+870 Find the amount of substance in 2,5 L of Potassium permanganate (KMnO4).
HINT: ρPotassium Permanganate=2.703 kg⋅L-1
$${\mathtt{n}} = {\frac{{\mathtt{m}}}{{\mathtt{M}}}}$$
You can use the Periodic table below to find the molar masses of the elements :
If you have no idea of where to start, ask for an hint; it will only cost you 13 points 
Just kidding
+19 $$\\m=\rho \times V=2.703 \times 2.5=6.7575 \textup{ kg}=6.7575\times 10^3 \textup{g}
\\\\M_{KMnO_4}=M_K+M_{Mn}+4\times M_O
\\=39.098+54.936+(4\times 15.999)
\\=158.03\textup{ g}\cdot \textup{mol}^{-1}
\\\\n=\frac{m}{M}=\frac{6.7575\times 10^3}{158.03}=\textcolor[rgb]{0,0,1}{42.7609\textup{ mol}}$$
+19 $$\\m=\rho \times V=2.703 \times 2.5=6.7575 \textup{ kg}=6.7575\times 10^3 \textup{g}
\\\\M_{KMnO_4}=M_K+M_{Mn}+4\times M_O
\\=39.098+54.936+(4\times 15.999)
\\=158.03\textup{ g}\cdot \textup{mol}^{-1}
\\\\n=\frac{m}{M}=\frac{6.7575\times 10^3}{158.03}=\textcolor[rgb]{0,0,1}{42.7609\textup{ mol}}$$
+870 Fine; MrGenius you've got
$$\textcolor[rgb]{1,0,0}{20/20}$$
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