Calculate the percent error in Avogadro’s number calculated by using the following experimental data. Pipette calibrated as 30 drops/mL Diameter of the watch glass: \$15.5 mathrmcm\$ Number of drops to finish monolayer: 23 Concentration of stearic acid solution: \$1.2 imes10^-4 mathrmg/mL\$

Updated: My assumed on this: I would take \$23/30=0.77 mathrmml\$. Then take \$0.77 imes1.2 imes10^-4=9.24 imes10^-5 mathrm g\$ of stearic acid. Then \$(9.24 imes10^-5)/284.48 mathrm g=3.25 imes10^-7 mathrmmol\$ of stearic acid. Then how carry out I discover the Avogadro"s number (molecules per mole)?

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edited Oct 9 "16 at 15:40
user7951
asked Oct 9 "16 at 14:39

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When a problem calls for calculations utilizing worths, always write the worths with the correct devices and bring the units via the calculation. Do not omit the devices while perdeveloping intermediate actions and perform not simply reintroduce units at the end of the calculation.

Nevertheless, your technique is correct. You calculated the volume of the solution as\$\$eginalignV_ extsolution&=frac2330 mathrmml^-1 ag1\<6pt>&=0.77 mathrmml ag2endalign\$\$the concentration as\$\$eginalignc&=frac nV_ extsolution=frac mMcdot V_ extsolution ag3\<6pt>&=frac gamma M ag4\<6pt>&=frac1.2 imes10^-4 mathrmg ml^-1284.48 mathrmg mol^-1 ag5\<6pt>&=4.2 imes10^-7 mathrmmol ml^-1 ag6endalign\$\$and the amount of stearic acid as\$\$eginalignn&=ccdot V_ extsolution ag7\<6pt>&=frac1.2 imes10^-4 mathrmg ml^-1284.48 mathrmg mol^-1cdotfrac2330 mathrmml^-1 ag8\<6pt>&=3.2 imes10^-7 mathrmmol ag9endalign\$\$

In addition, you have the right to calculate the area of the surface as\$\$eginalignA&=pi r^2 ag10\<6pt>&=fracpi d^24 ag11\<6pt>&=fracpi left(15.5 mathrmcm ight)^24 ag12\<6pt>&=189 mathrmcm^2 ag13endalign\$\$

For simplicity’s sake, you may assume that a solitary stearic acid molecule is a small block through a length \$l\$ and also a width \$w\$.

Depending on the supplied design, you might uncover that the proportion is roughly

\$\$l=6w ag14\$\$

Assuming that the surconfront is densely packed via stearic acid molecules, the number \$N\$ of molecules have the right to be estimated as\$\$N=frac Aw^2 ag15\$\$

The corresponding volume of stearic acid is

\$\$eginalignV&=Acdot l ag16\<6pt>&=Acdot 6w ag17endalign\$\$

You likewise recognize that\$\$eginalignV&=frac m ho ag18\<6pt>&=frac ncdot M ho ag19endalign\$\$

wright here \$ ho\$ is the thickness of stearic acid. Instead of making use of the thickness of solid stearic acid at room temperature, you may want to assume the thickness of liquid stearic acid close to its melting allude, which is

\$\$ ho=0.847 mathrmg cm^-3 ag20\$\$

Solving \$ ext(17)\$ for \$w\$ and also inserting the outcome into \$ ext(15)\$ yields

\$\$N=Acdotleft(frac6AV ight)^2 ag21\$\$

And inserting \$ ext(19)\$ for \$V\$ yields

\$\$N=Acdotleft(frac6Acdot honcdot M ight)^2 ag22\$\$

The Avogadro consistent is defined as\$\$N_mathrm A=frac Nn ag23\$\$

Inserting \$ ext(22)\$ for \$N\$ yields

\$\$eginalignN_mathrm A&=fracAcdotleft(frac6Acdot honcdot M ight)^2n ag24\<6pt>&=left(frac6 hoM ight)^2left(frac An ight)^3 ag25\<6pt>&=left(frac6 imes0.847 mathrmg cm^-3284.48 mathrmg mol^-1 ight)^2left(frac189 mathrmcm^23.2 imes10^-7 mathrmmol ight)^3 ag26\<6pt>&=6.6 imes10^22 mathrmmol^-1 ag27endalign\$\$

Even once taking the uncertainty of the provided values right into account, this result is significantly smaller sized than the literary works value of the Avogadro continuous. Therefore, you may conclude that a large component of the difference may be attributed to experimental error. I guess, you did not use the floating device approach to push the stearic acid molecules together, which provides certain that the molecules are densely packed and also lined up on the surchallenge of the water via polar heads on the surconfront and nonpolar tails sticking up ameans from the surconfront.