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How much would the boiling point of water increase if 4 mol of NaCl were added to 1kg of water?
To determine the boiling point elevation of water when 4 mol of NaCl is added to 1 kg of water, we can use the formula: ΔT_b = i * K_b * m, where ΔT_b is the boiling point elevation, i is the van 't Hoff factor (which is 2 for NaCl), K_b for water is approximately 0.512 °C kg/mol, and m is the molality. Since NaCl dissociates into 2 ions (Na⁺ and Cl⁻), i = 2, resulting in a total of 8 mol of particles (4 mol NaCl × 2). The molality (m) is 4 mol / 1 kg = 4 mol/kg. Thus, ΔT_b = 2 * 0.512 °C kg/mol * 4 mol/kg = 4.096 °C. Therefore, the boiling point of water would increase by approximately 4.1 °C.
What is the morality solution that has 3 mol of glucose in 6 kg if water?
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.
What would you need to do to calculate the molality of 10 mol of NaCI in 200 mol of water?
To calculate the molality of 10 mol of NaCl in 200 mol of water, you would use the formula for molality (m), which is defined as the number of moles of solute (NaCl) divided by the kilograms of solvent (water). First, convert the mass of water from moles to kilograms; since the molar mass of water is approximately 18 g/mol, 200 mol of water is about 3,600 g or 3.6 kg. Finally, plug the values into the formula: molality (m) = 10 mol NaCl / 3.6 kg water, yielding a molality of approximately 2.78 mol/kg.
What is the molarity of a solution that has 6 mol of CaCl2 in 3 kg of water?
The molarity is 2 mol/L.
What is the molality of a solution that has 6 mol of CaCI2 in 3 kg of water?
Molality (m) is calculated using the formula ( m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} ). In this case, there are 6 moles of CaCl₂ dissolved in 3 kg of water. Therefore, the molality is ( m = \frac{6 , \text{mol}}{3 , \text{kg}} = 2 , \text{mol/kg} ). Thus, the molality of the solution is 2 mol/kg.
What is the molarity of a solution that has mol of kcl in 0.800 kg in water?
4 mol/0.800 kg
What is the molality of a solution that has 6 mol of CaCl2 in 3 kg of water?
2 m
How much would the freezing point of water decrease if 4 mol of sugar were added to 1 kg of water?
7.44 C
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.
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).
How much would the boiling point of water increase if 4 mol of NaCl were added to 1kg of water?
To determine the boiling point elevation of water when 4 mol of NaCl is added to 1 kg of water, we can use the formula: ΔT_b = i * K_b * m, where ΔT_b is the boiling point elevation, i is the van 't Hoff factor (which is 2 for NaCl), K_b for water is approximately 0.512 °C kg/mol, and m is the molality. Since NaCl dissociates into 2 ions (Na⁺ and Cl⁻), i = 2, resulting in a total of 8 mol of particles (4 mol NaCl × 2). The molality (m) is 4 mol / 1 kg = 4 mol/kg. Thus, ΔT_b = 2 * 0.512 °C kg/mol * 4 mol/kg = 4.096 °C. Therefore, the boiling point of water would increase by approximately 4.1 °C.
What is the morality solution that has 3 mol of glucose in 6 kg if water?
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.
What is the molality of a solution made by dissolving 2 mole of NaOH in 6kg of water?
The molality of a solution is calculated by dividing the moles of solute by the mass of the solvent in kg. In this case, there are 2 moles of NaOH dissolved in 6 kg of water. Therefore, the molality would be 2 mol / 6 kg = 0.33 mol/kg.
What is the molality of a solution that has 6 mol of cacl2 in 3kg of 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.
What is the molarity of a solution that has 6 mol of CaCl2 in 3 kg of water?
The molarity is 2 mol/L.
What is the molality of a solution that has 6 mol of CaCI2 in 3 kg of water?
Molality (m) is calculated using the formula ( m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} ). In this case, there are 6 moles of CaCl₂ dissolved in 3 kg of water. Therefore, the molality is ( m = \frac{6 , \text{mol}}{3 , \text{kg}} = 2 , \text{mol/kg} ). Thus, the molality of the solution is 2 mol/kg.
What is the molality of a solution that has 6 mol of CACI2 on 3 kg of water?
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.
