Question

Exercițiu 2 va rog, am nevoie de toate cerințele
Mulțumesc ❤️

Answer 1

Formula concentrației molare:

$C_m=\frac{n}{V_s}$ , unde:

n - nr. de moli de subst. dizolvați în 1000mL soluție

Vs - volumul soluției

Relația dintre masă și nr. de moli:

$n=\frac{m}{M}$ , unde:

m - masa de substanță cunoscută

M - masa moleculară a substanței

Relația dintre masă și densitate:

$\rho=\frac{m}{V}$ , unde:

m - masa de subst./masa soluției cunoscute

V - volumul soluției

Rezolvare ex. 1:

a.

$n_{H_2SO_4}=2\ mol\\V_s=250\ cm^3\ sol.\\\\C_m=\frac{n}{V_s}=\frac{2\ mol}{0.25\ L}=8\ mol/L\ (M)$

b.

$m_{NaOH}=400\ g\\V_s=5L\ sol.\\\\M_{NaOH}=A_{Na}+A_O+A_H=23+16+1=40\ g/mol\\\\n=\frac{m}{M}=\frac{400\ g}{40\ g/mol}=10\ mol\ NaOH\\\\C_m=\frac{n}{V_s}=\frac{10\ mol}{5\ L}=2\ mol/L\ (M)$

c.

$m_{H_2SO_4}=196\ g\\m_s=200\ g\ sol.\\\rho=1.84\ g/cm^3\\\\{\rho}=\frac{m}{V_s}\implies V_s=\frac{m}{\rho}=\frac{200\ g}{1.84\ g/cm^3}=108.7\ cm^3=0.1087\ dm^3(L)\ sol.\\\\M_{H_2SO_4}=2A_H+A_S+4A_O=2+32+64=98\ g/mol\\\\n=\frac{m}{M}=\frac{196\ g}{98\ g/mol}=2\ mol\ H_2SO_4\\\\C_m=\frac{n}{V_s}=\frac{2\ mol}{0.1087\ L}=18.4\ mol/L\ (M)$

Rezolvare ex. 2:

a.

$V_s_{H_2SO_4}=0.7\ m^3=700\ dm^3(L)\\C_m=0.01\ mol/L\ (M)\\\\C_m=\frac{n}{V_s}\implies n=C_m*V_s=0.01\ mol/L\ *\ 700L=7\ mol\ H_2SO_4\\\\M_{H_2SO_4}=98\ g/mol\\\\n=\frac{m}{M}\implies m=n*M=7\ mol*98\ g/mol=686\ g\ H_2SO_4$

b.

$V_s_{H_2SO_4}=600\ L\\C_m=1.5\ mol/L\ (M)\\\\C_m=\frac{n}{V_s}\implies n=C_m*V_s=1.5\ mol/L\ *\ 600L=900\ mol\ H_2SO_4\\\\M_{H_2SO_4}=98\ g/mol\\\\n=\frac{m}{M}\implies m=n*M=900\ mol*98\ g/mol=88200\ g=88.2\ kg\ H_2SO_4$

c.

$V_s_{H_2SO_4}=250\ mL=0.25\ L\\C_m=2\ mol/L\ (M)\\\\C_m=\frac{n}{V_s}\implies n=C_m*V_s=2\ mol/L\ *\ 0.25L=0.5\ mol\ H_2SO_4\\\\M_{H_2SO_4}=98\ g/mol\\\\n=\frac{m}{M}\implies m=n*M=0.5\ mol*98\ g/mol=49\ g\ H_2SO_4$