Calculate the mass of ascorbic acid (Vitamin `C, C_6H_8O_6`) to be dissolved in 75 g of acetic acid to lower its melting point by `1.5^@C. K_f = 3.9 K kg "mol"^(-1)` .
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Lowering in melting point `(Delta T_f) = 1.5^@` Mass of solvent `(CH_3COOH), w_1 = 75 g` Molar mass of solvent `(CH_3COOH), M_1= 60 g "mol"^(-1)` Molar mass of solute `(C_6H_8O_6), M_2 = 72 + 8 + 96 = 176 g "mol"^(-1)` For acetic acid, `K_f = 3.9 K kg "mol"^(-1)` Applying the formula, ` M_2 = (1000 K_f w_2)/(w_1 Delta T_f) " or " w_2 = (M_2 xx w_1 xx Delta T_f)/(1000 xx K_f)` Substituting the values, we get ` w_2 = ((176 g "mol"^(-1) )(75g)(1.5K))/((1000 g kg^(-1) )(3.9 K kg "mol"^(-1) )) = 5.077 g `
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