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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 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 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.
What is the molality of a solution that has 6 mol of CaCI2 in 3kg of water?
Molality (m) is calculated using the formula ( m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} ). In this case, 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.
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).
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 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 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 molality of solution that has 3 mol of glucose in 6 Kg of water?
3mol/6kg
How much would the boiling point of water increase if 4 mol of NaCl were addes to 1 kg of water?
The boiling point elevation constant for water is 0.512 °C/kg/mol. When 4 mol of NaCl are added to 1 kg of water, the increase in boiling point would be 4 * 0.512 = 2.048 °C.
What is the molality of .630g of NaCl dissolved in 525g of water?
To find molality, first convert grams of NaCl to moles: 0.630g NaCl / 58.44 g/mol NaCl = 0.0108 mol NaCl. Next, calculate molality using moles of solute and mass of solvent: molality = moles of solute / kg of solvent = 0.0108 mol NaCl / 0.525 kg water = 0.0206 mol/kg.
How much would the freezing point of water decrease if 4 mole of Na Cl were added to 1 kg of water?
The freezing point depression constant for water is 1.86°C kg/mol. With 4 moles of NaCl added to 1 kg of water, the molality would be 4 mol / 1 kg = 4 m. The freezing point depression can be calculated as 1.86°C kg/mol * 4 m = 7.44°C.
