Determine the mass of 10.55 x 1029 molecules of cadmium chloride, CdCl,().
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
Answer : The final molarity of iodide anion in the solution is 0.0508 M.
Explanation :
First we have to calculate the moles of and
.
Molar mass of KI = 166 g/mole
and,
Now we have to calculate the limiting and excess reagent.
The given chemical reaction is:
From the balanced reaction we conclude that
As, 1 mole of KI react with 1 mole of
So, 0.0178 mole of KI react with 0.0178 mole of
From this we conclude that, is an excess reagent because the given moles are greater than the required moles and KI is a limiting reagent and it limits the formation of product.
Now we have to calculate the moles of
From the reaction, we conclude that
As, 1 mole of react to give 1 mole of
So, 0.0178 moles of react to give 0.0178 moles of
Thus,
Moles of AgI = Moles of anion = Moles of
cation = 0.0178 moles
Now we have to calculate the molarity of iodide anion in the solution.
Therefore, the final molarity of iodide anion in the solution is 0.0508 M.