**Problem:**A sample of an acid weighing 2.4 g is dissolved in 100 ml of water. This sample is titrated with 40 ml of sodium hydroxide (0.10 N). Find out the molecular mass of the acid, knowing that it is mono-equivalent.**Solution Explanation:**This problem involves a titration process where a known concentration of a base (sodium hydroxide) is used to determine the molecular mass of an acid. The acid is described as "mono-equivalent," meaning that one mole of the acid reacts with one mole of the base. The objective is to calculate the molecular mass of the acid using the given information.

An acid- base titration is a quantitative analysis which is done to determine the concentration of either an acid or base by neutralizing it with a known concentration of standard solution of an acid or base. Given: N1V1 (acid)=N2V2(base)N1= xN2= 0.10NV1= 100 mLV2= 40 mL x x 100 = 0.10 x 40x = 0.10 x 40100= 0.04 N Normality (N) =Weight of Solute (g)Equivalent weight × Volume (L)Equivalent weight = Wt. of solute(g) Normalityx Volume (L)= 2.40.04 x 0.1= 600 Equivalent wt = Molecular massNo. of equivalent As it is given that the acid is mono-equivalent, hence it will have 1 no. of equivalent. Therefore, MW= Eq x no. of equivalent= 600 x 1= 600g

Answered: 2- A sample of an acid 2.4 g dissolved…

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2- A sample of an acid 2.4 g dissolved in 100 ml. water. This sample was titrated with 40 ml sodium hydroxide (0.10 N). Find out the molecular mass of the acid knowing that it is mono-equivalent.

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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