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### AuthorTopic: determine the 20% mol retative to a chemical  (Read 666 times) !function(d,s,id){var js,fjs=d.getElementsByTagName(s)[0];if(!d.getElementById(id)){js=d.createElement(s);js.id=id;js.src="https://platform.twitter.com/widgets.js";fjs.parentNode.insertBefore(js,fjs);}}(document,"script","twitter-wjs"); (function() {var po = document.createElement("script"); po.type = "text/javascript"; po.async = true;po.src = "https://apis.google.com/js/plusone.js";var s = document.getElementsByTagName("script")[0]; s.parentNode.insertBefore(po, s);})();

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#### rockoff

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##### determine the 20% mol retative to a chemical
« on: May 16, 2018, 02:26:07 AM »

Hello,

I just wanted to check with you if what I calculated is right.
I am asked to add 10% sodium hydroxide molar relative to acetaldehyde with a 5 mol / L sodium hydroxide solution.
I know that we put 7 L of acetaldehyde (100%).

1 - Calculating how many mol of acetaldehyde has been put:
data on acetaldehyde:
volume (V): 7 L
density (d): 780 g / L
Molar mass (M): 44.05 g / mol

calculation of the mass in 7 L
d = m * L
m = 780 * 7 g
m = 5460 g

calculation of the number of moles in 7 L
n = m / M
n = 5460 / 44.05
n = 123.95 mol

2 - Calculating the volume of sodium hydroxide that must be added
20% of 123.93 mol => 24.7 mol.
The concentration of sodium hydroxide solution being 5 mol / L, it is necessary to add 24.7 / 5 => 4.95 L of this sodium hydroxide solution to have 20 mol% relative to acetaldehyde.

Are my calculations correct?

Thank you !
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#### mjc123

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##### Re: determine the 20% mol retative to a chemical
« Reply #1 on: May 16, 2018, 04:39:59 AM »

According to OP, acetaldehyde is 100%; NaOH is 5 mol/L.
But OP initially says he is asked to add 10 mol% NaOH, then calculates 20%. Which is it?
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#### rockoff

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##### Re: determine the 20% mol retative to a chemical
« Reply #2 on: May 16, 2018, 08:28:12 AM »

My bad, the correct % is 20%, thanks for the remark
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