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# Answer to Question #191378 in General Chemistry for Christine

Question #191378

A 4.0520g sample of HCl, sp. Gr 11.18,

required 44.15ml of 0.9035Mof Sodium

Hydroxide in titration, compute for

percent purity.

1
2021-05-11T05:51:12-0400

The balanced chemical equation for above is:

"NaOH+HCl\\Rightarrow NaCl+H_2O"

Theoritical mass of "HCl" given = 4.0520 gm

Percent purity "= \\dfrac{Experimental mass}{Theoritical mass} \\times100"

Now we have to find experimental mass,which is calculated by the reaction.

For "NaOH"

M=0.9035 M

V=44.15 ml

so by the formula of molarity

"M=\\dfrac {moles(n)}{volume(ml)} \\times1000"

"0.9035= \\dfrac{n_{NaOH}}{44.15}\\times1000"

"n_{NaOH}= 0.04" moles

Now we will calculate the experimental moles of HCl

From the reaction:

1 mole of NaOH gave 1 mole of HCl

hence 0.04 moles of NaOH will gave 0.04 moles of HCl

Hence the experimental mass of HCl "=moles \\times molar mass"

Experimental mass"=0.04\\times36.5=1.46" gm

Now percent purity"=\\dfrac{Experimental mass }{Theoriticalmass}\\times100"

"=\\dfrac{1.46}{4.0520}\\times100=36.03" %

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