1.250 cm3 of aqueous sodium sulphate solution contains 7.10 g of salt. Find the number of Na+ ions per dm3 of the solution.
2.A certain mass of glucose (C6H12O6) is added to 360 g of water so that the resulting solution contains molecules of water and glucose in the ratio 50:1.
(a)What is the mass of glucose added?
(b)How many molecules (for all kinds) does the solution contain?
3.The purple colour of KMnO4 solution is due to the MnO4- ions. When 5 cm3 of 0.0004 M KMnO4 solution are diluted to 1 dm3, the solution still appears pink to normal eyesight. Calculate how many MnO4- ions are there in 1 drop of the diluted solution (assuming there are approximately 20 drops in each cm3).
F4 Chem Molarity Help plz!!!
2013-02-12 12:58 am
回答 (1)
2013-02-12 4:07 am
✔ 最佳答案
1.No. of moles of Na2SO4 = 7.10/(23 + 32.1 + 16x4) = 0.0596mol
No. of moles of Na2SO4 in each dm³ soln = 0.0596 x (1/0.25) =0.238 mol
No. of Na⁺ ions per dm³ soln = 0.2384 x 2 x (6.02x 10²³) = 2.87 x 10²³
2.
(a)
No. of moles of H2O molecules = 360/(1x2 + 16) = 20 mol
No. of moles of C6H12O6 molecules = 20 x(1/50) = 0.4 mol
Mass of C6H12O6 = 0.4 x (12x6 + 1x12 + 16x6)= 72 g
(b)
Total no. of moles of molecules = 20 + 0.4 = 20.4 mol
Total no. of molecules = 20.4 x (6.02 x 10²³) = 1.23 x 10²⁵
3.
No. of moles of KMnO4 in 1 dm³ soln = 0.0004 x (5/1000) =2 x 10⁻⁶ mol
No. of moles of MnO4⁻ in 1 cm³ = (2 x 10⁻⁶) x (1/1000) = 2 x 10⁻⁹ mol
No. of moles of MnO4⁻ in 1 drop = (2 x 10⁻⁹) x (1/20) = 1 x 10⁻¹⁰ mol
No. of molecules of MnO4⁻ in 1 drop = (1 x 10⁻¹⁰) x (6.02 x 10²³) = 6.02 x 10¹³
2013-02-16 01:01:20 補充：
The answer of Q.1 should be amended as follows :
No. of moles of Na2SO4 = 7.10/(23x2 + 32.1 + 16x4) = 0.0500 mol
No. of moles of Na2SO4 in each dm³ soln = 0.0500 x (1/0.25) =0.200 mol
No. of Na⁺ ions per dm³ soln = 0.200 x 2 x (6.02x 10²³) = 2.41 x 10²³ ...... (answer)
2013-02-16 22:52:39 補充：
The answer of Q.1 should be amended as follows :
No. of moles of Na2SO4 = 7.10/(23x2 + 32.1 + 16x4) = 0.0500 mol
No. of moles of Na2SO4 in each dm³ soln = 0.0500 x (1/0.25) =0.200 mol
No. of Na⁺ ions per dm³ soln = 0.200 x 2 x (6.02x 10²³) = 2.41 x 10²³ ...... (answer)
參考: fooks, fooks
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