EBK EXPLORING CHEMICAL ANALYSIS
EBK EXPLORING CHEMICAL ANALYSIS
5th Edition
ISBN: 9781319416942
Author: Harris
Publisher: VST
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Chapter 1, Problem 1.15P

(a)

Interpretation Introduction

Interpretation:

The volume of 98 wt%H2SO4 which has molarity 18.0M, that has to be diluted to 1.00 L to make 1.00 MH2SO4 has to be determined.

Concept Introduction:

Molarity, is one of the parameters used to express concentration of a solution.  It is defined as the number of moles of the solute dissolved per liter of the solution.  It is expressed as,

  Molarity = number of moles of solutevolume of solution in L

Weight percent is one of parameters used to express concentration of a solute and it is defined as percentage of weight of the solute present with respect to weight of the solution.  It is expressed as,

    Weight percent of a solute = weight of soluteweight of solution × 100%

Dilution formula refers to a formula to obtain the quantity of concentrated solution to be withdrawn to make dilute solution of desired quantity.  The formula is given as,

Mconc.Vconc=Mdil.Vdil

Where,

    Mconc = molar concentration of concentrated solutionVconc = volume of concentrated solutionMdil = molar concentration of diluted solutionVdil = volume of diluted solution

(a)

Expert Solution
Check Mark

Answer to Problem 1.15P

The volume of 98 wt%H2SO4 which has molarity 18.0M, that has to be diluted to 1.00 L to make 1.00 MH2SO4 is calculated as 55.6 mL.

Explanation of Solution

Applying the dilution formula volume of 18.0M98 wt%H2SO4 that has to be withdrawn to make 1.00 L of 1.00 MH2SO4

The Dilution formula is given as,

Mconc.Vconc=Mdil.Vdil          (1)

Let ‘x’ be the volume of 18.0M (concentrated solution) 98 wt%H2SO4 that has to be withdrawn for diluting to 1.00L of 1.00 MH2SO4 (diluted)  solution.  Thus,

    Mconc = 18.0 MVconc = xMdil = 1.00 MVdil = 1.00L

Plugging the above values in the equation (1),

    18.0 M × x =  1.00 M × 1.00L     x = 1.00 M × 1.00 L18.0 M = 0.0556 L = 55.6 mL

(b)

Interpretation Introduction

Interpretation:

Density of 98 wt%H2SO4 has to be determined.

Concept Introduction:

Refer to part (a)

(b)

Expert Solution
Check Mark

Answer to Problem 1.15P

Density of 98 wt%H2SO4 is calculated as 1.80 g/mL.

Explanation of Solution

Given that molarity of 98 wt%H2SO4 is 18.0M.  The solution is diluted to volume of 1.00L and molarity 1.0M.

Calculate the number of moles of H2SO4 in 98 wt%H2SO4

    moles of H2SO4 Molarity of H2SO4×Volume of solution in L =18.0M×1.00L = 18.0 mol

Convert 18.0 mol to mass,

  Mass of H2SO4  =  Moles of H2SO4×Formula mass of H2SO4 = 18 mol×98.079 g/mol = 1.77×103 g H2SO4

The solution contains 98 wt%H2SO4, mass of H2SO4 per milliliter is 1.77g,

Density of solution is calculated as,

  density = massvolume

Plug the above values to calculate density,

    density = 1.77 g 0.980 g NaOH/ g solution = 1.80 g/mL

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