Question

a certain solution of salt water has a molality of 8.12 m. how many moles of solute are in 1 kg of water? mol how many moles of solute are in 5 kg of water? mol submit question question 24 0/1 pt 5 19 details calculate the molality of a sucrose solution made by dissolving 73.5 grams sucrose into 98.3 grams water. the molar mass of sucrose is 342.2 g/mol. 0.22 747.71 0.02 2.19 0 submit question question 25 0/3 pts 5 19 details what is the molarity of a solution that contains 129 g of nacl in 15 l of solution? m

Explanation:

First Part (Moles in 1 kg and 5 kg water)

Step1: Recall molality formula

Molality ($m$) is defined as moles of solute ($n$) per kilogram of solvent ($kg$), so $m=\frac{n}{kg}$. To find moles, rearrange to $n = m\times kg$.

Step2: Moles in 1 kg water

Given $m = 8.12\ m$, $kg = 1\ kg$. Substitute into $n = m\times kg$: $n=8.12\ mol/kg\times1\ kg = 8.12\ mol$.

Step3: Moles in 5 kg water

Given $m = 8.12\ m$, $kg = 5\ kg$. Substitute into $n = m\times kg$: $n=8.12\ mol/kg\times5\ kg = 40.6\ mol$.

Step1: Calculate moles of sucrose

Moles ($n$) = mass ($m$) / molar mass ($M$). Given $m = 73.5\ g$, $M = 342.2\ g/mol$. So $n=\frac{73.5\ g}{342.2\ g/mol}\approx0.2148\ mol$.

Step2: Convert water mass to kg

Mass of water is $98.3\ g = 0.0983\ kg$.

Step3: Calculate molality

Molality ($m$) = moles of solute / kg of solvent. So $m=\frac{0.2148\ mol}{0.0983\ kg}\approx2.19\ m$.

Step1: Calculate moles of NaCl

Molar mass of NaCl is $58.44\ g/mol$ (Na:22.99, Cl:35.45). Moles ($n$) = mass / molar mass: $n=\frac{129\ g}{58.44\ g/mol}\approx2.207\ mol$.

Step2: Recall molarity formula

Molarity ($M$) = moles of solute / liters of solution. Solution volume is $15\ L$.

Step3: Calculate molarity

$M=\frac{2.207\ mol}{15\ L}\approx0.147\ M$ (approximate, but following steps: $n=\frac{129}{58.44}\approx2.207$, $M=\frac{2.207}{15}\approx0.147$).

Answer:

For 1 kg: $\boldsymbol{8.12}$ mol
For 5 kg: $\boldsymbol{40.6}$ mol

Question 24 (Molality of sucrose solution)