Calculate the number of grams of HNO3 which must be added to 31.5 g of H20 to prepare a 0.950 m solution.

1.89g of HNO3

Explanation:

Data obtained from the question include:

Mass of water = 31.5 g

Molality of HNO3 = 0.950 m

Mass of HNO3 =.?

Next, we shall determine the number of mole of HNO3 in the solution.

Mass of water = 31.5 g = 31.5/1000 = 0.0315 Kg

Molality of HNO3 = 0.950 m

Number of mole of HNO3 =..?

Molality is simply defined as the mole of solute per unit kilogram of solvent (water) i.e

Molality = mole /kg of water

With the above formula, the mole of HNO3 can be obtained as follow:

Molality = mole /kg of water

0.950 = mole of HNO3 /0.0315

Cross multiply

Mole of HNO3 = 0.950 x 0.0315

Mole of HNO3 = 0.03 mole

Finally, we shall convert 0.03 mole of HNO3 to grams. This is illustrated below:

Molar mass of HNO3 = 1 + 14 + (16x3) = 63g/mol

Mole of HNO3 = 0.03 mole

Mass of HNO3 =..?

Mole = mass /molar mass

0.03 = mass of HNO3 /63

Cross multiply

Mass of HNO3 = 0.03 x 63

Mass of HNO3 = 1.89g

Therefore, 1.89g of HNO3 is needed to prepare the solution.

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