Answer:
![\rho_t=1025000 gmL^{-1}](https://tex.z-dn.net/?f=%5Crho_t%3D1025000%20gmL%5E%7B-1%7D)
Explanation:
From the question we are told that:
Density of acetic acid ![\rho_a=205 gL^{-1}](https://tex.z-dn.net/?f=%5Crho_a%3D205%20gL%5E%7B-1%7D)
Density of Water ![\rho_w=820 gL^{-1}](https://tex.z-dn.net/?f=%5Crho_w%3D820%20gL%5E%7B-1%7D)
Generally the equation for Solution Density is mathematically given by
![\rho_t= \rho_w+\rho_a](https://tex.z-dn.net/?f=%5Crho_t%3D%20%5Crho_w%2B%5Crho_a)
![\rho_t=205+820](https://tex.z-dn.net/?f=%5Crho_t%3D205%2B820)
![\rho_t=1025 gL^{-1}](https://tex.z-dn.net/?f=%5Crho_t%3D1025%20gL%5E%7B-1%7D)
![\rho_t=1025000 gmL^{-1}](https://tex.z-dn.net/?f=%5Crho_t%3D1025000%20gmL%5E%7B-1%7D)
Answer: 1.32
Explanation:
Given
85 gm of sulfur dioxide is present at STP
The molar mass of sulfur dioxide is ![32+2\times 16=64\ g/mol](https://tex.z-dn.net/?f=32%2B2%5Ctimes%2016%3D64%5C%20g%2Fmol)
The number of moles of sulfur dioxide is
![\Rightarrow n=\dfrac{85}{64}\\\\\Rightarrow n=1.32\ \text{mol}](https://tex.z-dn.net/?f=%5CRightarrow%20n%3D%5Cdfrac%7B85%7D%7B64%7D%5C%5C%5C%5C%5CRightarrow%20n%3D1.32%5C%20%5Ctext%7Bmol%7D)
1.24973017189471 is probably the answer to your equation
The answer to this question would be: lower molar concentration
Osmotic pressure is influenced by the number of ions and the concentration of the molecule in the solution. In NaCl, the molecule will split into 1 Na+ ion and 1 Cl- ion which results in 2 ions per compound. In MgCl2, the compound will split into 1 Mg2+ ion and 2 Cl- ion which results in 3 ions. Therefore, the osmotic pressure of MgCl2 will be 3/2 times of NaCl.
MgCl2 will need less concentration to achieve same osmotic pressure as NaCl. If the MgCl2 solution is isotonic with NaCl, the concentration of MgCl2 would be lower than NaCl
Answer:
<h3>The answer is 0.91 %</h3>
Explanation:
The percentage error of a certain measurement can be found by using the formula
![P(\%) = \frac{error}{actual \: \: number} \times 100\% \\](https://tex.z-dn.net/?f=P%28%5C%25%29%20%3D%20%20%5Cfrac%7Berror%7D%7Bactual%20%5C%3A%20%20%5C%3A%20number%7D%20%20%5Ctimes%20100%5C%25%20%5C%5C%20)
From the question
actual volume = 198.2 mL
error = 200 - 198.2 = 1.8
So we have
![P(\%) = \frac{1.8}{198.2} \times 100 \\ = 0.90817356205...](https://tex.z-dn.net/?f=P%28%5C%25%29%20%3D%20%20%5Cfrac%7B1.8%7D%7B198.2%7D%20%20%5Ctimes%20100%20%5C%5C%20%20%3D%200.90817356205...)
We have the final answer as
<h3>0.91 %</h3>
Hope this helps you