Answer to Question #150550 in Chemistry for Ananya

Question

how many grams of salt is necessary for preparing 150 gram of 4% sodium chloride solution?what is molarity of this solution?

Answer 1

Solution:

1) Mass Percent = (Grams of Solute / Grams of Solution) × 100%

Hence,

Grams of Solute = Grams of Solution × Mass Percent / 100%

Grams of Solute = 150 grams × 4% / 100% = 6 grams

Grams of Salt = Grams of Solute = 6 grams

2) Salt = sodium chloride = NaCl

Moles of NaCl = Mass of NaCl / Molar mass of NaCl

The molar mass of NaCl is 58.44 g mol^-1.

Hence,

Moles of NaCl = 6 grams / 58.44 g mol^-1 = 0.1027 moles

Molarity is the number of moles of solute per liter of solution, which can be calculated using the following equation:

Molarity = Moles Solute / Liters of Solution

Liters of Solution = Grams of Solution (assume that the density of the solution is 1).

Liters of Solution = 150 g = 150 mL = 0.15 L

Hence,

Molarity = 0.1027 moles / 0.15 L = 0.685 mol/L = 0.685 M

Molarity = 0.685 M

Answer:

1) 6 grams of salt is necessary for preparing solution.

2) The molarity of this solution is 0.685 M.