NTA Abhyas JEE Main2020ChemistrySome Basic Concepts of ChemistryPractice
Which of the following gives the molarity of a 17.0% by mass solution of sodium acetate, CH 3 COONa FM = 82 .0 amu in water? Given the density is 1 .09 g/mol .
Options
- A2.26 × 1 0 - 6 M
- B0.207 M
- C2.07 M
- D2.26 M
Correct answer
D. 2.26 M
Step-by-step solution
17 .0% solution of CH 3 COONa means that 100 g solution contain 17 g of CH 3 COONa . Moles of CH 3 COONa = 17 g 82 .0 g mol − 1 = 17 82 .0 mol Volume of 100 g solution = Mass of solution Density of solution = 100 g 1 .09 g mol − 1 = 100 1 .09 mL So molarity of solution = Moles of solute Liter of soln . =   17 82 .0 mol 100 1 .09 ×10 − 3   L =   2 .259   M   =   2 .26   M