Answer to Question #95635 in Cell Biology for Mayer

Question

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?

Answer 1

(i)

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

$$n(NaHCO_3) = c*V$$

$$n(NaHCO_3) = 2 M *1 L = 2mol$$

$$m(NaHCO_3) = n*M$$

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

(ii)

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

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

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

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

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

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