Question

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

Answer 1

The concentration of the KCl solution is 1.61 molal.

Step-by-step explanation:

The concentration of a solution can be described by its molality (m), which is the number of moles of solute per kilogram of solvent. To find the molality of the KCl solution, we need to calculate the number of moles of KCl and the amount of water in kilograms. Given that 1.43 mol of KCl is dissolved in 889 g of water, we convert the mass of water to kilograms (889 g = 0.889 kg). Now we can calculate the molality:

Molality (m) = moles of KCl / kilograms of water

Molality (m) = 1.43 mol / 0.889 kg = 1.61 molal

Answer 2

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

m = 1.43/0.889 = 1.61 molal