Avem:
C = 94%
p = 1,5 g/cm3
Ni se cere:
Cm = ?
Mai intai folosim formula:
[tex]C_m=\frac{m_d}{V_s*M_{HNO_3}}[/tex]
Din aceasta formula, noi putem afla masa molara a acidului azotic:
M HNO3 = A H + A N + 3 * A O = 1 + 23 + 3 * 16 = 24 + 48 = 72 g/mol
Folosim formula [tex]C = \frac{m_d}{m_s} * 100 =\ \textgreater \ m_d = \frac{C*m_s}{100} [/tex]
Folosim formula densitatii:
p = ms/Vs => Vs = ms/p
Atunci, in formula concentratiei molare avem:
[tex]C_M=\frac{\frac{C*m_s}{100}}{\frac{\rho}{m_s}*M_{HNO_3}} = \frac{C*m_s}{100}*\frac{\rho}{m_s*M_{H_{NO3}}} = \frac{C*\rho}{100*M_{HNO_3}} =\ \textgreater \ C_M = \frac{94*1,5}{100*72}\\ =\ \textgreater \ C_M = 0,01 \ M[/tex]
