Question

A stock solution was prepared by dissolving 0.593 g of pure ASA (molar mass = 180.2 g/mol) with NaOH solution and diluting with water in a 1000 mL volumetric flask. This was followed by pipetting 4.00 mL of the stock solution into flask A and 5.0 mL of the stock into flask B. Flasks A and B were each diluted to 20 mL mark of the volumetric flask. What are the concentrations, in molarity, of ASA in flasks A and B

Answer 1

First finding concentration a Stock solutim ASA . Given weight g ASA = 0.5939 molar mass = 180.2 mol mol 7 AS A = given mass Molar mass = 0.5939 -0.0033 Mo 180.2 moit Volumen solur'on = 100ml -IL - 1L=1000m2 Concentration of s tock - Volume (L) - = 0.0033M 0.0033 mol IL Flask-A Volume (V) y stock sol te ken = 4:00ML Molavity ČMp) 9 stock sol = 0.0033 M volume (V2) of diluted sol = 20m Molarity (M2) diluted sole using Molarity equation (stock) Mivi = Mal (Prepared) 0.0033 MXUML= M₂ x 20m 00033 M X 40C 20 me 0.00066 M - M2 - M2

concentran'm in Flask A = 6.6x10-4M SL formada Flask B Molosity (M1) of stock sol = 5ml volume (VD of stock sol = 5.oome volume (V2) of Prepared 80l = 20m M2 = ? (stock) miri = Mall ( prepared) M2x zome 0.00 33 mx5m2= = M₂ 0.0033 MX 5 2ome 0.0008 25 m = m2 Concentration in Flask B = 8.25x 10-4m