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Calculate the mass of ascorbic acid ( Vi...

Calculate the mass of ascorbic acid `(` Vitamin `C,C_(6)H_(8)O_(6))` to be dissolved in `75g` of acetic acid to lower its melting poit by `1.5^(@)C. K_(f)=3.9 K kg mol ^(-1)`

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To solve the problem of calculating the mass of ascorbic acid (Vitamin C, C₆H₈O₆) to be dissolved in 75g of acetic acid to lower its melting point by 1.5°C, we will use the formula for freezing point depression: \[ \Delta T_f = K_f \cdot m \] Where: - \(\Delta T_f\) = change in freezing point (in °C) - \(K_f\) = cryoscopic constant of the solvent (in K kg mol⁻¹) - \(m\) = molality of the solution (in mol/kg) ### Step 1: Calculate the molality (m) Rearranging the formula for molality gives us: \[ m = \frac{\Delta T_f}{K_f} \] Substituting the values: \[ \Delta T_f = 1.5 \, °C \] \[ K_f = 3.9 \, K \, kg \, mol^{-1} \] Calculating molality: \[ m = \frac{1.5}{3.9} \approx 0.3846 \, mol/kg \] ### Step 2: Calculate the number of moles of solute Molality is defined as the number of moles of solute per kilogram of solvent. We need to convert the mass of acetic acid from grams to kilograms: \[ \text{mass of acetic acid} = 75 \, g = 0.075 \, kg \] Now, using the definition of molality: \[ m = \frac{n}{\text{mass of solvent in kg}} \] Where \(n\) is the number of moles of ascorbic acid. Rearranging gives: \[ n = m \cdot \text{mass of solvent in kg} = 0.3846 \, mol/kg \cdot 0.075 \, kg \approx 0.028845 \, mol \] ### Step 3: Calculate the mass of ascorbic acid Now we need to convert the number of moles of ascorbic acid to mass using its molar mass. The molar mass of ascorbic acid (C₆H₈O₆) is calculated as follows: \[ \text{Molar mass} = (12 \times 6) + (1 \times 8) + (16 \times 6) = 72 + 8 + 96 = 176 \, g/mol \] Now, we can calculate the mass of ascorbic acid: \[ \text{mass of ascorbic acid} = n \cdot \text{molar mass} = 0.028845 \, mol \cdot 176 \, g/mol \approx 5.08 \, g \] ### Final Answer The mass of ascorbic acid to be dissolved in 75g of acetic acid to lower its melting point by 1.5°C is approximately **5.08 grams**. ---

To solve the problem of calculating the mass of ascorbic acid (Vitamin C, C₆H₈O₆) to be dissolved in 75g of acetic acid to lower its melting point by 1.5°C, we will use the formula for freezing point depression: \[ \Delta T_f = K_f \cdot m \] Where: - \(\Delta T_f\) = change in freezing point (in °C) ...
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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) .

Calculating the mass of ascorbic acid (C_(6)H_(8)O_(6)) to be dissolved in 74 g of acetic acid to lower its melting point by 1.5^(circ)C. K_(f) for CH_(3)COOH is 3.9 K kg nol^(-1) .

Knowledge Check

  • The mass of ascorbic acid (C_(6)H_(8)O_(6)) to be dissolved in 100g of acetic acid to lower its freezing point by 1.5^(@)C in g is : ( K_(f) for acetic acid is "4.0 K kg mol"^(-1) )

    A
    17.6
    B
    8.8
    C
    6.6
    D
    13.2
  • What is the [H^+] in a 0.10 N solution of ascorbic acid, C_6H_8O_6 ?

    A
    `8.0xx10^(-6) M`
    B
    `2.8xx10^(-3) M`
    C
    `4.0xx10^(-3) M`
    D
    `5.3xx10^(-3) M`
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