In the formula for water (H2O) what does the subscript 2 indicate?
1 answer:
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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
<u>Answer:</u> The molality of non-electrolyte is 24.69 m
<u>Explanation:</u>
We are given:
Mole fraction of saturated aqueous solution = 0.310
This means that 0.310 moles of non-electrolyte is present.
Moles of water (solvent) = 1 - 0.310 = 0.690 moles
To calculate the mass from given number of moles, we use the equation:
Moles of water = 0.690 moles
Molar mass of water = 18 g/mol
Putting values in above equation, we get:
To calculate the molality of solution, we use the equation:
Where,
= Moles of solute (non-electrolyte) = 0.310 moles
= Mass of solvent (water) = 12.42 g
Putting values in above equation, we get:
Hence, the molality of non-electrolyte is 24.69 m