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27 January, 17:04

What mass of solute is needed to prepare each of the following solutions? a. 3.00 L of 0.125 M K2SO4 b. 550 mL of 0.015 M NaF c. 700 mL of 0.350 M C6H12O6

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  1. 27 January, 20:03
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    A.

    M = 0.125 M

    V = 3.0 L

    M=n/V (L) ⇒ n = 0.125 mol/L * 3.0L = 0.375 mol

    MM of K2SO4

    K2 = 2 * 39 = 78 g/mol

    S = 32 g/mol

    O4 = 4*16 = 64 g/mol

    MM = 78g/mol + 32g/mol + 64g/mol = 174 g/mol

    mass = n*MM = 0.375 mol*174 g/mol = 65.25 g

    b.

    M = 0.015 M

    V = 0.55L

    n = 0.015 mol/L * 0.55L = 0.00825 mol

    MM NaF

    Na: 23g/mol

    F: 19 g/mol

    MM = 23g/mol + 19g/mol = 42g/mol

    mass = n * MM = 0.00825 mol * 42g/mol = 0.3465 g

    c.

    n = M*V (L) = 0.350 mol/L * 0,700L = 0.245 mol

    MM = 6*12 g/mol + 12*1g/mol + 6*16g/mol = 180 g/mol

    mass = n*MM = 0.245 mol * 180g/mol = 44.1 g
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