Amount of substance #2

Question

Amount of substance #2

0

1063

2

+870

Find the amount of substance in 2,5 L of Potassium permanganate (KMnO₄).

HINT: ρ_{Potassium 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

Good luck !

EinsteinJr May 10, 2015

Best Answer

2⁺⁰ Answers

#1

+19

+5

Best Answer

$\\m=\rho \times V=2.703 \times 2.5=6.7575 \textup{ kg}=6.7575\times 10^3 \textup{g} \\\\M_{KMnO_4}=M_K+M_{Mn}+4\times M_O \\=39.098+54.936+(4\times 15.999) \\=158.03\textup{ g}\cdot \textup{mol}^{-1} \\\\n=\frac{m}{M}=\frac{6.7575\times 10^3}{158.03}=\textcolor[rgb]{0,0,1}{42.7609\textup{ mol}}$

MrGenius May 10, 2015

0 Online Users

MrGenius · Answer 1 · 2015-05-10T02:35:00

$\\m=\rho \times V=2.703 \times 2.5=6.7575 \textup{ kg}=6.7575\times 10^3 \textup{g} \\\\M_{KMnO_4}=M_K+M_{Mn}+4\times M_O \\=39.098+54.936+(4\times 15.999) \\=158.03\textup{ g}\cdot \textup{mol}^{-1} \\\\n=\frac{m}{M}=\frac{6.7575\times 10^3}{158.03}=\textcolor[rgb]{0,0,1}{42.7609\textup{ mol}}$

.

EinsteinJr · Answer 2 · 2015-05-10T02:37:28

Fine; MrGenius you've got

$\textcolor[rgb]{1,0,0}{20/20}$

CONGRATULATIONS !

Here's your cookie:

Amount of substance #2

Good luck !

0 users composing answers..

Best Answer

2⁺⁰ Answers

CONGRATULATIONS !

0 Online Users

Top Users

Sticky Topics

Amount of substance #2

Good luck !

0 users composing answers..

Best Answer

2+0 Answers

CONGRATULATIONS !

0 Online Users

Top Users

Sticky Topics

2⁺⁰ Answers