# F4 Chem Molarity Help plz!!!

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).

Update:

1. molar mass of Na2SO4 不是 (23x2 + 32.1 + 16x4) 咩?

Rating

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)

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