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Calculate the number of molecules of exalic acid `(H_(2)C_(2)O_(4)2H_(2)O)` in 100 mL of 0.2 N oxalic acid solution. |
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Answer» Calculation of mass of oxlic acid `" Normality of solution"=("Mass of oxalic acid"//"Equivalent mass")/("Volume of solution in litres")` `"Normality mass of acid"= ("Molar mass")/("Basicity")=(2xx1+2xx12+4xx16+2xx18)/2=126/2` `63 g equiv^(-1)` Volume of solution =100mL=100/1000=0.1L `(0.2"equiv"L^(-1))=("Mass of oxalic acid")/(63"gequiv"^(-1)xx(0.2L))` `"Mass of oxalic acid"=(0.2equiv.L^(-1))xx(63 g equiv.^(-1))xx(0.1L)=1.26 g` Calculation of no. of molecules of oxalic acid. Molar mass of xalic acid =`126 g mol^(-1)` Now, 126 gof exalic acid has molecules=`(N_(0)=6.022xx10^(23))` 1.26 g of oxalic acid has molecules=`(6.022xx10^(23))/((126g))xx(1.26g)=6.022xx10^(21)` |
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