Mass fraction of the component

 

1.Glauber salt Na₂SO₄∙10H₂O is applied as laxative substance. How many grams of Na₂SO₄∙10H₂O are necessary to prepare 250 g of solution with mass fraction of Na₂SO₄ equal to 5%.

Data                                          Solution

m(sol-n) =250 g              M(Na₂SO₄) = 142 g/mol

ω(Na₂SO₄)= 5% = 0,05   M(Na₂SO₄∙10H₂O) = 322 g/mol

                                                                     m(s-n)∙5% Na₂SO₄

m(Na₂SO₄∙10H₂O) -?       1) m(Na₂SO₄) = ————————— =

                                                                                    100%

                                             =250 ∙ 0,05 = 12,5 g.

                                        

2) According to stoichiometric scheme:

                                             Na₂SO₄ → Na₂SO₄∙10H₂O

                                             υ(Na₂SO₄) = υ(Na₂SO₄∙10H₂O)

                                         m(Na₂SO₄)    m(Na₂SO₄∙10H₂O)

                                        ————— = ————————

                                         M(Na₂SO₄)     M(Na₂SO₄∙10H₂O)

                                     M(Na₂SO₄∙10H₂O) = 12,5∙ 322/142 = 28,4 g

The answer: 28,4 g

 

2. Zinc sulphate ZnSO₄ is applied as solution with mass fraction 0,25% used as eyes drops. How many grams of water mast we add to 25 g of solution with mass fraction ZnSO₄ equal to 2% to prepare eye drops?

Date:                                                Solution

m(s-n) =25 g                   According to the “rule of cross” let make the

ω₁(ZnSO₄)=2%     2%                  0,25%  

ω₂(ZnSO₄)=0,25%                      

                                  0,25%                  

m(H₂O) =?                                  

                       0%                1,75

                                    m(2% s-n) 0,25   1     25

                                    ———— = —— = — = ———  =>

                                     m(H₂O)   1,75  7 m(H₂O)

                                           

m(H₂O)=7∙25=175g                                                                       

                                                                        The answer: 175 g.

 

3. To balance the content of hydrochloric acid in gastric juice it’s solution is applied. How many milliliters of 24% HCl solution with density of 1,12 g/ml are necessary to prepare 500g of solution with mass fraction 5%.

Date:                                                       Solution:

ρ(s-n1) = 1,12 g/ml                        ω₂% · m(s-n 2)

ω₁(HCl) = 24 %            m(HCl) = —————— = 500 ∙ 0,05= 25g

m(s-n 2) = 500 g                                    100%

ω₂(HCl) = 5 %                             m(HCl) ·100%  m(HCl) ·100%   

                                   ω₁(HCl) = —————— = ——————  

V(s-n 1) -?                                    m(s-n 1)         V(s-n 1)∙ρ(s-n 1)

                                        

                                                       m(HCl)       25

                                   V(s-n 1) = ———— = ———— = 93 (ml)

                                                       ω₁· ρ(s-n1) 0,24·1,12

                                           The answer: 93 ml

 

4. Solve for mass fraction (in %) of HCl in solution prepared by mixing of 50 ml of 20%-solution and 20 ml of 10% solution.

Data:                                        Solution:

V(s-n 1) = 50 ml   1) m₁(HCl) = V(s-n 1) ∙ρ(s-n 1) ∙ ω₁(HCl)= 50∙1,1∙0,2 = 11g

ρ(s-n 1) = 1,1 g/ml  2) m₂(HCl) =V(s-n 2) ∙ρ(s-n 2) ∙ ω₂(HCl)= 20∙1,05∙0,1=2,1 g.

 ω₁(HCl) = 20%     

ρ(s-n 2) =1,05g/ml                  

ω₂(HCl)=10%                 3) m₁(HCl)+m₂(HCl) 11+2,1

                              ω₂(HCl)= —————— = —————∙100% =17,2%

ω₃(HCl)=?                                m(s-n1) + m(s-n2) 50∙1,1+20∙1,05

The answer: 17,2%

 

Molarity

5. To determine the temporary recalcification of blood plasma, CaCl₂ solution with molarity of 0,025 mol/l is applied. How many grammes of CaCl₂ is required to prepare 250 ml of this solution?

