Question

discribe the preparation of 900ml of 3M hno3- from a commercial reagent that is 69% hno3- and has a specefic gravity 1.42 solution

Answer 1

Question #58852, Chemistry, Other

Describe the preparation of 900 ml of 3 M HNO₃ from a commercial reagent that is 69% HNO₃ and has a specific gravity 1.42 solution.

Answer:

$$C_M = \frac{v}{V}$$ $$v = \frac{m}{M}$$

M (HNO₃) = 63 g/mol

The amount of HNO₃ contained in 900 ml of 3 M HNO₃ solution is:

$$v = C_M \cdot V$$ $$v(HNO_3) = 3 \cdot 0.9 = 2.7 \text{ mol}$$

This amount of matter corresponds to the mass of 69% HNO₃ solution:

m = \frac{vM}{\% $$m(HNO_3)_{69\%} = \frac{2.7 \cdot 63}{0.69} = 246.5 \text{ g}$$

Taking into account the specific gravity of this solution, its volume is:

$$\rho = \frac{m}{V}$$ $$V = \frac{m}{\rho}$$ $$V(HNO_3)_{69\%} = \frac{246.5}{1.42} = 173.61 \text{ ml}$$

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