The molarity of the solutions are as follows:
- solution B has the highest molarity
- solutions A, D and F have the same molarity
- solutions A and C are mixed together have a lower molarity than B
- solution F and D will have the same molarity
- Volume of water required to be evaporated is 8.3 mL
<h3>What is molarity of a solution?</h3>
The molarity of a solution is the amount in moles of a substance present in a given volume of solution.
From the image of the solution given:
- solution B has the highest molarity
- solutions A, D and F have the same molarity
- when solutions A and C are mixed, the resulting solution have a lower molarity than B
- solution F and D will have the same molarity after 75 mL and 50 mL of water are added to each respectively
- the molarity of B is 12/50 = 4/16.7. Volume of water required to be evaporated = 25 - 16.7 = 8.3 mL
Therefore, the molarity of the solutions depends on the moles of substance present per given volume of solution.
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Answer:
i dont know what you're asking here. But if you're asking what it is then:
Explanation:
Aluminum Hydroxide + Hydrogen Chloride = Aluminum Chloride + Water
Al and Oh3: Aluminum Hydroxide
H and Cl: Hydrogen Chloride
Al and Cl3: Aluminum Chloride
H2O: water
btw the first letter of every element should be capitalized. While the second and theird are lowercased.
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We know that:
Molar Mass H2O: 18 g/mol
<span>Molar Mass of Eugenol: 164 g/mol </span>
<span>Boiling point of H2O: 100 degrees C </span>
<span>Boiling point of Eugenol: 254 degrees C </span>
<span>Density of water: 1.0 g/mL </span>
<span>Density of Eugenol: 1.05 g/mL </span>
<span>Using formula:
V= [mole fraction x molar mass] / density </span>
<span>mH20: 0.9947 * 18
= 17.9046 / 1 g/mL
= 17.9046 </span>
<span>morg: 0.0053 * 164
= 0.8692/ 1.05 g/mL
= 0.8278 </span>
<span>V% = Vorg/(Vorg + VH2O) * 100 </span>
<span>(0.8278/18.7324) * 100 = 4.419% </span>
Yotal volume = 30 mL; therefore,
<span>0.0442 = (volume eugenol/30) </span>
<span>(m eug/mH2O) = (peug*164/pH2O*18) </span>
<span>(m eug/30) = (4*164/760*18) </span>
<span>m eug = about 1.44g and </span>
<span>
volume = mass/density
= 1.44/1.05
= about 1.37 mL </span>
Here is the complete question.
Benzalkonium Chloride Solution ------------> 250ml
Make solution such that when 10ml is diluted to a total volume of 1 liter a 1:200 is produced.
Sig: Dilute 10ml to a liter and apply to affected area twice daily
How many milliliters of a 17% benzalkonium chloride stock solution would be needed to prepare a liter of a 1:200 solution of benzalkonium chloride?
(A) 1700 mL
(B) 29.4 mL
(C) 17 mL
(D) 294 mL
Answer:
(B) 29.4 mL
Explanation:
1 L = 1000 mL
1:200 solution implies the 
in 200 mL solution.
200 mL of solution = 1g of Benzalkonium chloride
1000 mL will be 
200mL × 1g = 1000 mL × x(g)
x(g) = 
x(g) = 0.2 g
That is to say, 0.2 g of benzalkonium chloride in 1000mL of diluted solution of 1;200 is also the amount in 10mL of the stock solution to be prepared.
∴ 
y(g) = 
y(g) = 5g of benzalkonium chloride.
Now, at 17% concentrate contains 17g/100ml:
∴ the number of milliliters of a 17% benzalkonium chloride stock solution that is needed to prepare a liter of a 1:200 solution of benzalkonium chloride will be;
= 
z(mL) = 
z(mL) = 29.41176 mL
≅ 29.4 mL
Therefore, there are 29.4 mL of a 17% benzalkonium chloride stock solution that is required to prepare a liter of a 1:200 solution of benzalkonium chloride
