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What mass of solid NaOH $(97.0 \% \mathrm{NaOH} \text { by mass) is required to prepare } 1.00 \mathrm{L} \text { of a } 10.0 \% \text { solution of } \mathrm{NaOH}$ by mass? The density of the 10.0$\%$ solution is 1.109 $\mathrm{g} / \mathrm{mL}$ .
114 $\mathrm{g}$
Chemistry 102
Chemistry 101
Chapter 3
Composition of Substances and Solutions
Solutions
Composition
Carleton College
University of Central Florida
Rice University
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this question we asked to determine the mass off so suited my drug side and the souls of my drugs that has 97% sodium hydroxide by mass that is needed to prepare a 1.0 leader off a 10% solution off sodium hydroxide by mass. We've been told that the density which have represented by the Greek variable rule of this 10% solution is 1.109 grams and new leaders, In order to determine the mass of the solution, you have to first find the volume off the solution. So the volume that we're gonna be the praying so it will prepare at one point their leader off a 10% solution. So our mass off the solution going to be given by density, which is 1.109 grams Premier leader, right times one leader. So one leader is 1000 milliliters it and for 10% bowling by volume solution. Essentially, we can represent that by 10 milliliters, right? Bye. 100 new leaders. So that's 10% walling by bony, right? And keep into account that Okay, Yeah. So we can take into account the 97% after we've calculated this mass, because this mass is going to give us the mass off. I were, um, 100% solution. I'm sorry. Yeah. So if you saw for this, you're going to get a valley off 100 10.9 grams. Hey, And in order to determine the mass off the sodium hydroxide was just 97% mass. Well, we have to do ISS. So a mass off suited my dark side for the solution for which we want to have the 97% certain hydroxide by mass is going to be quoted the mass we calculated 110.9 grams divided by are 97% just 97 minute by 100. This is going to give us a value most 114. 23 grams.
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