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Answer 1 · 2016-03-14T19:48:29

$0.20 M$ $0.20M$

The molarity ( $C_{M}$ $C_M$ ) of a solute (oxalic acid) in an aqueous solution is found by:

$C_{M} = \frac{n}{V}$ $C_M=n/V$

where, $n$ $n$ is the number of mole of oxalic acid, $V = 0.250 L$ $V=0.250L$ is the volume of the solution.

The number of mole $n$ $n$ is found by: $n = \frac{m}{M M}$ $n=m/(MM)$

where, $m = 6.3 g$ $m=6.3g$ is the mass of oxalic acid, and $M M = 126 \frac{g}{m o l}$ $MM=126g/(mol)$ is the molar mass of the oxalic acid.

$\Rightarrow n = \frac{m}{M M} = \frac{6.3 g}{126 \frac{g}{m o l}} = 0.050 m o l$ $=>n=m/(MM)=(6.3cancel(g))/(126cancel(g)/(mol))=0.050mol$

Thus, $C_M=n/V=(0.050mol)/(0.250L)=0.20M$

Here is a video on how to prepare a solution.
Lab Demonstration | Solution Preparation & Dilution.