The given is 18gr per 90 mL so that's 1 gram per (90/18) mL or 1 gr per 5mL There is 1000 mL per one L so it's 1 gr per .005 L
The molarity of a 5% solution of NH3 in water depends on the density and molecular weight of NH3. Without this information, it is not possible to calculate the molarity.
Glucose is the solute; water is the solvent.
The molarity is 2 mol/L.
A 30% glucose solution is purely glucose and water, though it is actually impossible to keep other contaminants out of it. To create a 30% solution of glucose, you take a fixed volume of water and add 30% of that value of glucose to the water. The amount of glucose is in grammes. For example, 3g of glucose would be added to 10ml of water.
The molarity is 6 mol/l.
The molarity of a solution with 3 mol of glucose in 6 kg of water cannot be determined without knowing the volume of the solution. Molarity is defined as the amount of solute (in mol) divided by the volume of the solution in liters. Without the volume, the molarity of the solution cannot be calculated.
No, glucose itself does not contain electrolytes. Electrolytes are typically ions such as sodium, potassium, chloride, and bicarbonate that can conduct electricity in solution. If electrolytes are needed, they would have to be added separately to a glucose solution.
To find the molarity of the sugar solution, we first need to convert the mass of glucose to moles. The molar mass of glucose (C6H12O6) is 180 g/mol. So, 80 g of glucose is equal to 80/180 = 0.444 moles. Next, calculate the molarity by dividing moles of glucose by liters of solution (0.75 L), Molarity = 0.444 moles / 0.75 L = 0.592 M.
To determine the molarity, we need the volume of the solution. Assuming the density of water is 1 kg/L, we can convert the mass of water to volume (6 kg = 6000 mL). Since glucose doesn't affect the volume significantly, we consider the volume of water only. With 3 moles of glucose in 6000 mL of water, the molarity is 0.5 M (3 moles / 6000 mL).
To prepare a 10% glucose solution, you would mix 10 grams of glucose with enough water to make a total solution volume of 100 ml. This means the final solution would contain 10 grams of glucose and 90 ml of water.
The molarity of a 5% solution of NH3 in water depends on the density and molecular weight of NH3. Without this information, it is not possible to calculate the molarity.
The balloon will contain a mixture of the 10% and 5% glucose solutions. Since water can pass through but not glucose, the glucose concentration inside the balloon will decrease over time as water moves from the lower concentration in the beaker to the higher concentration in the balloon through osmosis.
No, adding water to a solution does not change the molarity of the solute. The molarity of a solution is calculated using the amount of solute and the volume of the solution, so diluting with water only changes the volume, not the amount of solute present.
The molarity of a solution is calculated by dividing the number of moles of solute by the volume of the solution in liters. In this case, the molarity of the solution with 4 mol of NaOH dissolved in 2 L of water would be 2 M.
The question, as worded, is a little ambiguous. Rather, the question you should be asking is “What is the molarity of a 125 ml aqueous solution containing 10.0g of acetone?” Acetone is roughly 58 grams per mole. Therefore, a 125 mil solution with 10 g of acetone would contain roughly 0.17 moles, and the molarity would be roughly 1.4See the Related Questions for more information about how to calculate the molarity of a solution
Glucose is the solute; water is the solvent.
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.