1. Oblicz stężenie molowe 40% roztworu kwasu azotowego V o gęstości 1.5g/cm3
2. Jakie jest stężenie roztworu 30% kwasu solnego o gęstości 1.151g/cm3?
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1.
Cm = ??
Cp = 40%
d = 1,5 g/cm3
M HNO3 = 1g/mol + 14g/mol + 48g/mol = 63g/mol
Cm = (Cp*d):(M*100%)
Cm = (40%*1,5):(63*100%)
Cm = 60:6300
Cm ~ 0,001 mol/dm3
2.
Cm = ??
Cp = 30%
d = 1,151 g/cm3
M HCl = 1g/mol + 35,5g/mol = 36,5g/mol
Cm = (Cp*d):(M*100%)
Cm = (30*1,151):(36,5*100)
Cm = 0,01 mol/dm3
Pozdrawiam ;)