Question

A 1.05 g sample of a monoprotic acid is dissolved in water and titrated with 0.320 M NaOH. What is the molar mass of the acid if 19.0 mL of the NaOH solution is required to neutralize the sample? molar mass: molar mass: 8/mol g/mol about us can

Answer 1

Dante equ HA + NaOH Monoprotic Sudium acid hydroxide → Na Salt + 4,0 Water Molarity of Nach & Mole of NaOH Volume of Naoy on in litre ? 0.320 M s 0.019) mole of NaOH. 0.019 L wix 1L loro mi mole of NaOH = 00320 mole x 0.0194 (M = mole) Co s 0.00608 mole According to balanced equation, 1 mole of 4A (monoprotic acid) react with I mole of nach Similarly 0.00608 nole of Naon will react with 0.00608 mole of HA. Mole of HA - 0.00608 mole - Given mass of HA Molar mass of HA 105_q_ Molar mass of HA -1.05 9 0.00608 mole 172-69_9/mole Molar mass of HA Molar mass of HA 172.69 g/mol T (monoprotic and) or . 172g/mol (approx)

Well, I gave you exact value as well as value after rounding off (172.7). See , what question want - exact value or rounded off value