250 mL NaOH solution, 25. 0 g of NaOH is required. This is calculated by multiplying the molarities of the solution (0. 1 N) and the volume of solution in liters (0. 250 L) and then multiplying the result by the molar mass of NaOH (40.
0 g/mol). Therefore, the total amount needed is:
(0.1 N)(0.250 L)(40.0 g/mol) = 25.0 g of NaOH
What is the weight of NaCl required to prepare 0.5 N solution 250?
If you want to prepare a 0. 5 N solution of sodium chloride (NaCl), you will need to know the molar mass of NaCl which is 58. 44 grams per mole. Since molarity is defined as moles of solute per liter of solution, you will need to calculate the amount of sodium chloride (in grams) required to prepare 250 liters of 0.
5 N NaCl solution.
To calculate this, multiply the molar mass of NaCl with the volume of the solution (250 liters) and the desired molarity (0.5N):
(58.44 g/mol) x (250 L) x (0.5 N) = 8720 grams of NaCl
Therefore, you will need a total of 8720 grams of NaCl to prepare a 0.5 N solution of 250 liters.
What is meant by 0.5 M solution?
A 0. 5 M solution is a term commonly used in chemistry to describe a solution with a specific concentration. The letter “M” stands for molar, which is a unit of concentration that measures the amount of solute (in moles) per unit volume of solution (in liters).
In a 0. 5 M solution, 0. 5 moles of a solute is present in every liter of solution. A 0. 5 M solution can be used to describe any type of solution, no matter its composition. For example, a 0. 5 M solution of sodium chloride would contain 0.
5 moles of sodium chloride per liter of solution, while a 0. 5 M solution of acetic acid would contain 0. 5 moles of acetic acid per liter of solution.
What is the concentration of 0.5 M?
The concentration of 0. 5 M is equal to 0. 5 moles of solute per liter of solution. This means that, if you had a liter of 0. 5 M solution, it would contain 0. 5 moles of the solute. For example, if the solute was sodium chloride, it would contain 0.
5 moles of sodium chloride. Concentrations are usually written using a letter abbreviation and the molarity, such as 0. 5 M NaCl.