Question #32a07

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Question #32a07

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Answer 1 · 2016-03-08T23:33:11

$C = 0.2$ $C=0.2$ mol/ $d m^{3}$ $dm^3$

Explanation:

the molar concentration C is defined as : $C = \frac{n}{V}$ $C=n/V$
n : amount of solute and V : Volume

we know that $n = \frac{m}{M}$ $n=m/M$
m : masse of the solute and **M : Molar mass **

the concentration will be $C = \frac{m}{M \cdot V}$ $C=m/(M*V)$
in your exemple $V = 500 c m^{3} = 0.5 d m^{3}$ $V=500 cm^3 = 0.5dm^3$

M for Sodium Hydroxide NaOH is
$M = M (N a) + M (O) + M (H)$ $M=M(Na)+M(O)+M(H)$
$= 22.98976 + 15.9994 + 1.00794 = 39.9971$ $=22.98976+15.9994+1.00794=39.9971$ g/mol

$m = 4 g$ $m=4g$
so $C = \frac{4}{39.9971 \cdot 0.5} = 0.2$ $C=4/(39.9971*0.5)=0.2$ mol/ $d m^{3}$ $dm^3$

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Question #32a07

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