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9 March, 12:57

Commercial nitric acid comes in a concentration of 16.0 mol/L. The density of this solution is 1.42 g/mL

Calculate the percent (w/w) of nitric acid, HNO3, in this solution.

How many milliliters of the concentrated acid have to be taken to prepared 250 g of a solution that is 10.0% (w/w) HNO3?

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  1. 9 March, 13:13
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    71.0% (w/w)

    24.8 mL

    Explanation:

    First, let's calculated the molar mass of the acid, knowing the molar masses of its elements:

    H = 1 g/mol

    N = 14 g/mol

    O = 16 g/mol

    So, the molar mass is: 1x1 + 1x14 + 3x16 = 63.0 g/mol

    In 1 L of the solution, there is 16.0 mol of the acid, so the mass will be the number of moles multiplied by the molar mass:

    m = 16.0x63.0 = 1008.0 g

    The density is 1.42 g/mol, so the total mass of the solution is the density multiplied by the volume. For 1 L (1000 mL):

    mt = 1.42x1000 = 1420.0 g

    The percent of nitric acid is its mass divided by the total mass multiplied by 100%

    (1008.0/1420.0) x100% = 71.0% (w/w)

    To prepare a solution by dilution of the other solution, we can use the equation:

    C1m1 = C2m2

    Where C is the concentration, m is the mass, 1 is the initial solution and 2 the final solution. So:

    71%xm1 = 10%x250

    m1 = 35.21 g

    The volume is the mass divided by the density:

    V = 35.21/1.42

    V = 24.8 mL
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