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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.
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A solution contains 180g of glucose C6H12O6 and 162 g of water What is the mole fraction of glucose?
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.
A chemist forms a solution by mixing 0.50 mol glucose and 2.0 kg water?
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.
What is the molality of a solution that as 3 mol of glucose in 6 kg of water?
Molality is calculated by dividing the number of moles of solute by the mass of the solvent in kilograms. In this case, the solution has 3 moles of glucose dissolved in 6 kg of water. Therefore, the molality (m) is 3 mol / 6 kg = 0.5 mol/kg. Thus, the molality of the solution is 0.5 m.
The molecular mass of glucose C6H12O6 is 180 g To make a 0.5 M solution of glucose you should do what?
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.
What is the molality of a solution that has 3 mol of glucose in 6kg of water?
Molality (m) is calculated using the formula ( m = \frac{\text{moles of solute}}{\text{kg of solvent}} ). In this case, there are 3 moles of glucose and 6 kg of water. Therefore, the molality of the solution is ( m = \frac{3 , \text{mol}}{6 , \text{kg}} = 0.5 , \text{mol/kg} ). Thus, the molality of the solution is 0.5 mol/kg.
What is the morality of a solution that has 3 mol of glucose in 6kg of water?
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.
What is the molality of a solution that has 3 mol of glucose in 6 kg of water?
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).
A solution contains 180g of glucose C6H12O6 and 162 g of water What is the mole fraction of glucose?
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.
What is the molality solution that has 3 mol of glucose in 6 kg if water?
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.
A chemist forms a solution by mixing 0.50 mol glucose and 2.0 kg water?
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.
What is the molality of a solution that as 3 mol of glucose in 6 kg of water?
Molality is calculated by dividing the number of moles of solute by the mass of the solvent in kilograms. In this case, the solution has 3 moles of glucose dissolved in 6 kg of water. Therefore, the molality (m) is 3 mol / 6 kg = 0.5 mol/kg. Thus, the molality of the solution is 0.5 m.
What is the molality of a solution of 90 grams of glucose C6H12OSUB6 dissolved in 800 grams of water?
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.
The molecular mass of glucose C6H12O6 is 180 g To make a 0.5 M solution of glucose you should do what?
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.
What is the molality of a solution that has 3 mol of glucose in 6kg of water?
Molality (m) is calculated using the formula ( m = \frac{\text{moles of solute}}{\text{kg of solvent}} ). In this case, there are 3 moles of glucose and 6 kg of water. Therefore, the molality of the solution is ( m = \frac{3 , \text{mol}}{6 , \text{kg}} = 0.5 , \text{mol/kg} ). Thus, the molality of the solution is 0.5 mol/kg.
How much glucose powder in 0.5 L of water for a 1 molar solution?
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.
What is the molality of solution that has 3 mol of glucose in 6 Kg of water?
3mol/6kg
How much glucose powder in .5 of water for a 1 molar solution?
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.
