(a) How many moles of H,SO4 are there in 250 mL of a 0.800 mol/L sulfuric acid solution? (b) What mass of acid is in this solution?
(a) How many moles of H,SO4 are there in 250 mL of a 0.800 mol/L sulfuric acid solution? (b) What mass of acid is in this solution?
Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Chapter1: Introduction
Section: Chapter Questions
Problem 1.1P
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Transcribed Image Text:الاسم حيدرطالب يس
القسم والمرحلة هندسة تقساتالوقر والطاقة الرول
1) CarbondisulPhide, CS,, is aliquid havinPadensity of 1.26 J/mL-
What Volume is ocufied by 5. 50 X 10 molecyles of CS2 2
22
Solutions
S-1.26 %mt
Nm-5.50xl" molccules
22
We Know,
Molar mass of CS2 (Mu) = 76.139 Vmot
Avagadro's number (NA) = 6.022*1 molecwes
mol
Volume occufied (V)
V- Nm * Mw
NA * S
V 5.50 *lomolecles *76.139%ol
6.022*103m0)lecules 126 ,
22
V= 5.51 mL = Volume occufied
![Question 1.1
What is the molarity when 0.75 mol CaSO, (calcium sulphate) is dissolved in 2.50 L of solution?
molarity = moles of solute /volume of solution
molarity = [CaSOa] = c = n/V 0.75 mol / 2.50 L = 0.30 mol/L = 0.30 M
Question 1.2
Sea water contains roughly 28.0 g of NaCl per liter. what is the molarity of sodium chloride in sea water?
molarity = moles of solute / volume of solution
molar mass of NaCl = 23.0 + 35.5 = 58.5 g/mol
moles of NaCl = mass/molar mass = 28.0 / 58.5 = 0.479 mol
molarity = [NaCI] = c = n/V = 0.479 mol /1L = 0.479 mol/L = 0.479 M
Question 1.3
5.95g of potassium bromide (KBr) was dissolved in water, and make a 400 ml solution.
(a) Calculate its molarity. (b) What is the concentration in grams per L?
(a) moles = mass / molar mass, (KBr = 39.1 + 79.9 119 g/mol)
mol KBr = 5.95/119 0.0500 mol
400 ml = 400/1000 = 0.400 L
molarity = moles of solute / volume of solution
molarity of KBr solution = 0.0500/0.400 = 0.125 mol/L
(b) concentration = mass / volume, volume = 400 / 1000 = 0.400 L
concentration = 5.95 /0.400 = 14.9 g/L
Question 1.4
(a) How many moles of H,SO, are there in 250 mL of a 0.800 mol/L sulfuric acid solution?
(b) What mass of acid is in this solution?
(a) molarity = moles of solute / volume of solution, rearranging equation for the sulfuric acid
mol H2SO4 = molarity H2SO4 x volume of H2SO, inL
mol H2SO4 = 0.800 x 250/1000 = 0.200 mol](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4ab122e4-af2c-448b-b018-d82263f3f93f%2Febf94de2-64dd-4010-a441-217a74f3d372%2Fkf0tc6o_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Question 1.1
What is the molarity when 0.75 mol CaSO, (calcium sulphate) is dissolved in 2.50 L of solution?
molarity = moles of solute /volume of solution
molarity = [CaSOa] = c = n/V 0.75 mol / 2.50 L = 0.30 mol/L = 0.30 M
Question 1.2
Sea water contains roughly 28.0 g of NaCl per liter. what is the molarity of sodium chloride in sea water?
molarity = moles of solute / volume of solution
molar mass of NaCl = 23.0 + 35.5 = 58.5 g/mol
moles of NaCl = mass/molar mass = 28.0 / 58.5 = 0.479 mol
molarity = [NaCI] = c = n/V = 0.479 mol /1L = 0.479 mol/L = 0.479 M
Question 1.3
5.95g of potassium bromide (KBr) was dissolved in water, and make a 400 ml solution.
(a) Calculate its molarity. (b) What is the concentration in grams per L?
(a) moles = mass / molar mass, (KBr = 39.1 + 79.9 119 g/mol)
mol KBr = 5.95/119 0.0500 mol
400 ml = 400/1000 = 0.400 L
molarity = moles of solute / volume of solution
molarity of KBr solution = 0.0500/0.400 = 0.125 mol/L
(b) concentration = mass / volume, volume = 400 / 1000 = 0.400 L
concentration = 5.95 /0.400 = 14.9 g/L
Question 1.4
(a) How many moles of H,SO, are there in 250 mL of a 0.800 mol/L sulfuric acid solution?
(b) What mass of acid is in this solution?
(a) molarity = moles of solute / volume of solution, rearranging equation for the sulfuric acid
mol H2SO4 = molarity H2SO4 x volume of H2SO, inL
mol H2SO4 = 0.800 x 250/1000 = 0.200 mol
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