Suatu larutan NaNO3 kadarnya 8,5 % berat dalam air Ar Na=23, N= 14, O=16 tentukan molaritas larutan dan fraksi mol NaNO3

Diket :

@ massa larutan = 100 gr

@ massa NaNO₃ = 8,5% x 100 = 8,5 gr

@ Mr NaNO₃ = 85 gr/mol

@ massa air = 100 – 8,5 = 91,5 gr

@ volume air = 91,5 mL = 0,0915 Liter

Jawab :

@ [NaNO₃] = mol NaNO₃ / Volume air = ( 8,5 / 85 ) / 0,0915 = 1,09289 M

@ X NaNO₃ = mol NaNO₃ / ( mol NaNO₃ + mol air ) = (8,5/85) / ({8,5/85} + {91,5/18}) = 0,0192926

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