Question

the molal freezing point depression constant ( k_f = 7.11 , ^circ\text{c kg mol}^{-1} ) for a certain substance ( x ). when 8.7 g of urea (((nh_2)_2co)) are dissolved in 350. g of ( x ), the solution freezes at ( -1.8 , ^circ\text{c} ). calculate the freezing point of pure ( x ). be sure your answer has the correct number of significant digits.

Explanation:

Step 1: Calculate moles of urea

Molar mass of urea \((NH_2)_2CO\) is \(60.06\ g/mol\). Moles of urea \(n=\frac{mass}{molar\ mass}=\frac{8.7\ g}{60.06\ g/mol}\approx0.1448\ mol\).

Step 2: Calculate molality of the solution

Mass of solvent \(X\) is \(350.0\ g = 0.3500\ kg\). Molality \(m=\frac{n}{mass\ of\ solvent\ (kg)}=\frac{0.1448\ mol}{0.3500\ kg}\approx0.4137\ mol/kg\).

Step 3: Use freezing point depression formula

Freezing point depression \(\Delta T_f = K_f\times m\times i\). Urea is a non - electrolyte, so \(i = 1\). \(\Delta T_f=7.11\ ^{\circ}C\cdot kg/mol\times0.4137\ mol/kg\times1\approx2.942\ ^{\circ}C\).
Let the freezing point of pure \(X\) be \(T_{f}^0\) and the freezing point of the solution be \(T_f=- 1.8\ ^{\circ}C\). We know that \(\Delta T_f=T_{f}^0 - T_f\) (since the solution freezes at a lower temperature than the pure solvent). So \(T_{f}^0=T_f+\Delta T_f\).
\(T_{f}^0=-1.8\ ^{\circ}C + 2.942\ ^{\circ}C=1.142\ ^{\circ}C\approx1.1\ ^{\circ}C\) (considering significant figures, the given values: \(8.7\ g\) (2 sig figs), \(350.0\ g\) (4 sig figs), \(K_f = 7.11\) (3 sig figs), \(T_f=-1.8\ ^{\circ}C\) (2 sig figs). The least number of sig figs in the given data for calculation of \(\Delta T_f\) related to the final answer is 2 from \(8.7\ g\) and \(- 1.8\ ^{\circ}C\), but let's check the calculation again. Wait, maybe I made a mistake in sig figs. Let's recalculate with more precision.

Wait, let's redo the steps with more accurate calculations:

1. Moles of urea: \(n=\frac{8.7\ g}{60.055\ g/mol}\) (more accurate molar mass of urea) \(n=\frac{8.7}{60.055}\approx0.14486\ mol\)
2. Molality \(m=\frac{0.14486\ mol}{0.350\ kg}\) (since \(350.0\ g = 0.3500\ kg\), but 8.7 has two sig figs, 350.0 has four, \(K_f\) has three, \(-1.8\) has two. So when multiplying/dividing, the result should have two sig figs? Wait, no: the formula \(\Delta T_f=K_f\times m\times i\). \(m=\frac{n}{kg\ solvent}\), \(n=\frac{mass}{molar\ mass}\). The mass of urea is 8.7 g (two sig figs), molar mass of urea is 60.06 g/mol (four sig figs), mass of solvent is 350.0 g (four sig figs, so 0.3500 kg). So \(m=\frac{8.7\ g}{60.06\ g/mol\times0.3500\ kg}=\frac{8.7}{60.06\times0.3500}\ mol/kg\). \(60.06\times0.3500 = 21.021\), \(\frac{8.7}{21.021}\approx0.4138\ mol/kg\) (three sig figs from 8.7? Wait 8.7 is two sig figs. So \(m\) should have two sig figs: \(0.41\ mol/kg\)? Wait, no, the rule is that when multiplying or dividing, the number of significant figures in the result is equal to the number of significant figures in the least precise measurement. 8.7 has two, 60.06 has four, 0.3500 has four. So the least is two, so \(m = 0.41\ mol/kg\) (rounded to two sig figs). Then \(\Delta T_f=7.11\ ^{\circ}C\cdot kg/mol\times0.41\ mol/kg\times1 = 2.9151\ ^{\circ}C\approx2.9\ ^{\circ}C\) (two sig figs). Then \(T_{f}^0=T_f+\Delta T_f=-1.8\ ^{\circ}C + 2.9\ ^{\circ}C = 1.1\ ^{\circ}C\). But if we consider that 350.0 has four sig figs, 8.7 has two, \(K_f\) has three, \(-1.8\) has two. Let's do the calculation without rounding intermediate steps:

