Question

a certain substance x has a normal freezing point of 0.2 °c and a molal freezing point depression constant ( k_f = 5.86 , ^circ\text{c·kg mol}^{-1} ). calculate the freezing point of a solution made of 11.93 g of urea ( left( (\text{nh}_2)_2\text{co}
ight) ) dissolved in 700. g of x.
be sure your answer has the correct number of significant digits.

Explanation:

Step1: Calculate moles of urea

Molar mass of urea \((\text{NH}_2)_2\text{CO}\): \(2\times(14 + 2\times1)+12 + 16=60.06\space g/mol\)
Moles of urea: \(\frac{11.93\space g}{60.06\space g/mol}\approx0.1986\space mol\)

Step2: Calculate molality (m)

Mass of solvent \(X\) in kg: \(700. \space g = 0.700\space kg\)
Molality \(m=\frac{\text{moles of solute}}{\text{kg of solvent}}=\frac{0.1986\space mol}{0.700\space kg}\approx0.2837\space mol/kg\)

Step3: Calculate freezing point depression (\(\Delta T_f\))

Using \(\Delta T_f = K_f\times m\times i\) (urea is non - electrolyte, \(i = 1\))
\(\Delta T_f=5.86\space^{\circ}\text{C}\cdot\text{kg}\cdot\text{mol}^{-1}\times0.2837\space mol/kg\times1\approx1.663\space^{\circ}\text{C}\)

Step4: Calculate freezing point of solution

Normal freezing point \(T_f^0 = 0.2\space^{\circ}\text{C}\)
Freezing point of solution \(T_f=T_f^0-\Delta T_f = 0.2\space^{\circ}\text{C}- 1.663\space^{\circ}\text{C}=- 1.463\space^{\circ}\text{C}\) (Rounding to correct significant digits, considering the given values: 11.93 (4 sig figs), 700. (3 sig figs), \(K_f = 5.86\) (3 sig figs), \(T_f^0 = 0.2\) (1 sig fig? Wait, 0.2 could be 1 or maybe 1 if it's written as 0.2. Wait, 11.93 is 4, 700. is 3, \(K_f\) is 3, \(T_f^0\) is 0.2 (1 decimal place, 1 sig fig? No, 0.2 has 1 significant figure? Wait, no, 0.2 is one significant figure? Wait, maybe the 0.2 is 0.20? No, the problem says 0.2 \(^{\circ}\text{C}\). Wait, but 11.93 is 4, 700. is 3 (the decimal after 700 means it's 3 sig figs), \(K_f\) is 3. So when we calculate, the least number of sig figs in multiplication/division is 3 (from 700. and \(K_f\)), and in subtraction, the decimal places: \(T_f^0\) has 1 decimal place, \(\Delta T_f\) has 3 decimal places (1.663), so the result should have 1 decimal place? Wait, no, let's recalculate with more precision.

Wait, moles of urea: \(11.93\div60.06 = 0.1986347\space mol\)
Molality: \(0.1986347\div0.700 = 0.2837639\space mol/kg\)
\(\Delta T_f=5.86\times0.2837639\times1 = 5.86\times0.2837639 = 1.662866\space^{\circ}\text{C}\)
\(T_f = 0.2-1.662866=- 1.462866\space^{\circ}\text{C}\)

Now, considering significant figures:
- Mass of urea: 11.93 g (4 sig figs)
- Mass of solvent: 700. g (3 sig figs)
- \(K_f\): 5.86 (3 sig figs)
- Freezing point of pure solvent: 0.2 \(^{\circ}\text{C}\) (1 sig fig? Or maybe it's 0.20? The problem says 0.2, so 1 decimal place, 1 significant figure? But that seems odd. Wait, maybe the 0.2 is 0.20 \(^{\circ}\text{C}\) (2 sig figs). But the problem says "0.2 \(^{\circ}\text{C}\)". Alternatively, maybe the answer should be rounded to three significant figures. Let's check the calculation again.

Wait, the formula for freezing point depression is \(\Delta T_f=K_f\times m\), where \(m\) is molality (moles of solute per kg of solvent). Urea is a non - electrolyte, so \(i = 1\).

