Moles and Molarity For acid-base chemistry purposes, it is more appropriate to measure solute concentration in moles, or individual particles (e.g., atoms, molecules), per unit volume rather than mass per unit volume. This is because atoms react with each other in known proportions in a way...
To find the mass of a compound, first the number of moles of the compound must be determined using the molarity and volume. Molarity is calculated by...Become a member and unlock all Study Answers Start today. Try it now Create an account Ask a question Our ...
What is the molarity of a solution containing 0.325 moles of solute in 250 mL of solution? A solution was made by dissolving 12.5 g of Na3PO4 in 125.00 mL of water. The volume of the resulting solution was 128 mL. Calculate the ...
-so, now I find the #moles using Br's mm (79.904 g/mol), and get 1.95 mol So there is my answer, 1.95 mol/L is the molarity in the 5% solution. Does this seem correct? It really comes down the whole 5% Bromine by volume assumption that I made...I think. Any help appreciated...
I found that the molality of the solution is 0.085 m and there is 0.0156 kg of solvent. Calculate the moles of solute present in the solution. You make a solution using a solute (5g) in water (total volume 100ml). The molar mass of the solute is 110 ...
NaOH is a base solution. There are many ways to express the concentration of NaOH solution. Some of NaOH solution concentration expressions are percentage mass (%m/m), molarity, molality, percentage volume (%v/v), and many more.Answer and Explanation: Fir...
Complete following table ("dent"is diethylenetriamine, and "bipy" is 2.2-bipyridyl) Complete the following table: | |Mass|Mol|Volume| |Solute|Solute|Solute|Solution|Molarity |KNO3|23.5 g|___|135.0 mL|___ |NaHCO3|___|___|200.0 mL|0.140 M |C12H22O11|56.38 g|___|___|_0.1 Complete ...
What is the pH of a 1.0 L buffer solution that consists of 0.45 M CH3COOH and 0.35 M CH3COONa after the addition of 0.100 moles of NaOH? Assume no volume change resulting from the addition of the strong base. (Ka...