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What is the molarity of 2.0 L sodium hydroxide solution containing 10.0 grams solute?
The molarity is 0,125 M.
What is the molarity of an aqueous solution containing 22.5 grams of sucrose in 35.5 milliliters of solution?
To find the molarity, first convert the grams of sucrose to moles using its molar mass. Sucrose has a molar mass of 342.3 g/mol. Then, convert the milliliters of solution to liters. Finally, divide the moles of solute by the liters of solution to get the molarity.
What is the molarity of a solution that contains 1.1 of lithium in 0.5 liters of solution?
If 1,1 is grams the molarity is 0,317.
How does the amount of solute moles and grams relate to the density and molarity of the solution?
Increasing the amount of the solute in the solution the molarity and the density of this solution increases.
What is the molarity of 67.94 mL aqueous solution containing 2.822 g of KI?
To find the molarity, first calculate the number of moles of KI using its molar mass (166 g/mol). Then divide the moles of KI by the volume of the solution in liters (67.94 mL = 0.06794 L) to get the molarity. The molarity of the solution is approximately 6.24 M.
What is the molarity of 2.0 L sodium hydroxide solution containing 10.0 grams solute?
The molarity is 0,125 M.
What is the molarity of a 2.0 L sodium hydroxide solution containing 10.0 grams of a solute?
.13 (Plato)
What is the molarity of a solution containing 2000 grams of H2SO4 in 2500 mL of water?
The molarity of a solution containing 2 000 grams of H2SO4 in 2500 mL of water is 8,156.
What is the molarity of an aqueous solution containing 22.5 grams of sucrose in 35.5 milliliters of solution?
To find the molarity, first convert the grams of sucrose to moles using its molar mass. Sucrose has a molar mass of 342.3 g/mol. Then, convert the milliliters of solution to liters. Finally, divide the moles of solute by the liters of solution to get the molarity.
What is the molarity of a solution that contains 1.1 of lithium in 0.5 liters of solution?
If 1,1 is grams the molarity is 0,317.
What is the molarity of a solution made by dissolving 25 grams of Mg in excess NaOH solution?
The molarity of the solution is 0.5 M.
What is the molarity of a kf(aq) solution containing 116grams of kf in 1.00 liter solution?
The molar mass of KF is 58.1 g/mol. Given that there are 116 grams of KF in the solution, this corresponds to 2 moles of KF. Therefore, the molarity of the solution is 2 M.
What is the molarity of a 2.0-liter solution containing 58 grams of NaCl?
To calculate the molarity, you first need to convert the mass of NaCl to moles using its molar mass. The molar mass of NaCl is approximately 58.5 g/mol. Then, divide the moles of NaCl by the volume of the solution in liters to get the molarity. In this case, molarity = (58 grams / 58.5 g/mol) / 2.0 L.
How does the amount of solute moles and grams relate to the density and molarity of the solution?
Increasing the amount of the solute in the solution the molarity and the density of this solution increases.
What is the molarity of a CaCl2 solution containing 330 grams of CaCl2 in 1 liter of solution?
First, calculate the molar mass of CaCl2, which is 110.98 g/mol. Next, calculate the number of moles of CaCl2 in 330 grams using the formula moles = mass / molar mass. This gives you 2.97 mol of CaCl2. Finally, divide the moles of CaCl2 by the volume of the solution in liters to get the molarity, which is 2.97 M.
What is the molarity of a solution that contains 31.0 grams of HCl in 500cm3 of solution?
To calculate the molarity of the solution, first convert the mass of HCl to moles using its molar mass. Then, divide the number of moles by the volume of the solution in liters (500 cm3 = 0.5 L) to get the molarity.
What is the molarity of 67.94 mL aqueous solution containing 2.822 g of KI?
To find the molarity, first calculate the number of moles of KI using its molar mass (166 g/mol). Then divide the moles of KI by the volume of the solution in liters (67.94 mL = 0.06794 L) to get the molarity. The molarity of the solution is approximately 6.24 M.
