I would appreciate it if anyone could help me out with theproblems on molarity,

Question

I would appreciate it if anyone could help me out with theproblems on molarity, thank you!

A 9.50 % by mass solution of acetone (C3H6O) in water has adensity of 0.9849 g/mL at 20°C. What is the molarity of this solution? A) 0.621 M B) 1.61 M C) 1.66 M D) 1.71 M E) 16.9 M

Which of the following gives the molarity of a 17.0% by masssolution of sodium acetate, CH3COONa (molar mass = 82.0 g/mol) in water? Thedensity of the solution is 1.09 g/mL. A) 2.26 × 10–6 M B) 0.207 M C) 2.07 M D) 2.26 M E)2.72 M I would appreciate it if anyone could help me out with theproblems on molarity, thank you!

A 9.50 % by mass solution of acetone (C3H6O) in water has adensity of 0.9849 g/mL at 20°C. What is the molarity of this solution? A) 0.621 M B) 1.61 M C) 1.66 M D) 1.71 M E) 16.9 M

Which of the following gives the molarity of a 17.0% by masssolution of sodium acetate, CH3COONa (molar mass = 82.0 g/mol) in water? Thedensity of the solution is 1.09 g/mL. A) 2.26 × 10–6 M B) 0.207 M C) 2.07 M D) 2.26 M E)2.72 M Which of the following gives the molarity of a 17.0% by masssolution of sodium acetate, CH3COONa (molar mass = 82.0 g/mol) in water? Thedensity of the solution is 1.09 g/mL. A) 2.26 × 10–6 M B) 0.207 M C) 2.07 M D) 2.26 M E)2.72 M

Explanation / Answer

(1) 9.5 % by mass means 9.5 g of acetone dissolved in100 g of solution Given that density d = 0.9849 g / mL Density d = mass m / Volume V So Volume = 100 g / 0.9849 g/mL                  = 101.53 mL                  =0.10153 L We know that Molarity M = ( mass / Molar mass ) / Volumeof the solution in L Molar mass of acetone C3H6O is = 3 * 12 + 6 * 1 + 16 = 58g / mol So , Molarity M = ( 9.5 g / 58 g/mol ) / 0.10153L                           = 1.61 M (2) 17 % by mass means 17 g of sodium acetate dissolved in 100 g of solution Given that density d =1.09 g / mL Density d = mass m / Volume V So Volume = 100 g / 1.09 g/mL                  = 91.74 mL                  = 0.09174 L We know that Molarity M = ( mass / Molar mass ) / Volumeof the solution in L Molar mass of sodium acetate   CH3COONa is = 12 +3* 1 + 12 +2* 16 + 23 = 82 g / mol So , Molarity M = ( 17 g / 82 g/mol )/ 0.09174L                           =  2.26 M Given that density d =1.09 g / mL Density d = mass m / Volume V So Volume = 100 g / 1.09 g/mL                  = 91.74 mL                  = 0.09174 L We know that Molarity M = ( mass / Molar mass ) / Volumeof the solution in L Molar mass of sodium acetate   CH3COONa is = 12 +3* 1 + 12 +2* 16 + 23 = 82 g / mol So , Molarity M = ( 17 g / 82 g/mol )/ 0.09174L                           =  2.26 M

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