Molar mass of urea: \(C = 12.01, O = 16.00, N = 14.01, H = 1.008\)
\((\text{NH}_2)_2\text{CO}=2\times(14.01 + 2\times1.008)+12.01+16.00=2\times(14.01 + 2.016)+28.01=2\times16.026 + 28.01 = 32.052+28.01 = 60.062\space g/mol\approx60.06\space g/mol\)

Moles of urea: \(n=\frac{11.93\space g}{60.06\space g/mol}=0.1986\space mol\) (as before)

Mass of solvent in kg: \(700\space g = 0.700\space kg\) (3 sig figs)

Molality \(m=\frac{0.1986\space mol}{0.700\space kg}=0.2837\space mol/kg\) (3 sig figs from 0.700)

\(\Delta T_f=K_f\times m=5.86\space^{\circ}\text{C}\cdot\text{kg}\cdot\text{mol}^{-1}\times0.2837\space mol/kg = 1.663\space^{\circ}\text{C}\) (3 sig figs from \(K_f\) and \(m\))

Freezing point of solution: \(T_f=T_f^0-\Delta T_f=0.2\space^{\circ}\text{C}-1.663\space^{\circ}\text{C}=- 1.463\space^{\circ}\text{C}\)

Now, the initial freezing point is 0.2 (1 decimal place, 1 sig fig? But 11.93 is 4, 700. is 3, \(K_f\) is 3. Maybe the 0.2 is considered as 0.20 (2 sig figs). But if we take the least number of decimal places in subtraction: 0.2 has 1 decimal place, 1.663 has 3 decimal places. So the result should have 1 decimal place. But - 1.463 rounded to 1 decimal place is - 1.5? Wait, no, 0.2 is 0.2 (one significant figure? No, 0.2 has one significant figure? Wait, no, 0.2 is one significant figure, 0.20 is two. The problem says 0.2 \(^{\circ}\text{C}\), so one significant figure? But that seems inconsistent with the other values. Alternatively, maybe the problem expects us to use the given values as is, without worrying too much about sig figs, and just calculate the value. Let's see:

\(T_f = 0.2-5.86\times\frac{11.93}{60.06\times0.700}\)

Calculate \(\frac{11.93}{60.06\times0.700}=\frac{11.93}{42.042}\approx0.2837\)

\(5.86\times0.2837\approx1.663\)

\(0.2 - 1.663=-1.463\approx - 1.5\) (if 0.2 is one sig fig) or - 1.46 (if 0.2 is two sig figs, but 0.2 is one). Wait, maybe the 0.2 is 0.20 (two sig figs). Let's check the problem statement again: "normal freezing point of 0.2 \(^{\circ}\text{C}\)". So it's 0.2, one decimal place, one significant figure? No, significant figures: leading zeros don't count, so 0.2 has one significant figure. But 11.93 has four, 700. has three, \(K_f\) has three. This is a bit confusing. But maybe the intended answer is around - 1.5 \(^{\circ}\text{C}\) or - 1.46 \(^{\circ}\text{C}\). Let's do the calculation precisely:

\(n=\frac{11.93}{60.055}\) (more accurate molar mass of urea: \(2\times(14.0067 + 2\times1.00794)+12.0107+16.00=2\times(14.0067 + 2.01588)+28.0107=2\times16.02258+28.0107 = 32.04516+28.0107 = 60.05586\space g/mol\))

\(n=\frac{11.93}{60.05586}\approx0.1986\space mol\)

\(m=\frac{0.1986}{0.700}=0.2837\space mol/kg\)

\(\Delta T_f=5.86\times0.2837 = 1.662582\)

\(T_f=0.2 - 1.662582=-1.462582\approx - 1.46\space^{\circ}\text{C}\) (if we consider 0.2 as 0.20, two sig figs, but 0.2 is one. Alternatively, maybe the problem expects three sig figs because 11.93 (4), 700. (3), \(K_f\) (3) – the least is 3, so the answer should have 3 sig figs. - 1.46 \(^{\circ}\text{C}\) (three sig figs: 1, 4, 6)).

Answer:

\(-1.46\) (or \(-1.5\) depending on sig fig interpretation, but the precise calculation gives approximately \(-1.46\space^{\circ}\text{C}\))