Question

1.23 m sugar (c₁₂h₂₂o₁₁) solution (density of solution = 1.12 \\(\frac{g}{ml}\\)). be sure your answer has the correct number of significant digits.

Explanation:

Step1: Determine the molar mass of sucrose ($\ce{C_{12}H_{22}O_{11}}$)

The molar mass of $\ce{C}$ is $12.01\ \text{g/mol}$, $\ce{H}$ is $1.008\ \text{g/mol}$, and $\ce{O}$ is $16.00\ \text{g/mol}$.
Molar mass of $\ce{C_{12}H_{22}O_{11}}$ = $12\times12.01 + 22\times1.008 + 11\times16.00$
= $144.12 + 22.176 + 176.00$
= $342.296\ \text{g/mol}$

Step2: Assume we have 1 L of the solution

Molarity ($M$) = 1.23 M, so moles of sucrose = $1.23\ \text{mol}$ (since $M = \frac{n}{V}$, $n = M\times V$ and $V = 1\ \text{L}$)

Step3: Calculate the mass of sucrose

Mass of sucrose ($m_{\text{sucrose}}$) = moles $\times$ molar mass = $1.23\ \text{mol} \times 342.296\ \text{g/mol}$ ≈ $421.024\ \text{g}$

Step4: Calculate the mass of the solution

Density of solution ($
ho$) = $1.12\ \text{g/mL}$, volume of solution ($V$) = $1\ \text{L} = 1000\ \text{mL}$
Mass of solution ($m_{\text{solution}}$) = $
ho\times V$ = $1.12\ \text{g/mL} \times 1000\ \text{mL}$ = $1120\ \text{g}$

Step5: Calculate the mass of solvent (water)

Mass of solvent ($m_{\text{solvent}}$) = mass of solution - mass of sucrose = $1120\ \text{g} - 421.024\ \text{g}$ ≈ $698.976\ \text{g} = 0.698976\ \text{kg}$

Step6: Calculate molality ($m$)

Molality ($m$) = $\frac{\text{moles of solute}}{\text{mass of solvent (in kg)}}$ = $\frac{1.23\ \text{mol}}{0.698976\ \text{kg}}$ ≈ $1.76\ \text{m}$ (Wait, but the given value is 1.76 m, maybe the question is to find molality from molarity? Wait, maybe I misread. Wait, the problem says "1.23 M sugar solution... density 1.12 g/mL. Be sure your answer has the correct number of significant digits." Wait, maybe the question is to calculate molality from molarity. Wait, let's recheck.

Wait, maybe the initial calculation was wrong. Let's do it properly.

Wait, let's start over. Let's find molality from molarity.

1. Molarity ($M$) = 1.23 mol/L (so in 1 L solution, 1.23 mol solute)
2. Mass of solution: $1\ \text{L} = 1000\ \text{mL}$, density = 1.12 g/mL, so mass = $1000\ \text{mL} \times 1.12\ \text{g/mL} = 1120\ \text{g}$
3. Mass of solute (sucrose): $1.23\ \text{mol} \times 342.3\ \text{g/mol} = 421\ \text{g}$ (approx)
4. Mass of solvent: $1120\ \text{g} - 421\ \text{g} = 699\ \text{g} = 0.699\ \text{kg}$
5. Molality ($m$) = $\frac{\text{moles of solute}}{\text{kg of solvent}} = \frac{1.23\ \text{mol}}{0.699\ \text{kg}} \approx 1.76\ \text{m}$

Wait, but the problem shows a calculation with 1.76 m, maybe the question is to confirm or calculate molality. But the given value is 1.76 m, which matches the calculation. So the answer is 1.76 m (with correct significant digits, 1.23 has 3, 1.12 has 3, so 1.76 has 3).

Answer:

$\boxed{1.76\ \text{m}}$