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.
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.
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.
Molality (m) is calculated using the formula ( m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} ). For a solution with 6 moles of CaCl₂ dissolved in 3 kg of water, the molality would be ( m = \frac{6 \text{ mol}}{3 \text{ kg}} = 2 \text{ mol/kg} ). Therefore, the molality of the solution is 2 mol/kg.
To determine the boiling point of a solution of glucose in water, we first calculate the molality of the solution. With 0.10 moles of glucose in 200 mL of water (approximately 0.2 kg), the molality is 0.5 mol/kg. The boiling point elevation can be calculated using the formula ΔT_b = i * K_b * m, where K_b for water is 0.512 °C kg/mol. Thus, the boiling point elevation is approximately 0.26 °C, raising the boiling point of water from 100 °C to about 100.26 °C.
To find the molality of the solution, you can use the formula: molality (m) = moles of solute / mass of solvent (in kg). In this case, there are 6 moles of CaCl2 and 3 kg of water. Therefore, the molality is 6 mol / 3 kg = 2 m. Thus, the molality of the solution is 2 molal.
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).
3mol/6kg
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.
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.
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.
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 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.
4 mol/0.800 kg
2 m