Lead + Nitric Acid discplacment

When 2.69 g 2.69 g of a nonelectrolyte solute is dissolved in water to make 345 mL 345 mL of solution at 26 °C, 26 °C, the solut

Gre4nikov [31]

Answer:

The molar concentration of this solution is 0.0463 mol/L

Explanation:

Step 1 : Data given

Mass of a nonelectrolyte solute = 2.69 grams

Volume of water = 345 mL = 0.345 L

Temperature = 26.0°CC = 273 + 26 = 299 K

The osmotic pressure = 863 torr

⇒ 863torr /760 = 1.13553 atm

Step 2: Calculate the molar concentration of this solution

Π = i*M*R*T

⇒with Π = the osmotic pressure = 1.13553 atm

⇒with i = the van't Hoff factor of the nonelectrolyte solute = 1

⇒with M = the molar concentration = TO BE DETERMINED

⇒with R = the gas constant = 0.08206 L*atm/mol*K

⇒with T = the temperature = 299 K

1.13553 atm = 1 * M * 0.08206 L*atm/mol*K * 299 K

M = 1.13553 / (0.08206*299)

M = 0.0463 mol/L

The molar concentration of this solution is 0.0463 mol/L

5 0

3 years ago

A) Calculate the average density of the following object (assume it is a perfect sphere). SHOW ALL YOUR WORK (formulas used, num

-BARSIC- [3]

Answers:

A) 2040 kg/m³; B) 58 600 km

Explanation:

A) Density

$V = \frac{ 4}{3 }\pi r^{3} = \frac{ 4}{3 }\pi\times (\text{1150 km})^{3} = 6.37 \times 10^{9} \text{ km}^{3}$

$\text{Density} = \frac{\text{mass}}{\text{volume}} = \frac{1.3\times 10^{22} \text{ kg} }{6.37 \times 10^{9} \text{ km}^{3}}\times (\frac{\text{1 km}}{\text{1000 m}})^{3} = \text{2040 kg/m}^{3}$

<em>B) Radius</em>

$\text{Volume} = \frac{\text{mass}}{\text{density}} = \frac{5.68\times 10^{26} \text{ kg} }{687 \text{ kg/m}^{3} }= 8.268 \times 10^{23} \text{ m}^{3}$

$V = \frac{ 4}{3 }\pi r^{3}$

$r^{3} = \frac{3V }{4 \pi }\$

$r= \sqrt [3]{ \frac{3V }{4 \pi } }$

$r= \sqrt [3]{ \frac{3\times 8.268 \times 10^{23} \text{ m}^{3}}{4 \pi } }= \sqrt [3]{ 1.974 \times 10^{23} \text{ m}^{3}}= 5.82 \times 10^{7} \text{ m}=\text{58 200 km}$