Definitions of solution, solute, and solvent. How molarity is used to quantify the concentration of solute, and how to calculate molarity.
Log in johannmlmstn 9 years agoPosted 9 years ago. Direct link to johannmlmstn's post “Hi there, I was just wo...” Hi there, • (54 votes) Daniel Stoken 8 years agoPosted 8 years ago. Direct link to Daniel Stoken's post “I believe you're correct....” I believe you're correct. There was likely a typographic error in the example. We see in the previous step the conversion was done correctly (50 mL = .050 L) so we have 0.02401 mol / .050 L. A quick check with the calculator shows that this is in fact 0.48 mol/L or 0.48 M. (33 votes) Anson Chan 8 years agoPosted 8 years ago. Direct link to Anson Chan's post “I was told in school that...” I was told in school that molarity should be moles/dm^3, but is this different from moles/litres? • (5 votes) Esther Dickey 8 years agoPosted 8 years ago. Direct link to Esther Dickey's post “A liter is equal to a cub...” A liter is equal to a cubic decimeter, so it is the same. (31 votes) Dawen 8 years agoPosted 8 years ago. Direct link to Dawen's post “So this isn't quite the r...” So this isn't quite the right place for my question, but I can't find the right place for the life of me... If someone could maybe point me to a video/article on converting between concentration units, especially molarity to ppt or ppm, that'd be great. In the mean time, I've been asked to take a known molarity of a solution and convert it into parts per thousand. I tried Google and I /think/ I got the right formula but I'm not positive, so can someone check it for me please? So what I did was start with my given molarity as mol/L. I assumed there wouldn't be enough solute to drastically affect density and so I changed 1 L to 1000g, so I now have mol/1000g. Then I multiply by the molar mass of the solute (NaOH - 39.998) so I'm now g NaOH/1000g solution. Then I multiply the whole thing by 1000 to get ppt, right? Sort of like calculating a percent? • (5 votes) Ernest Zinck 8 years agoPosted 8 years ago. Direct link to Ernest Zinck's post “You did it almost perfect...” You did it almost perfectly. (8 votes) Rachel Silverman 8 years agoPosted 8 years ago. Direct link to Rachel Silverman's post “in hint one how do you kn...” in hint one how do you know there is .1L of solute? • (4 votes) Philomath 8 years agoPosted 8 years ago. Direct link to Philomath's post “There must have been a ty...” There must have been a typo. I think in the description they meant 0.100L instead of 0.100mL. (4 votes) Abigail Baricevich 8 years agoPosted 8 years ago. Direct link to Abigail Baricevich's post “How would you find the mo...” How would you find the molarity of SO2 if you have it dissolved in 100 grams of water at 85 degrees Celcius? • (2 votes) Hazelle R. Dela Cruz 6 years agoPosted 6 years ago. Direct link to Hazelle R. Dela Cruz's post “Assuming that you do not ...” Assuming that you do not know the amount of SO2 that was dissolved to prepare the solution, you may try to invoke Henry's Law and determine the concentration of SO2 in the headspace (just above) of the solution. For a primer on Henry's Law, you can check out this article: You can also check these links below for sample procedures on determining the amount of SO2 vapor (<- what causes acid rain!): Finally, you can check this link, so you can convert your determined SO2 vapor concentration to SO2 molarity in water: Yeah, this is some detective work (and a lot of hard work!). (2 votes) Jeff Sellers 7 years agoPosted 7 years ago. Direct link to Jeff Sellers's post “Question: Is this just c...” Question: Is this just coincidence, or does this make sense... • (2 votes) RogerP 7 years agoPosted 7 years ago. Direct link to RogerP's post “What you suggest is fine ...” What you suggest is fine just as long as the concentrations of the two solutions are the same. But if, say, the Pb(NO3)2 solution was twice the strength of the KI solution then you would only need 0.1 L of each to get the same number of moles. (2 votes) FoxFace 8 years agoPosted 8 years ago. Direct link to FoxFace's post “I understood what molarit...” I understood what