\(n=\frac{8.7}{60.055}=0.14486\)

\(m=\frac{0.14486}{0.3500}=0.41389\)

\(\Delta T_f = 7.11\times0.41389\times1=2.943\)

\(T_{f}^0=-1.8 + 2.943 = 1.143\approx1.1\ ^{\circ}C\) (if we take two sig figs from - 1.8 and 8.7) or 1.14 \(^{\circ}C\) (if we take three sig figs from \(K_f\) and 350.0). The problem says "Be sure your answer has the correct number of significant digits." Let's check the given values:

- Mass of urea: 8.7 g (2 significant digits)
- Mass of solvent: 350.0 g (4 significant digits)
- \(K_f\): 7.11 \(^{\circ}C\cdot kg/mol\) (3 significant digits)
- Freezing point of solution: - 1.8 \(^{\circ}C\) (2 significant digits)

When calculating \(\Delta T_f=K_f\times m\times i\), \(m=\frac{n}{kg\ solvent}\), \(n=\frac{mass\ of\ urea}{molar\ mass\ of\ urea}\). The molar mass of urea is a constant with more than enough significant digits. So the limiting factors for significant digits are the mass of urea (2), \(K_f\) (3), mass of solvent (4), and freezing point of solution (2). The least number of significant digits in the values that affect the final answer (mass of urea and freezing point of solution) is 2. But let's see, the formula is \(T_{f}^0=T_f+\Delta T_f\), and \(\Delta T_f = K_f\times\frac{\frac{mass\ of\ urea}{molar\ mass\ of\ urea}}{mass\ of\ solvent\ (kg)}\times i\).

Let's calculate \(\Delta T_f\) precisely:

\(mass\ of\ urea = 8.7\ g = 0.0087\ kg\) (wait no, mass of solvent is 350.0 g = 0.3500 kg, mass of solute is 8.7 g = 0.0087 kg, but molality is moles of solute per kg of solvent.

Moles of urea: \(n=\frac{8.7\ g}{60.06\ g/mol}=0.144855\ mol\)

Molality \(m=\frac{0.144855\ mol}{0.3500\ kg}=0.41387\ mol/kg\)

\(\Delta T_f=7.11\ ^{\circ}C\cdot kg/mol\times0.41387\ mol/kg\times1 = 7.11\times0.41387=2.943\ ^{\circ}C\)

Now, \(T_{f}^0=T_f+\Delta T_f=-1.8\ ^{\circ}C + 2.943\ ^{\circ}C = 1.143\ ^{\circ}C\). Now, considering the significant digits: the freezing point of the solution is - 1.8 (two sig figs), the mass of urea is 8.7 (two sig figs), \(K_f\) is 7.11 (three sig figs), mass of solvent is 350.0 (four sig figs). The rule for addition/subtraction: the number of decimal places in the result is equal to the number of decimal places in the least precise measurement. \(-1.8\) has one decimal place, \(2.943\) has three. So when we add them, the result should have one decimal place? Wait, no: \(-1.8 + 2.943=1.143\). The number with the least number of decimal places is - 1.8 (one decimal place), so we round to one decimal place: \(1.1\ ^{\circ}C\). But if we consider that the mass of urea is two sig figs, \(K_f\) three, solvent four, solution freezing point two. Maybe the answer is expected to be around \(1.1\ ^{\circ}C\) or \(1.14\ ^{\circ}C\). Wait, maybe I made a mistake in the formula. The freezing point depression is \(\Delta T_f=T_{f}^0 - T_f\), so \(T_{f}^0=T_f+\Delta T_f\). Yes, that's correct because the solution freezes at a lower temperature than the pure solvent, so the pure solvent's freezing point is higher than the solution's.

Wait, let's check the calculation again. Maybe the mass of solvent is 350 g (not 350.0 g)? The problem says "350. g of X", maybe the decimal is a typo? If it's 350 g (three sig figs), then:

\(m=\frac{0.144855\ mol}{0.350\ kg}=0.4139\ mol/kg\)

\(\Delta T_f=7.11\times0.4139 = 2.943\)

\(T_{f}^0=-1.8 + 2.943 = 1.143\approx1.1\ ^{\circ}C\) (if two sig figs) or \(1.14\ ^{\circ}C\) (if three sig figs from \(K_f\) and 350 g (three sig figs)).

Answer:

\(1.1\ ^{\circ}C\) (or \(1.14\ ^{\circ}C\) depending on significant figure considerations, but likely \(1.1\ ^{\circ}C\) or \(1.1\) (with correct sig figs))