Answer to Question #95635 in Cell Biology for Mayer

Question #95635
How many grams of NaHCO3 (sodium bicarbonate, or baking soda) would you need to make a 2 molar (2M) solution?

How many grams of C12H22O11 (sucrose) would you need to make a 0.5 molar (1M) solution?
1
Expert's answer
2019-10-01T08:06:55-0400


(i)

Let's say the solution volume is 1 L. Then:

n(NaHCO3)=cVn(NaHCO_3) = c*V


n(NaHCO3)=2M1L=2moln(NaHCO_3) = 2 M *1 L = 2mol

m(NaHCO3)=nMm(NaHCO_3) = n*M

m(NaHCO3)=2mol84g/mol=168g.m(NaHCO_3) = 2mol*84g/mol = 168 g.

(ii)

Let's say the solution volume is 1 L. Then:

n(C12H22O11)=cVn(C_{12}H_{22}O_{11}) = c * V

If solution 0.5 molar, then n(C12H22O11)=0.5moln(C_{12}H_{22}O_{11}) = 0.5 mol.

Then m(C12H22O11)=nM=0.5mol342g/mol=171gm(C_{12}H_{22}O_{11}) = n*M = 0.5mol*342g/mol = 171 g.

If solution is 1M, then n(C12H22O11)=1moln(C_{12}H_{22}O_{11}) = 1 mol.

Then m(C12H22O11)=342g.m(C_{12}H_{22}O_{11}) = 342g.


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