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How much silver nitrate should be used in preparing 50 mL of a solution such that 5 mL diluted to 500 mL will yield a 1 : 1000 solution?
1 : 1000 means 1 g of silver nitrate in 1000 mL of solution
1000 mL = 1 g 500 mL X g
X = 0.5 g of silver nitrate in 500 mL of diluted solution (1 : 1000), which is also the amount in 5 mL of the stronger (stock) solution, since the 50 mL and the 5 mL are the same strength.
5mL/50ml = 0.5g/Xg
X = 5 g, answer
How much (in milliliters) of a 17% w/v concentrate of benzalkonium chloride should be used in preparing 300 mL of a stock solution such that 15 mL diluted to 1 L will yield a 1 : 5000 solution?
1 L = 1000 mL
1 : 5000 means 1 g of benzalkonium chloride in 5000 mL of solution.
X = 0.2 g of benzalkonium chloride in 1000 mL of diluted solution (1 : 5000), which is also the amount in 15 mL of the stronger (stock) solution to be prepared, because the amount of drug has not changed but simply been diluted.
X = 4 g of benzalkonium chloride.
A 17% w/v concentrate contains 17 g/100 mL. Therefore:
17 g = 100 mL 4g XmL
X = 23.5 mL, answer
5mL =0.5g 50mL Xg
X = 5 g, answer
How much silver nitrate should be used in preparing 50 mL of a solution such that 5 mL diluted to 500 mL will yield a 1 : 1000 solution?
1 : 1000 means 1 g of silver nitrate in 1000 mL of solution
1000 mL = 1 g 500 mL X g
X = 0.5 g of silver nitrate in 500 mL of diluted solution (1 : 1000), which is also the amount in 5 mL of the stronger (stock) solution, since the 50 mL and the 5 mL are the same strength.
5mL/50ml = 0.5g/Xg
X = 5 g, answer
How much (in milliliters) of a 17% w/v concentrate of benzalkonium chloride should be used in preparing 300 mL of a stock solution such that 15 mL diluted to 1 L will yield a 1 : 5000 solution?
1 L = 1000 mL
1 : 5000 means 1 g of benzalkonium chloride in 5000 mL of solution.
X = 0.2 g of benzalkonium chloride in 1000 mL of diluted solution (1 : 5000), which is also the amount in 15 mL of the stronger (stock) solution to be prepared, because the amount of drug has not changed but simply been diluted.
X = 4 g of benzalkonium chloride.
A 17% w/v concentrate contains 17 g/100 mL. Therefore:
17 g = 100 mL 4g XmL
X = 23.5 mL, answer
5mL =0.5g 50mL Xg
X = 5 g, answer
Stock solutions contain a standardized concentration of the solute in a solution. To prepare 50 mL of solution with a ratio of 1:1000 solution, 5 gms of the drug will be required.
What is dilution?
Dilution is the property of the solution with a less amount of solute concentration than the stock solution. A dilute solution is prepared by the addition of more solvent.
The addition of the solvent in a solution decreases the molar concentration which leads to dilution.
The ratio of 1:1000 means that 1 gm of the drug is present in 1000 mL (1L) of solution.
1000 mL = 1 gm × 500 mL × X gm
Here, the X represents the 0.5 gm of drug dissolved in 500 mL.
The amount of drug used is calculated as,
5 mL ÷ 50 ml = 0.5 g ÷ X g
X = 0.5 × 50 ÷ 5
= 5 gm
Therefore, 5 gm of the drug is required to prepare 50 mL of solution.
Learn more about dilution here:
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