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# The density of 3M solution of $Na_2S_2O_3$ is 1.25 g $mL^{-1}$ Calculate the $\%$ by weight of $Na_2S_2O_3$.

$(a)\;37\qquad(b)\;39.9\qquad(c)\;37.92\qquad(d)\;40$

Given, $Na_2S_2O_3$ has molarity = $3mol\;litre^{-1}$
Moles of $Na_2S_2O_3$ = 3
$\therefore$ Weight of $Na_2S_2O_3 = 3\times158 = 474g$
and, Volume of the solution = 1 litre = 1000mL
$\therefore$ Weight of the solution = $1000\times1.25$ = 1250 g
$\therefore$ weight of water = 1250 - 474 = 776 g
$\%$ by weight of $Na_2S_2O_3$ =$\large\frac{Wt.\;of\; Na_2S_2O_3}{Wt.\; of \;solution}\times100$ $= \large\frac{474}{1250}\times100 = 37.92$
edited Mar 18, 2014