Question 21 Molar mass of NaCl = 23 + 35.5 = 58.5 gm/mol number of moles of NaCl

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Question 21

Molar mass of NaCl = 23 + 35.5 = 58.5 gm/mol

number of moles of NaCl = mass/molar mass = 35/58.5 = 0.5982 moles

molality = number of moles of solute/mass of solvent in Kg = 0.5982/0.5 = 1.196m

Change in Temperature = i * Kb * m = 2 * Kb * (1.196) =2.3931 Kb

Molar mass of C12H22O11 = 342 gm/mol

number of moles of C12H22O11 = mass/molar mass = 15/342 = 0.04385 moles

molality = number of moles of solute/mass of solvent in Kg = 0.04385/0.025 = 1.754m

Change in Temperature = i * Kb * m = 1 * Kb * (1.196) =1.754 Kb

Hence the NaCl solution will have higher boiling point than sucrose

Question 25

6HCl + 2Al ------ 2AlCl3 + 3H2

number of moles of HCl = Volume of solution (in L) * Molarity(M)

=> 35/1000 * 2

=> 0.07 moles

6 moles of HCl will react with 2 moles of Al

Number of moles of Al required = 0.07/3 = 0.02333333 moles

Molar mass of Al = 27 gm/mol

Mass of Al required =0.0233333 mol * 27 gm/mol = 0.63 grams

Question 50

Fe is getting oxidized from 0 oxidation state to +3 oxidation state (Fe is getting oxidized)

O is getting reduced from 0 oxidation state to -2 oxidation state (O is getting reduced)

Explanation / Answer

Please help with these problems D Question 21 2 pts Show all work on this problem. Which solution will have a higher boiling point: A solution containing 15 grams of sucrose (C12H22011) in 25 grams of water or a solution containing 35 grams of sodium chloride in 500 grams of water? HTML Editort

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