Formula concentrației molare:
[tex]C_m=\frac{n}{V_s}[/tex] , unde:
n - nr. de moli de subst. dizolvați în 1000mL soluție
Vs - volumul soluției
Relația dintre masă și nr. de moli:
[tex]n=\frac{m}{M}[/tex] , unde:
m - masa de substanță cunoscută
M - masa moleculară a substanței
Relația dintre masă și densitate:
[tex]\rho=\frac{m}{V}[/tex] , unde:
m - masa de subst./masa soluției cunoscute
V - volumul soluției
Rezolvare ex. 1:
a.
[tex]n_{H_2SO_4}=2\ mol\\V_s=250\ cm^3\ sol.\\\\C_m=\frac{n}{V_s}=\frac{2\ mol}{0.25\ L}=8\ mol/L\ (M)[/tex]
b.
[tex]m_{NaOH}=400\ g\\V_s=5L\ sol.\\\\M_{NaOH}=A_{Na}+A_O+A_H=23+16+1=40\ g/mol\\\\n=\frac{m}{M}=\frac{400\ g}{40\ g/mol}=10\ mol\ NaOH\\\\C_m=\frac{n}{V_s}=\frac{10\ mol}{5\ L}=2\ mol/L\ (M)[/tex]
c.
[tex]m_{H_2SO_4}=196\ g\\m_s=200\ g\ sol.\\\rho=1.84\ g/cm^3\\\\{\rho}=\frac{m}{V_s}\implies V_s=\frac{m}{\rho}=\frac{200\ g}{1.84\ g/cm^3}=108.7\ cm^3=0.1087\ dm^3(L)\ sol.\\\\M_{H_2SO_4}=2A_H+A_S+4A_O=2+32+64=98\ g/mol\\\\n=\frac{m}{M}=\frac{196\ g}{98\ g/mol}=2\ mol\ H_2SO_4\\\\C_m=\frac{n}{V_s}=\frac{2\ mol}{0.1087\ L}=18.4\ mol/L\ (M)\\[/tex]
Rezolvare ex. 2:
a.
[tex]V_s_{H_2SO_4}=0.7\ m^3=700\ dm^3(L)\\C_m=0.01\ mol/L\ (M)\\\\C_m=\frac{n}{V_s}\implies n=C_m*V_s=0.01\ mol/L\ *\ 700L=7\ mol\ H_2SO_4\\\\M_{H_2SO_4}=98\ g/mol\\\\n=\frac{m}{M}\implies m=n*M=7\ mol*98\ g/mol=686\ g\ H_2SO_4[/tex]
b.
[tex]V_s_{H_2SO_4}=600\ L\\C_m=1.5\ mol/L\ (M)\\\\C_m=\frac{n}{V_s}\implies n=C_m*V_s=1.5\ mol/L\ *\ 600L=900\ mol\ H_2SO_4\\\\M_{H_2SO_4}=98\ g/mol\\\\n=\frac{m}{M}\implies m=n*M=900\ mol*98\ g/mol=88200\ g=88.2\ kg\ H_2SO_4[/tex]
c.
[tex]V_s_{H_2SO_4}=250\ mL=0.25\ L\\C_m=2\ mol/L\ (M)\\\\C_m=\frac{n}{V_s}\implies n=C_m*V_s=2\ mol/L\ *\ 0.25L=0.5\ mol\ H_2SO_4\\\\M_{H_2SO_4}=98\ g/mol\\\\n=\frac{m}{M}\implies m=n*M=0.5\ mol*98\ g/mol=49\ g\ H_2SO_4[/tex]
