Question

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

required 44.15ml of 0.9035Mof Sodium

Hydroxide in titration, compute for

percent purity.

Answer 1

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$=\dfrac{Experimental mass }{Theoriticalmass}\times100$

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