1.The solubility of NaOH is 129 g NaOH/100 g water at 40°C. If a solution contai

Question

1.The solubility of NaOH is 129 g NaOH/100 g water at 40°C. If a solution contains 70 g NaOH in 50 g of water at 40°C, is the solution saturated or unsaturated?

2. What is the percent concentration (m/m) of a KBr solution prepared by combining 15.0 g of KBr with 150. g of water?

3. A bottle of wine contains 9.0% alcohol (v/v). How much alcohol is contained in 175 mL of wine?

4. What is the molarity of a MgSO4 solution that contains 18.4 g of MgSO4 in 350 mL of solution?

5. A student wants to prepare a 1.5 M NaCl solution. How many liters of solutions can be made if the student has 26.9 g of NaCl?

6. How many grams of Li2S are present in 450 mL of a 0.75 M Li2S solution?

7. A student needs to make 250 mL of 1.4 M Fe(NO3)3 by dilution. How many milliliters of a 5.0 M Fe(NO3)3 solution should the student use to prepare the solution?

8. How many milliliters of 2.0 M HCl is needed to completely react with 0.83 g of Mg?

Mg(s) + 2 HCl(aq) ® MgCl2(aq) + H2(g)

9. How many milliliters of 3.0 M Na3(PO4)2 are needed to react with 38.0 mL of 2.5 M MgCl2?

2 Na3PO4(aq) + 3 MgCl2(aq) ® 6 NaCl(aq) + Mg3(PO4)2(s)

10. How many grams of Mg3(PO4)2can be produced when 82.0 mL of 3.5 M MgCl2reacts with excess sodium phosphate? (Refer to the balanced chemical equation in question #9)

11. A solution is made up of 86.3g ethylene glycol (C2H6O2) in 143.2g of water. Calculate the freezing point of this solution (Tf = m X Kf, For water Kf = 1.86 oC/m)

12. Calculate the boiling point of the ethylene glycol/water solution from question #11

Explanation / Answer

1) S = 129g/100g of water = 129g/100g = 1.29g/g of water for 50g of water we have thant

the solubility is 50g x 1.29g/ml = 64.5g the solution is saturade

2) 15g of KBr + 150g of H2O = 165g of solution %KBr = 15g/165g x 100 = 9.1%

3) 175ml x 9ml of alcohol / 100ml of vine = 15.75ml

4) MW (MgSO4) = 120.34g/mol n (MgSO4) = 18.4g/ 120.34g/mol = 0.15mol and 350ml become to l is 350ml x 1l/1000ml = 0.35l =) M = 0.15mol/0.35l = 0.43M

5) 26.9g of NaCl become to mol MW NaCl = 58.44g/mol n = 26.9g/58.44g/mol = 0.46mol

M = n/V and M = 1.5M =) V = n/M = 0.45mol /1.5mol/l = 0.3l = 300ml

6) MW Li2S = 45.95g/mol n = Mx V where M = 0.75 and V = 450ml or 0.45l n = 0.75mol/l x 0.45l = 0.34mol of Li2S and n = m/MW =) m = nxMW = 0.34mol x 45.95g/mol = 15.62g

7) whit the conservation of the masa C1 V1 = C2 V2 where C1 = 1.4M V1 = 250ml and C2 = 5M V2 =?

V2 = C1xV1/C2 = 1.4Mx250ml /5M = 70ml V2 = 70ml

8) n(Mg) = m/MW wher the MW (Mg) = 24.31g/mol =) n = 0.83g/24.31g/mol = 0.034mol (Mg) the stequimetric relationn says than 1 mol of Mg reacction whit 2mol of HCl therefore i need 2x0.034mol = 0.068mol of HCl

C1xV1 = n HCl where C1 = 2M V1 = nHCl/C1 = 0.068mol/2mol/l = 0.034l l =) 0.034l x 1000ml/l = 34ml

9) nMgCl2 = C1 x V1 =  0.038l x 2.5mol/l = 0.095mol the estequiometric relation says than 2 mol of Na3PO4(aq) reacction whit 3 MgCl2(aq) for 0.095mol of MgCl2 need 2/3 n MgCl2 in mol of Na3PO4 =) 0.095x2/3 = 0.063mol of Na3PO4 as C2 x V2 = nNa3PO4 where C2 = 3M =) 0.063mol = C2V2 =) V2 = 0.063mol/3mol/l = 0.021l = 21ml of solution of Na3PO4

10) nMgCl2 = C1xV1 = 0.082l x3.5 mol/l = 0.287molMgCl2 and we have than

3mol of MgCl2(aq) make 1mol of Mg3(PO4)2(s) the n of Mg3(PO4)2 formed are n MgCl2/3 =  0.287mol/3 = 0.096mol of Mg3(PO4)2 MW = 262.9g/mol =) n = m/MW =) m = nxMW = 0.096molx262.9g/mol = 24.24g of Mg3(PO4)2

11) the MW of ethylene glycol = 62.07g/mol the molalidad of solution is  

1000x86.3g ethylene glycol / (143.2gx62.07g/mol) = 9.71mol/Kg as Tf = m X Kf =)

Tf = 9.71m X 1.86 oC/m = 18.1C as the fusion point for water is 0C for the solution is -18.1C

12) for the boiling point the Kb = 0,52 ºC Kg/mol   Tb = m X Kb =)
Tb = 9.71mol/Kg X 0,52 ºC Kg/mol = 5.05C as the boiling point for water is 100C and for the solution is 105.05C

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