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A sample contains both naoh and nacl. 0.500 g of this sample was dissolved in water to make a 20.0 ml solution and then this solution was titrated by 0.500 mol/l hcl solution. If 22.1 ml of hcl was used to reach the end point, what is the mass % of naoh in the sample?

Respuesta :

[tex]Solution:

moles of HCL =\frac{molarity\times volume}{1000}\\

=\frac{0.500\times22.1}{1000}\\

=0.0110\\

NAOH +HCL -------------->Nacl+H_{2}O\\

from the reaction NAOH and HCL react with 1:1\\

so moles of NAOH are needed = 0.0110\\

\frac{weight}{molar mass}=0.0110\\

weight=40\times0.0110\\

     =0.44g\\

% naoH = \frac{mass of NaoH}{total mass}\times 1000\\

(0.44\times0.5)\times100\\

88%[/tex]

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