A solution is prepared by adding 1.43 mol of kcl to 889 g of water. the concentration of kcl is ________ molal. a solution is prepared by adding 1.43 mol of kcl to 889 g of water. the concentration of kcl is ________ molal. 1.27 × 103 1.61 1.61 × 10-3 0.622 622
A solution is prepared by adding 1.43 mol of potassium chloride (kcl) to 889 g of water. The concentration of kcl is 1.61 molal. mol of Kcl (potassium chloride)= 1.43 water = 889 g the formula for calculating molality is: molality = moles of solute/kilograms of solvent 1kg = 1000g so, 889g = 0.889kg
