Question

A laboratory procedure calls for making 590.0 mL of a 1.1 M KNO3 solution. How much KNO3 in grams is needed?

Answer 1

590 mL = 590 cm³= 0,59 dm³

C = n/V
n = 1,1M × 0,59 dm³
n = 0,649 mol
_____________________________

M KNO₃ = 39g+14g+16g×3 = 101 g/mol

1 mole -------- 101g
0,649 --------- X
X = 101×0,649
X = 65,549g KNO₃

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