Question #9080e

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Answer 1 · 2017-10-23T18:54:06

#14.84g# of #Na_2CO_3# is required.

Firstly, the gram molecular mass #(M_m)# of sodium carbonate #(Na_2CO_3)# is:

#M_m=2(23)+12+3(16)g#

#:.M_m=(46+12+48)g#

#:.color(red)(M_m=106g)#.

Now, the dissociation reaction of #Na_2CO_3# is #rarr#

#Na_2CO_3rightleftharpoons2Na^++CO_3^(2-)#

Hence from the above equation, to get #0.8M# of #Na^+# in the solution, the #Na_2CO_3# solution should be #0.4M#.

#:.color(red)(M=0.4M)#.

Let, the mass of #Na_2CO_3# required be #x# grams.

Now, molarity #(M)# is the number of moles of solute per #1000ml# of solution.

Now given that the total volume #(V)# of the solution is #350ml#.

#:.color(red)(V=350ml)#.

Now,

#:.color(red)(M=x/(M_m)xx1000/V)#.

Now substituting the values in the formula we get:

#0.4=x/106xx1000/350#

#:.x=(0.4xx106xx350/1000)# grams.

#:.color(red)(x=14.84g)#.

Therefore, #color(red)(14.84)# of anhydrous #Na_2CO_3# is required.

Hope it Helps:)

Answer 2 · 2017-10-23T18:59:01

#(0.80mol " " Na^(+))/(1L) * (0.350L)/1 * (Na_2CO_3)/(2Na^(+)) * (106g)/(Na_2CO_3) approx 14.8g#

We want #0.8mol# of sodium ions per #L# of solution.

We want to produce #350mL# of solution.

Two sodium ions will dissociate upon hydration per mole of sodium carbonate.

Thus, we only need "half" the mass of sodium carbonate.