3mol/6kg
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.
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).
The molality of a solution is calculated by dividing the moles of solute by the mass of the solvent in kilograms. In this case, the molality of the CaCl2 solution would be 2 mol/kg, as 6 mol of CaCl2 dissolved in 3 kg of water results in a molality of 2 mol/kg.
The molality of a solution is calculated by dividing the moles of solute by the mass of the solvent in kilograms. In this case, the molality would be 0.5 mol/kg since there are 3 moles of glucose in 6 kg of water.
2 m is the molarity of a solution that has 6 mol of CaCl2 in 3 km of 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.
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).
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 molality of a solution is calculated by dividing the moles of solute by the mass of the solvent in kilograms. In this case, the molality of the CaCl2 solution would be 2 mol/kg, as 6 mol of CaCl2 dissolved in 3 kg of water results in a molality of 2 mol/kg.
4 mol over 0.800 kg
The molality of a solution is calculated by dividing the moles of solute by the mass of the solvent in kilograms. In this case, the molality would be 0.5 mol/kg since there are 3 moles of glucose in 6 kg of water.
2 m is the molarity of a solution that has 6 mol of CaCl2 in 3 km of water.
4 mol/0.800 kg
2 m
To find the molality, we first calculate the moles of Na2SO4: 10.0g Na2SO4 * (1 mol Na2SO4 / 142.04g Na2SO4) = 0.0705 moles Na2SO4. Then, molality is calculated as moles of solute (Na2SO4) / kilograms of solvent (water): 0.0705 mol / 1.000 kg = 0.0705 mol/kg, which is the molality of the solution.
To calculate the molality (m) of a solution, use the formula: [ m = \frac{\text{moles of solute}}{\text{mass of solvent (kg)}} ] In this case, there are 6 moles of calcium carbide (CaC₂) and 3 kg of water. Thus, the molality is: [ m = \frac{6 , \text{mol}}{3 , \text{kg}} = 2 , \text{mol/kg} ] Therefore, the molality of the solution is 2 mol/kg.
Molality = moles of solute / kilograms of solventm = 5.10 / 4.47m = 1.14 molal