Asked Dec 23, 2019
  • What is the concentration of ion or molecule in the solution of a mixture of 45.0 mL of 0.272 M NaCl and 65.0 mL of 0.0247 M (NH4)2CO3. Assume that the volume are additive.



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Solution of mixture of 45.0 mL of 0.272 M NaCl and 65.0 mL ...


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The dissociation equation is given below, NaCl→Na* + CI One mole sodium chloride gives one sodium ion and one chlorine ion. Therefore, The concentration of ions or molecule in the solution of 45.0 mL of 0.272 M NaCl is calculated as follows, Total volume of the solution=45mL+65mL=110mL 0.272 Mof NaCl x45mL. 110 mL Concentartion of NaCl=- Concentartion of NaCl= 0.111M Therefore, [ Na"]=0.111M [cr]=0.11M


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