Question

Question 8 How many grams of NaF are in 24.5 mL of a 52.5% by volume (m/v) Naf solution? 0.4678 0.129 g 12.98 46.78 Question 9 What is the molarity of a solution prepared by dissolving 27.2 g of Ca(NO3)2 Into 289 ml. of water? (MM of Ca(NO3)2 = 164.09 g/mol) 0.000574 M 94.1 M 0.574 M 0.0479 M

Answer 1

D Given volues, volume of solution = 2415 ml my % = 52.5 Y. my weight of slute & 100 volume of solution weight of solute non 52.5 = - X100 24.5 100 coelight of solute = 52.5x24.5 = 12.86259 I weight of solute = 12.86 grams | . Answer is ③ option. Ginen volues, weight of ca(NO3)2 = 27:29 volume = 280ml Molecular weight of ca(NO3)2 = 164. og glimda nolarity (M) = weight of solute 1000 molecular weight & volume, (luml) a x -

27. 29 100D х Molarity = 164.09 g/mol 289 0.16576269 * 3.4602076 = 0.57357 molllit rolarity = 0.574M . Answer is ③ option,