(i)
Let's say the solution volume is 1 L. Then:
n(NaHCO3)=c∗V
n(NaHCO3)=2M∗1L=2mol
m(NaHCO3)=n∗M
m(NaHCO3)=2mol∗84g/mol=168g. (ii)
Let's say the solution volume is 1 L. Then:
n(C12H22O11)=c∗VIf solution 0.5 molar, then n(C12H22O11)=0.5mol.
Then m(C12H22O11)=n∗M=0.5mol∗342g/mol=171g.
If solution is 1M, then n(C12H22O11)=1mol.
Then m(C12H22O11)=342g.
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