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8 March, 20:36

A student wants to make a 0.150 M aqueous solution of silver nitrate, AgNO3, and has a bottle containing 11.27 g of silver nitrate. What should be the final volume of the solution

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  1. 8 March, 20:51
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    V = 0.44 L

    Explanation:

    Moles is denoted by given mass divided by the molecular mass,

    Hence,

    n = w / m

    n = moles,

    w = given mass,

    m = molecular mass.

    From the question,

    w = 11.27 g

    m (for AgNO₃) = 169.87 g/mol

    Hence, the moles can be calculated as -

    n = w / m

    n = 11.27 g / 169.87 g/mol

    n = 0.066 mol

    Molarity -

    Molarity of a substance, is the number of moles present in a liter of solution.

    M = n / V

    M = molarity

    V = volume of solution in liter,

    From the question,

    M = 0.150 M

    n = 0.066 mol (calculated above)

    The final volume of the solution can be calculated by using the above equation,

    M = n / V

    0.150 M = 0.066 mol / V

    V = 0.44 L
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