Question #971c4

1 Answer
Ernest Z.
Jul 15, 2017

You need 9.9 mg of ethidium bromide.

Explanation:

Step 1. Calculate the moles of EtBr

#"Moles" = 0.100 color(red)(cancel(color(black)("L"))) × (2.5 × 10^"-4"color(white)(l)"mol EtBr")/(1 color(red)(cancel(color(black)("L")))) = 2.50 × 10^"-5" color(white)(l) "mol EtBr"#

Step 2. Calculate the mass of EtBr

#" Mass of EtBr" = 2.50 × 10^"-5" color(red)(cancel(color(black)("mol EtBr"))) × "394.3 g EtBr"/(1 color(red)(cancel(color(black)("mol EtBr")))) = "0.0099 g EtBr" = "9.9 mg EtBr"#

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