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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%.
m(sol-n) =250 g M(Na2SO4) = 142 g/mol ω(Na2SO4)= 5% = 0,05 M(Na2SO4∙10H2O) = 322 g/mol
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?
m(s-n) =25 g According to the “rule of cross” let make the
ω2(ZnSO4)=0,25%
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%.
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%
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.
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%
ω2(HCl)=10% 3) m1(HCl)+m2(HCl) 11+2,1
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?
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) =
The answer: 0,6938 g
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
Data: Solution: ω (NaCl) =0,85% The 1st version: ρ(s-n) = 1,005 g/ml Assume that V(s-n) =1000 ml= 1L, then:
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.
V(s-n) = 1,5 L The reaction proceeds according to:
m(Na2CO3) =? Na2CO3+ H2SO4 =Na2SO4 + H2O + CO2 feq = 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.
ω1%(CaCl2) = 10%= 0,1 CaCl2 + 2AgNO3= 2 AgCl + Ca(NO3)2 ρ(s-n 1)=1,04 g/ml feq = 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) =>
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.
ω1% (H2SO4) =50,1% Reaction equation: ρ(s-n 1)=1,4 g/ml H2SO4 + 2KOH = K2SO4 + 2 H2O feq(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:
C(1/Z Al2(SO4)3) =? feq (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)
C(1/2 Na2CO3) = 0,15 mol/l feq (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.
ω (HNO3) = 40% Assume that m(s-n) = 100g, then m(HNO3) = m(s-n)∙ ω (HNO3) = 100·0,4 =40 g.
υ(HNO3) m(HNO3) 40 b(HNO3) = ———— = ——————— = ——— = 10,6 (H2O) M(HNO3)•m(H2O) 63•0,06 The answer:10,6 Mol/kg
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Data Solution
m(s-n)∙5% Na2SO4
Date: Solution
ω1(ZnSO4)=2% 2% 0,25% 
0,25% 
m(H2O) =? 
0% 1,75
Date: Solution:
ω1(HCl) = —————— = —————— 
Data: Solution:
ρ(s-n 2) =1,05g/ml 
ω2(HCl)= —————— = —————∙100% =17,2%
Data: Solution:
M(CaCl2) -? = 0,025∙ 111∙ 0,25 =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.
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.
m(s-n) = ρ(s-n) · V(s-n) = 1,005·1000=1005 g
Data: Solution:
C(1/ZNa2CO3)=0,15 mol/l
Data: Solution:
C2(1/2CaCl2)•V(s-n2)•M(1/2 CaCl2)
Data: Solution:
Al2(SO4)3 + 6KOH = 2 Al(OH)3 + 3K2SO4
Data: Solution:
Data: Solution:
b(HNO3) =? m(H2O) = 100 –40 = 60 g =0,06(kg)
