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BototoyTheGreat24
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Answer & Explanation
Solution= solvent+ solute
Solute: The substance which is present in lesser amounts in the solution
Solvent: The substance which is present in higher amounts in the solution.
Step-by-step explanation
Image transcription text
Molarity = number of moles of the solute Volume of the solution in Litres Solute: The compound which is present in lesser amount the solution. Solvent: The compound which is present in higher amount in the solution. Solution = solute + solvent Molarity = number of moles of Ally Volume of the solution in Litres. 1 1 = 10 md 1ml = 10 L humber of moles of = Molarity x volume of the solution in All3 Litres 0. 85 M X 1250 X 10 L = 1. 0625 moles
Image transcription text
Molar mass of Alchy = 133.34 9 mol mass of ALug = humber of moles of X molar mass of All Alus 2 (1 - 06 25 mol ) X 133 . 34 9 = 141.673759 mass of Adds required = 141.673759 K ,soy - 2 k+ + soy Oxidation state of soy is ( - 2 ) n factor is '2' Molarity = Normality n - factor Molarity = 9.25 M Molarity = 4.625 M
Image transcription text
Number of moles of = Molarity x volume of the solution in K, soy Litres 4-625 M X 0.5L 2.3125 moles mass of K, soy = moles of K, soy x Molar mass of K, soy 2.3125 mod x 174. 259 9 mol = 402 . 9739 3759 ( " ) % = mass of the solute X 100 mass of the solution Here solute is ca(OH),. mass of (a (OH ), 13-2 = X loo 3509 mass of ca (OH), = 13 . 2X 350 loo mass of ca (OH), = 46.20 9.
Image transcription text
(2) mass of the solute ing' X 100 volume of the solution in me ( * ) = mass of CH2OH X loo volume of the Solution in me 22 X 100 100 ( w 1 % = 22 %%. Volume of the solute in and X loo Volume of the solution in md Here the solute is CH, OH Volume of CH3OH X loo Volume of the solution in and ( ) : = lomd X loo = 25%% yo ml