a) nNaCl = CM .V = 2,5.0,9 = 2,25 (mol)
→ mNaCl = 2,25.(23 + 35,5) = 131,625 (g)
b) mMgCl2=50.4%/100%=2(g)
c) nMgSO4 = 0,1.0,25 = 0,025 (mol)
→ mMgSO4 = 0,025.(24 + 64 + 32) = 3 (g)
a) `n_(NaCl) = CM .V = 2,5.0,9 = 2,25 (mol)`
`→ m_(NaCl) = 2,25.(23 + 35,5) = 131,625 (g)`
b) `m_(MgCl_2) = (50,4%)/(100%) = 2g`.
c) `n_(MgSO4) = 0,1.0,25 = 0,025 (mol)`
`→ m_(MgSO4) = 0,025.(24 + 64 + 32) = 3 (g)`
a) nNaCl =CM .V=2,5.0,9=2,25(mol)nNaCl =CM .V=2,5.0,9=2,25(mol)
→mNaCl=2,25.(23+35,5)=131,625(g)→mNaCl=2,25.(23+35,5)=131,625(g)
b) mMgCl2=50,4%100%=2gmMgCl2=50,4%100%=2g.
c) nMgSO4=0,1.0,25=0,025(mol)nMgSO4=0,1.0,25=0,025(mol)
→mMgSO4=0,025.(24+64+32)=3(g)
a) nNaCl = CM .V = 2,5.0,9 = 2,25 (mol)
→ mNaCl = 2,25.(23 + 35,5) = 131,625 (g)
b) mMgCl2=50.4%/100%=2(g)
c) nMgSO4 = 0,1.0,25 = 0,025 (mol)
→ mMgSO4 = 0,025.(24 + 64 + 32) = 3 (g)
