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OpenStudy (zaxoanl):
if you have an antifreeze solution that is 2.35M ethylene glyco (C2H6O2 -molar mass-62.1g/mol) in water. How many mL of this solution would you need if you want to have 93.0 g of ethylene glycol?
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OpenStudy (joannablackwelder):
2.35 M means 2.35 mol/L.
OpenStudy (joannablackwelder):
Can you find how many moles of ethylene glycol is in 93g?
OpenStudy (joannablackwelder):
@zaxoanl?
OpenStudy (zaxoanl):
93g / 62.1g i got 1.5 mol ethylene
OpenStudy (zaxoanl):
then i use 2.35M = 1.5mol/L i got .64L .64*1000= 640mL not sure if that is correct
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OpenStudy (joannablackwelder):
I get really close to that. Probably just rounding difference.
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