Question #60376

#"NaOH":#

This one is easy.

The molar mass of #"NaOH"# is 40.00 g/mol.

Simply transfer 40.00 g of #"NaOH"# to a 1 L volumetric flask.

Add enough water to dissolve the #"NaOH"#, then make up to the mark with distilled water.

If you need to know the concentration to more than 1 significant figure, you must standardize the solution by titrating an aliquot with an acid of known concentration.

Lisinopril:

The structure of Lisinopril is

Its molar mass is 405.49 g/mol.

Per the US Pharmacopoeia, Lisinopril tablets must contain between 90 % and 110 % of the labelled amount of #"C"_21"H"_31"N"_3"O"_5#.

Preparing the solution:

Calculate the mass of Lisinopril needed to prepare I L of a 1 nmol/L solution.

For easier writing, let's symbolize Lisinopril as #"Lis"#.

#"Mass of Lis" = 1 color(red)(cancel(color(black)("L solution"))) × (1 ×10^"-9" color(red)(cancel(color(black)("mol Lis"))))/(1 color(red)(cancel(color(black)("L solution")))) × "405.49 g Lis"/(1 color(red)(cancel(color(black)("mol Lis"))))#
#= 4.05 × 10^"-7"color(white)(l) "g Lis"#

So, we want a solution that contains #4.05 × 10^"-7" color(white)(l)"g of Lis"# in 1 L of solution.

It looks like a serial dilution is in order.

How about this?

Step 1: Dissolve 40 mg of #"Lis"# in 100 mL of water.

#"[Lis]" = (40 × 10^"-3" color(white)(l)"g")/"0.100 L" = "0.40 g/L"#

Step 2: Dilute 1 mL of this solution to 1 L.

We can use the formula

#color(blue)(bar(ul(|color(white)(a/a)c_1V_1= c_2V_2color(white)(a/a)|)))" "#

Then #c_2 = c_1 × V_1/V_2#

#"[Lis]" = "0.40 g/L" × (1 color(red)(cancel(color(black)("mL"))))/(1000 color(red)(cancel(color(black)("mL")))) = 4.0 × 10^"-4"color(white)(l) "g/L"#

Step 3: Dilute 1 mL of this solution to 1 L.

#"[Lis]" = 4.0 × 10^"-4" color(white)(l)"g/L" × (1 color(red)(cancel(color(black)("mL"))))/(1000 color(red)(cancel(color(black)("mL")))) = 4.0 × 10^"-7"color(white)(l) "g/L"#

Check:

#"[Lis]" = (4.0 × 10^"-7" color(red)(cancel(color(black)("g Lis"))))/(1 "L") × "1 mol"/(405.4 color(red)(cancel(color(black)("g Lis")))) = 9.9 × 10^"-10" color(white)(l)"mol/L"#
#= 0.99 × 10^"-9" color(white)(l)"mol/L" = "0.99 nmol/L"#

Close enough!

If you need more precision, you will have to analyze the solution according to the methods of the US Pharmacopoeia.