Answer:
V = 15.6 L
Explanation:
Hello there!
In this case, according to the definition of molarity in terms of the moles of solute divided by the volume of the solution, it is possible for us to write:
![M=n/V](https://tex.z-dn.net/?f=M%3Dn%2FV)
Thus, given the moles and concentration of the solution, we can find the volume as shown below:
V=n/M
Therefore, we plug in the given data to obtain:
V = 2.5mol /(0.16mol/L)
V = 15.6 L
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Answer:
the volume of a set of keys is 6.75ml
Explanation:
Plant cells have several structures not found in other eukaryotes. In particular, organelles called chloroplasts allow plants to capture the energy of the Sun in energy-rich molecules; cell walls allow plants to have rigid structures as varied as wood trunks and supple leaves; and vacuoles allow plant cells to change size.
<u>Answer:</u> The molar mass of the unknown protein is 6387.9 g/mol
<u>Explanation:</u>
To calculate the concentration of solute, we use the equation for osmotic pressure, which is:
![\pi=iMRT](https://tex.z-dn.net/?f=%5Cpi%3DiMRT)
or,
![\pi=i\times \frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}\times RT](https://tex.z-dn.net/?f=%5Cpi%3Di%5Ctimes%20%5Cfrac%7B%5Ctext%7BMass%20of%20solute%7D%5Ctimes%201000%7D%7B%5Ctext%7BMolar%20mass%20of%20solute%7D%5Ctimes%20%5Ctext%7BVolume%20of%20solution%20%28in%20mL%29%7D%7D%5Ctimes%20RT)
where,
= osmotic pressure of the solution = 0.0766 atm
i = Van't hoff factor = 1 (for non-electrolytes)
Mass of protein = 100. mg = 0.100 g (Conversion factor: 1 g = 1000 mg)
Molar mass of protein = ?
Volume of solution = 5.00 mL
R = Gas constant = ![0.0821\text{ L atm }mol^{-1}K^{-1}](https://tex.z-dn.net/?f=0.0821%5Ctext%7B%20L%20atm%20%7Dmol%5E%7B-1%7DK%5E%7B-1%7D)
T = temperature of the solution = ![25^oC=[25+273]K=298K](https://tex.z-dn.net/?f=25%5EoC%3D%5B25%2B273%5DK%3D298K)
Putting values in above equation, we get:
![0.0766atm=1\times \frac{0.100\times 1000}{\text{Molar mass of protein}\times 5}\times 0.0821\text{ L. atm }mol^{-1}K^{-1}\times 298K\\\\\pi=\frac{1\times 0.100\times 1000\times 0.0821\times 298}{0.0766\times 5}=6387.9g/mol](https://tex.z-dn.net/?f=0.0766atm%3D1%5Ctimes%20%5Cfrac%7B0.100%5Ctimes%201000%7D%7B%5Ctext%7BMolar%20mass%20of%20protein%7D%5Ctimes%205%7D%5Ctimes%200.0821%5Ctext%7B%20L.%20atm%20%7Dmol%5E%7B-1%7DK%5E%7B-1%7D%5Ctimes%20298K%5C%5C%5C%5C%5Cpi%3D%5Cfrac%7B1%5Ctimes%200.100%5Ctimes%201000%5Ctimes%200.0821%5Ctimes%20298%7D%7B0.0766%5Ctimes%205%7D%3D6387.9g%2Fmol)
Hence, the molar mass of the unknown protein is 6387.9 g/mol