Data:                                                              Solution:

C(CaCl₂) = 0,025 mol/l  V(s-n) = 250 ml = 0,25 l

V(s-n) = 250 ml         m(CaCl₂) =C (CaCl₂)∙ M(CaCl₂) ∙ V(s-n) =

M(CaCl₂) -?                        = 0,025∙ 111∙ 0,25 =0,6938 g.

                                                   The answer: 0,6938 g

 

6. How many milliliters of H₂SO₄ solution with density of 1,26 g/ml (34,6%) are required to prepare 3 l of 0,12 M solution.

 Data:                                                              Solution:

ρ(s-n 1) = 1,26 g/ml m₁(H₂SO₄) = m₂(H₂SO₄)

ω₁(H₂SO₄) = 34% 1)V(s-n)∙ρ(s-n1)^. ω₁(H₂SO₄)=C₂(H₂SO₄)∙M(H₂SO₄)∙V(s-n2)

V(s-n) = 3 L        2) V(s-n 1) ∙1,26∙ 0,346 = 0,12∙98 ∙3 =>              

(H₂SO₄) = 0,12 mol/L                       0,12∙ 98 ∙ 3

  V(s-n1) =?      V(s-n 1) = —————— = 81(ml)

                                                 1,26 ∙ 0,346

                                             The answer: 81 ml

 

7. Solution of NaCl with mass fraction 0,85% (ρ = 1,005 g/ml) is called physiological solution, it is applied for intravenous infusions. Determine the molarity of this solution.

Data:                                                              Solution:

ω (NaCl) =0,85% The 1^st version:

ρ(s-n) = 1,005 g/ml Assume that V(s-n) =1000 ml= 1L, then:

                             m(s-n) = ρ(s-n) · V(s-n) = 1,005·1000=1005 g

C(NaCl) =?             ω₁(NaCl) =0,0085

                                                        υ(NaCl)      m(NaCl)

                                    C(NaCl) = ———— = ——————— =

                                                         V(s-n)       M((NaCl)∙V(s-n)

                                      m (s-n)∙ω(NaCl) 1005 ∙ 0,0085

                                    = ——————— = ————— = 0,146 mol/L

                                            M(NaCl)∙V(s-n)   58,5 ∙ 1

 

  The second version:

                                 According to formula from the table 3 of appendix

                                     10·ρ(s-n)·ω(NaCl)  10 •1,005•0,85

                       C(NaCl)= ———————— = —————— =0,146 (mol/L)

                                               M(NaCl)            58,5

                                               The answer: 0,146 (mol/L) or 0,146 M.

 

Normality

 

8. How many grammes of Na₂CO₃ are required to prepare 1,5 L of solution  with molar concentration of equivalent equal to 0,15 mol/l. This solution is prepared for complete reacting.

Data:                                                         Solution:

V(s-n) = 1,5 L      The reaction proceeds according to:

C(1/ZNa₂CO₃)=0,15 mol/l

m(Na₂CO₃) =?   Na₂CO₃+ H₂SO₄ =Na₂SO₄ + H₂O + CO₂

                                                      f_eq = 1/2.

                        M(1/2 Na₂CO₃)=1/2 M(Na₂CO₃) =106/2 = 53 g/mol

                         m(Na₂CO₃)=C(1/2Na₂CO₃)∙V(s-n)∙M(1/2 Na₂CO₃)=

                                   =0,15∙1,5∙53 = 11,925 g

                                                             The answer: 11,925 g.

 

9. How many ml of CaCl₂ solution with mass fraction of 10% (ρ= 1,04 g/ml) are required to prepare 2 L of solution with molarity of equivalent 0,05 mol/l. This solution is prepared for the reaction proceeding completely.

