1.

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) .

Answer»

Solution : 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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