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Mol.Molyar massa. Modda miqdoriga doir masalalar yechish
1-masala.Quyidagi moddaning nisbiy molekulyar massasini hisoblang: Ca(OH)₂, Na₂SO₄, Ba(NO₃)₂, H₃PO₄.
Yechish: Moddaning nisbiy molekulyar massasini hisoblang.
Mr (NA₂SO₄)=2Ar(Na)+Ar(S)+4Ar(O)=2x23+2x16+2x1=40+32+22+74m. a. b.
Mr (Na₂SO₄)=2Ar(NA)+Ar(S)+4Ar(O)+2x23+32+4x16=32+64=142 m.a.b.
Mr(Ba(NO₃)₂)=Ar(Ba)+2Ar(N)+6Ar(O)=137+2x14+6x16=137+28+96=261.m.a.b.
Mr(H₃PO₄)=3Ar(H)+Ar(P)+4Ar(O)=3x1+31+4x16=31+64=98
2-masala. 49 g sulfat kislota necha molni tashkil qilishini hisoblang.
Yechish. 1-usul. Moddaning nisbiy molekulyar massasini hisoblaymiz:
Mr(H₂SO₄)=2Ar(H)+Ar(S)+4AR(O)=2x1+32+4x16=32+64=98.m.a.b.
V=m/M
Tenglamadan foydalaniib 49 g sulfat kislota necha molni tashkil qilishini hisoblaymiz:
V=m/M=49/98=0.5 mol.
2-usul. Masalani proporsiya usulida ham yechish mumkin.
Dastlab sulfat kislotaning nisbi molekulyar massasini hisoblaymiz:
Mr(H₂SO₄)=2Ar(H)+Ar(S)+4Ar(O)=2x1+32+4x16=2+32+64=98m.a.b.
98g H₂SO₄---------1 mol
49 gH₂SO₄---------x mol x mol=49 g 1 mol/98=0.5 mol.
3-masala.Gramm-mol massasi 80 g bo‘lgan CuO ning 0.3 mol miqdorida qancha gramm bo‘ladi?
Dastlab sulfat kislotaning nisbiy molekulyar massasini hisoblaymiz:
Mr(CuO)=Ar(Cu)+Ar(O)=64+16=80 m.a.b. yoki gramm/mol.
V=m/M
Tenglamadan foydalanib mis oksidining massasini (m) hisoblaymiz.
m=M=0.3 mol 80 g/1mol=124 gramm.
Modda miqdorini aniqlash.
1. 18 gr ugleroddagi modda miqdorini aniqlang.
Yechish: C ning nisbiy massasi=12 ga teng.
N(C)=18:12=1.5 J:1.5 mol.
2. 66 ggr uglerod (IV) oksiddagi modda miqdorini aniqlang.
Yechish:CO2=44.
N(CO2)=66/44=1.5 J:1.5mol.