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What volume of each of these solutions would contain 58.44 g of NaCl?a. 0.10 M NaClb. 3.0 M NaClc. 5.5 M NaCl

a. 10 $\mathrm{L}$b. 0.33 $\mathrm{L}$c. 0.18 $\mathrm{L}$

Chemistry 102

Chemistry 101

Chapter 15

Toxins in Solution

Section 3

Preparing Solutions

Solutions

Composition

Chemical reactions and Stoichiometry

University of Central Florida

Rice University

University of Maryland - University College

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in problem five. What volume? Off each off. The solution contained 58 point 44 g NSL. So this question is taken from the topic Preparing solution. So party is 0.10 Mueller and a C l. So as we know that the majority is basically the number off moles off salute divided by the volume off solution in later. And we know that the number off mole is equal to moss, our molecular mass. So the molecular mass off S E l is approximate 58.5 g Permal So And we find out the volume in this case or morality is, uh, equal toe number off moles. That means, uh, mass, our molecular mass in tow volume in later. So basically this formula we can use to calculate the volume. So McGarity is given in the question that is 0.10 and mass, uh, is also given. That is 58 point four and molecular masses 58.5, uh, into volume that we calculate. So on solving this we get volume and later physical toe 10 later. So, in case off the three Mueller and HCL solution is there that main polarity is three Moeller and on the same method, we can calculate its volume. So McGarity is three and this is 58.4. That is the mask our molecular masses 58.5 into volume in later. So on solving this again we get volume is 0.33 later and coming toe the party that is the 5.5 Moeller n a c l solution. So again, same method we can approach. So 5.5 physical to 58.4 upon 58.5 and two volume in later. So on solving we get volume is 0.18 later. So, uh, this these are over volume, so volume off solution one is basically 10 liter and volume off solution toe is 0.33 liters and volume of solution three is 0.18 leaders. So these are the volume. So based on data, that is 58.44 g off NSL. That when the masses given in the question and we used this formula to calculate the volume

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