Oblicz stęzenie procentowe kwasu siarkowego (VI) otrzymanego przez rozpuszczenie 5,6 dm3 tlenku siarki (VI) w 100 cm3 wody.
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M(H₂SO₄)=2*1g/mol+32g/mol+4*16g/mol=98g/mol
M(SO₃)=32g/mol+3*16g/mol=80g/mol
mH₂O=100cm³*1g/cm³=100g
22,4dm³ ---------------- 80g
5,6dm³ ------------------- xg
xg=5,6dm³ * 80g / 22,4dm³
xg=20g
SO₃+H₂O=H₂SO₄
20g-----------xg
80g-----------98g
xg=20g*98g/80g
xg=24,5g
ms=24,5g
mr=ms+mH₂O=24,5g+100g=124,5g
Cp=?
Cp=ms/mr * 100%
Cp=24,5g/124,5g * 100%
Cp=19,7%