Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and beginning April 20th, 2021 (Eastern Time) the Yahoo Answers website will be in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.
1.Normal saline solution contains 0.90 g of sodium chloride per 100. mL of solution. What is the molar concentration (M) of sodium chloride in normal saline?
2.How would you prepare 250. mL of 0.300 M LiNO3 solution, starting from the solid solute? Do a calculation AND write one sentence describing how you would prepare the solution.
- Roger the MoleLv 72 months agoFavorite Answer
(0.90 g NaCl) / (58.4430 g NaCl/mol) / (0.100 L) = 0.154 mol/L = 0.154 M NaCl
(0.250 L) x (0.300 mol LiNO3/L) x (68.9459 g LiNO3/mol) = 5.17 g LiNO3
Supposing the solution to be made is aqueous:
Weigh out 5.17 grams of lithium nitrate, put it in a 250-mL volumetric flask, add enough water to dissolve the salt with swirling, add more water until the solution reaches the flask's mark, and mix thoroughly.
- jacob sLv 72 months ago
.90 g x. 1mol/ 58.44 g = .015 mol NaCl
.015 mol /.1L = .15M
250 ml = 0.25 L of 0.300 M LiNO3
0.25 L x (.300 mol/L) = .075 moles LiNO3
Amount of solid LiNO3 required in grams
0.075 moles LiNO3 x 68.946 g/mol LiNO3 = 5.17 g LiNO3
Please Rate my work!!