molarity is quite well......but what is normality, formality and molarity? If we have molarity why are they even needed then? • (2 votes) tyersome 7 years agoPosted 7 years ago. Direct link to tyersome's post “With any luck, like most ...” With any luck, like most people, you will be able to safely ignore normality and formality. Molality is moles / mass of solvent (SI unit: mol/kg) -- for use see: https://en.wikipedia.org/wiki/Molality#Usage_considerations Normality is explained here: https://en.wikipedia.org/w/index.php?title=Equivalent_concentration&redirect=no Formality is more or less totally ignored and often when we say molarity we actually mean formality see: http://www.chemiasoft.com/chemd/node/25 A good discussion of most of these is here: https://socratic.org/questions/what-is-molarity-molality-and-normality (2 votes) Sevillano, Aida 4 years agoPosted 4 years ago. Direct link to Sevillano, Aida's post “how do you find the volum...” how do you find the volume when given the mass and M value • (1 vote) Astic 3 years agoPosted 3 years ago. Direct link to Astic's post “We know that the formula ...” We know that the formula to calculate the molarity of a substance is M = n/V (n = moles, and V = volume of the solution). (1 vote) Artemis 8 years agoPosted 8 years ago. Direct link to Artemis's post “What is the difference be...” What is the difference between molarity and molality? • (1 vote) Philomath 8 years agoPosted 8 years ago. Direct link to Philomath's post “Molarity is (mol of solut...” Molarity is (mol of solute)/(L of solution). (4 votes) miARNr 6 years agoPosted 6 years ago. Direct link to miARNr's post “Question1 :In a solution ...” Question1 :In a solution with 2 species "A" and "B" ,with "A" having a greater number of moles but the "B" having a bigger molecular mass in such a way that it exceeds the mass of "A", who is the solvent ? • (2 votes) cali24 4 years agoPosted 4 years ago. Direct link to cali24's post “For Question 2, I believe...” For Question 2, I believe that the substance you are using as the base is the solvent. For example, if you have 50 g of water and 50 g of salt, then the solvent would be the water, as you put the salt IN the water, not the water IN the salt. (1 vote)Want to join the conversation?
I was just wondering shouldnt the answer in example 1 be 0.48 mol/Litre
and NOT 4.8 M? I get the same answer to the last step before the answer, but when i do the calculation i get 0.48 mol/litre.
thank you so much.
A concentration of 1 g NaOH/1000 g solution is 1 g per 1000 g or one part per thousand (1 ppt) — no need to multiply by 1000.
In the same way, a concentration of 1 g per 100 g is one part per hundred (1 %).
https://www.thoughtco.com/henrys-law-example-problem-609500
https://www.law.cornell.edu/cfr/text/40/appendix-A-2_to_part_50
https://www.astm.org/Standards/D2914.htm
https://www.ems.psu.edu/~brune/m532/m532_ch5_aqueous_phase.htm
In the equation, we have 1 Pb(NO3)2 + 2 KI...we have twice as many KI as Pb(NO3)2. Since we have 0.1L of 1Pb(NO3)2, can I just multiply the 0.1 L x 2, since we use twice as much KI as we do Pb(NO3)2? (Or if the equation happened to have 4KI, could we simply multiply 0.1L x 4)? Thanks for the help!
Rearranging the formula to make 'V' the subject allows us to figure out that V = n/M.
When given the mass in Analytical Chemistry, we should always seek to covert the mass (given in any units) first into grams (if it is, then do not worry about this). We should then convert these grams into moles, to do so we require the molar mass of the solute, and dividing the given mass (in grams) by the molar mass provides us with the moles of the substance.
Therefore, we have everything we need, we have calculated the moles, and we are already given the Molarity (M). Seek to substitute these values into their respective position within the rearranged equation above- V = n/M, calculating this value will output the volume.
Molality is (mol of solute)/(kg of solvent).
More on the difference here: https://www.khanacademy.org/science/health-and-medicine/lab-values/v/molarity-vs-molality
Question 2 : when 2 species are in the same amount , what determines who is the solvent ?
