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11 June, 14:14

The concentration of the stock solution she needs is 100 milli molar (mm) and she needs to make 1.2 milli liters (ml) solution of the drug-a. the drug is available in a salt form with a molecular weight of 181.6 grams / mole. what is the amount (quantity in grams) of drug-a will she have to weigh-out in order to make the stock solution

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  1. 11 June, 15:54
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    Molarity is defined as the number of moles of solute in 1 L of solution

    molarity of stock solution to be prepared - 100 x 10⁻³ mol/L

    volume of stock solution to be prepared - 1.2 mL

    Therefore number of moles in 1.2 mL - 100 x 10⁻³ mol/L x 1.2 x 10⁻³ L

    number of moles of drug - 1.2 x 10⁻⁴ mol

    mass of drug required - 1.2 x 10⁻⁴ mol x 181.6 g/mol = 21. 8 mg

    21.8 g of drug is required to make the stock solution
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