How do you prepare 0.01 M from 65 percent HNO3?

Answer 1

The molar mass of HNO3 is 63 gram/mol and of water is 18 gram/mol. Lets assume that the density of the HNO3 + water mixture is equal to the density of water at room temperature is 1000 gram/L.

0.01 M is equal to 0.01 mol HNO3 per liter HNO3 + water. Multiplying the molar mass of HNO3 by the molarity shows this is equal to 0.63 gram HNO3 per liter HNO3 + water. Using the density of water shows that this is equal to 0.000063 gram HNO3 per gram water. This is therefore 0.0065 percent by weight. (The assumption that the density of the mixture is approximately equal to that of pure water seems justified because the amount of HNO3 in the mixture is very low)

Assuming we start with an initial 65 percent by weight of HNO3 (10.3 M using the same type of calculation), water will need to be added to dilute to 0.0065 percent (0.01M). To find the required amount of water that is added a total and component mass balance can be used.

The total mass balance is given by:

Mt = M0 + Ma

Here, M0 is the initial amount of water + HNO3, Ma is the added amount of (pure) water and Mt is the total mass of the water + HNO3 after adding more water.

The component mass balance over HNO3 is given by:

xtMt = x0M0 + xaMa= x0M0

Where xt is the final weight fraction of HNO3 in water, x0 is the initial weight fraction in water and xa is the weight fraction of HNO3 in pure water (which is logically equal to zero).

Lets say we have a starting mass M0 = 100 gram HNO3 + water with the initial weight fraction x0 = 0.65 gram HNO3 per gram and we want a final weight fraction xt = 0.000065, rearranging for the total weight after having added water to reach this dilution gives:

Mt = x0M0/xt = 0.65x100/0.000065 = 1000000 gram HNO3 + water

The amount of water which needs to be added then simply follows from:

Ma = Mt - M0 = 1000000 - 100 = 999900 gram water