Data:                                                         Solution:

ω₁%(CaCl₂) = 10%= 0,1   CaCl₂ + 2AgNO₃= 2 AgCl + Ca(NO₃)₂

ρ(s-n 1)=1,04 g/ml           f_eq = ½; M(1/2 CaCl₂)= 1/2·111= 55,5(g/mol)

V(s-n 2) = 2L                    m(CaCl₂) = V(s-n 1)·ρ(s-n 1)·ω₁(CaCl₂)=

C₂(1/zCaCl₂) = 0,05 mol/l =C₂(1/2 CaCl₂)·M(1/2 CaCl₂)·V(s-n2) =>

                                                 C₂(1/2CaCl₂)•V(s-n2)•M(1/2 CaCl₂)

V(s-n 1) =?                V(s-n1) = ——————————————  =

                                                            ρ(s-n 1) ∙ ω₁(CaCl₂)

                                                   0,05∙2∙55,5

                                              = ————— = 53,35(ml)

                                                    1,04 ∙ 0,1

                                            The answer: 53,35 ml.

 

10. Determine the normality of 50,1% H₂SO₄ solution with density 1,4 g/ml. Solution is applied for reaction, proceeding completely.

Data:                                                         Solution:

ω₁% (H₂SO₄) =50,1%      Reaction equation:

 ρ(s-n 1)=1,4 g/ml             H₂SO₄ + 2KOH = K₂SO₄ + 2 H₂O

f_eq(H₂SO₄) = ½; M(1/2 H₂SO₄)= 98/2= 49(g/mol)

C(1/Z H₂SO₄)=?           Assume that V(s-n) = 1000 ml =1L

                     

                         

                   Then  m(s-n) = ρ(s-n)· V(s-n) =1,4·1000 = 1400 (g)

                                           ν(1/z H₂SO₄)       m (H₂SO₄)

              C(1/Z H₂SO₄)= —————— = ———————

                                                  V(s-n)   M(1/2H₂SO₄)∙V(s-n)

 

                        m(s-n)∙ ω₁(H₂SO₄)       1400 ∙ 0,501

                = —————————— = ————— = 14,2 mol/l

                    M(1/2 H₂SO₄))· V(s-n)          49·1

                                   

 The second version: using the formula from table 3 of appendix:

 

                         10∙ ρ · ω₁(H₂SO₄) 10∙1,4∙50,1

 C(1/Z H₂SO₄) = ——————— = ————— =14,2 mol/l

                           M (1/2H₂SO₄)             49

 

                         The answer: 14,2 mol/l

 

11. Determine the normality of 0,5 M Al₂(SO₄)₃ solution destined for reaction of alluminium hydroxide formation.

Data:                                                         Solution:

C(Al₂(SO₄)₃) =0,5 M                             Reaction equation:

                                   Al₂(SO₄)₃ + 6KOH = 2 Al(OH)₃ + 3K₂SO₄

C(1/Z Al₂(SO₄)₃) =?                    f_eq (Al₂(SO₄)₃) = 1/6

                                       C(1/6Al₂(SO₄)₃) = 6•C(Al₂(SO₄)₃)= 6∙0,5=3 Mol/l

                                   

The answer:3 mol/l (3N)

 

12. Determine the titre of 0,15 N Na₂CO₃ solution destined for full ion-exchange reaction.(look at task №8)

 

Data:                                                         Solution:

C(1/2 Na₂CO₃) = 0,15 mol/l             f_eq (Na₂CO₃) =1/2;

                                                         C(1/2 Na₂CO₃)∙ M(1/2 Na₂CO₃)

t(Na₂CO₃)=?           t(Na₂CO₃)= ————————————— =

                                                                           1000

                                                      0,15∙53

                                                = ———— = 0,007950 g/ml

                                                       1000

                                                  The answer: 0,007950 g/ml

 

Molality

 

13. Calculate the molality of 40% HNO₃- solution.

Data:                                                         Solution:

ω (HNO₃) = 40%         Assume that m(s-n) = 100g, then

                                m(HNO₃) = m(s-n)∙ ω (HNO₃) = 100·0,4 =40 g.

b(HNO₃) =?             m(H₂O) = 100 –40 = 60 g =0,06(kg)

                                υ(HNO₃)       m(HNO₃)            40

           b(HNO₃) = ———— = ——————— = ——— = 10,6

                                  (H₂O)  M(HNO₃)•m(H₂O)   63•0,06

The answer:10,6 Mol/kg