Question

Question 1 (0.25 points) What is the molar concentration of sucrose (C12H22011) in a solution made by dissolving 33.95g of sucrose in enough water to make 500.0mL of solution? Use the Periodic Table in Appendix A of your textbook and round to an appropriate number of sig figs. Your Answer: Answer units Question 2 (0.25 points) What is the molar concentration (M) of potassium acetate in a solution made by dissolving 10.83g of potassium acetate in enough water to make 200.0mL of solution? Use the Periodic Table in Appendix A of your textbook and round to an appropriate number of sig figs. Your Answer: Answer units Question 3 (0.25 points) What is the molar concentration (M) of alanine (C3H302N) in a solution made by dissolving 16.19g of alanine in enough water to make 295.0mL of solution?

Answer 1

Molar concentration is also called molasity. solute dissalve per Molarity (M) It is the number of moles of lite of solution. ie; M= mules of solute volume in litres Unit of Molarity is mol I Question-1 : To find moles of sucorse C C12H2.2011): : weight moles = 2 moler mass Given weight = 33.95g Molar mass of C2H220, 12 x 12 + 12x! + || X16 = 342 g/mol с H Put in 33.95 moles - = 0.099 mol 3428/mol Greven volume = 500 ml = 500 T000 L =0.5L v Now substitute the values of moles & volume in ear ② M= 0.099 mol 0.5 L 0 198 moli can be wome [: M=mol moj 2] off to 0.2 mola Ans: 0.2 mol L

Question 2 : To find moles of Potassium acetate ( CHşcook): Given weight = 10.83 g Molar mass of CH₂ cook 988/mol Put in moles 10.83g 989/mol 0.11 moe ✓ Volume = 2ooml 200 L = 0.2L Ooo Now substitute the valen of moles & volume in ea ① M= 0.11 mol 0.2 L 0.55 molil = Ans 0.55 mol i * 0.55 can be sound off to 0.6 moti

Question- To find moles of alanine (C347 BN): Given Weight = 16:19 g molar mass of C3HY ON = 89 almal Put in 2 moles 16.199 = = 0.182 mol u 89 g/mol Given volume = 295 ml = v 295 1000 L = 0.295 L Now substitute these values in eq 0 0.182 mol M= = 0.62 mola 0.295 Ans 0.62 molt
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