To calculate the molality (m) of a solution, you use the formula: ( m = \frac{\text{moles of solute}}{\text{kg of solvent}} ). In this case, with 3 moles of glucose in 6 kg of water, the molality would be ( m = \frac{3 , \text{mol}}{6 , \text{kg}} = 0.5 , \text{mol/kg} ). Therefore, the molality of the solution is 0.5 mol/kg.
To calculate the mole fraction of glucose, first find the moles of each component by dividing its mass by its molar mass. The molar mass of glucose (C6H12O6) is 180 g/mol. The moles of glucose is 180g / 180 g/mol = 1 mol. The moles of water is 162g / 18 g/mol = 9 mol. The total moles in the solution is 1 + 9 = 10 mol. The mole fraction of glucose is 1 mol / 10 mol = 0.1.
To create the solution, the chemist dissolves 0.50 moles of glucose in 2.0 kg of water. The molar mass of glucose (C₆H₁₂O₆) is approximately 180 g/mol, so 0.50 moles of glucose weighs about 90 grams. This results in a solution where the concentration of glucose can be calculated, and the total mass of the solution is approximately 2.0 kg of water plus 0.090 kg of glucose. Thus, the final mass of the solution is about 2.09 kg.
Dissolve 90 g of glucose in a small volume of water, and then add more water until the total volume of the solution is 1 L.
To prepare a 1 molar solution, you would need to dissolve 180 grams of glucose powder in enough water to make a final volume of 0.5 liters.
To make a 1 molar solution of glucose (C6H12O6) in 1 liter of water, you would need to dissolve 180 grams of glucose. This is because the molar mass of glucose is 180 g/mol, so 1 mole of glucose weighs 180 grams.
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.
The molality of the solution is 0.5 mol/kg. This is calculated by dividing the number of moles of solute (3 mol glucose) by the mass of solvent in kilograms (6 kg water).
To calculate the mole fraction of glucose, first find the moles of each component by dividing its mass by its molar mass. The molar mass of glucose (C6H12O6) is 180 g/mol. The moles of glucose is 180g / 180 g/mol = 1 mol. The moles of water is 162g / 18 g/mol = 9 mol. The total moles in the solution is 1 + 9 = 10 mol. The mole fraction of glucose is 1 mol / 10 mol = 0.1.
The molality of the solution is calculated by dividing the moles of solute by the mass of the solvent in kg. In this case, there are 3 mol of glucose in 6 kg of water. Therefore, the molality of the solution is 0.5 mol/kg.
To create the solution, the chemist dissolves 0.50 moles of glucose in 2.0 kg of water. The molar mass of glucose (C₆H₁₂O₆) is approximately 180 g/mol, so 0.50 moles of glucose weighs about 90 grams. This results in a solution where the concentration of glucose can be calculated, and the total mass of the solution is approximately 2.0 kg of water plus 0.090 kg of glucose. Thus, the final mass of the solution is about 2.09 kg.
first determine the number ofmoles dissolved in given solution then .5 moles moles dissolved in 800g. as comparison with 1000g of water, we know 100g of water dissolve only.1 moles of a glucose so we .7moles of glucose dissolve in 800g.
Dissolve 90 g of glucose in a small volume of water, and then add more water until the total volume of the solution is 1 L.
To make a 1 molar solution of glucose in 0.5 L of water, you would need to dissolve 90.1 grams of glucose powder. This is because the molar mass of glucose (C6H12O6) is approximately 180.2 g/mol, and for a 1 molar solution in 0.5 L of water, you would need 1 mole of glucose, which is 180.2 grams.
3mol/6kg
To prepare a 1 molar solution, you would need to dissolve 180 grams of glucose powder in enough water to make a final volume of 0.5 liters.
The mole fraction of NaCl in the solution is calculated by dividing the moles of NaCl by the total moles of NaCl and water. In this case, the mole fraction of 18 mol NaCl in a solution of 100 mol water is 0.15 (18 mol NaCl / (18 mol NaCl + 100 mol water)).
2 moles in 2 liters means 1 mole in 1 liter - so it its molarity is 1.