Question

All the questions refer to an experimental drug (RDRPkill), designed to treat COVID-19. The drug is a specific inhibitor of the coronavirus specific RNA-directed RNA-polymerase.

Although it does not dissolve in pure water, RDRPkill is soluble in 10% (w/v) ethanol or pure dimethyl sulfoxide (DMSO). It is supplied in vials that generally contain 5 mg of powder. Liquid is transferred in and out of these vials by puncturing the rubber seal at the top with a syringe.

6.   Because everyone in the world is using lung cell culture medium, there is a shortage. But, no problem, you can make a homemade version using basic components. The final concentration of the components in culture medium are 130 mM sodium chloride, 2.5 mM potassium chloride, 25 mM sodium bicarbonate, and 100 μg/mL BSA. You need to prepare a 10 x concentrate of culture medium. You have the following stock solutions: 5 M sodium chloride, 0.5 M potassium chloride, 2.5 M sodium bicarbonate, and 30% (w/v) BSA. How would you make up 200 mL of this 10 x concentrate?

7. It turns out that RDRP-kill may need to be stabilised in the incubations by a reducing agent. A commonly used reducing agent is β-mercaptoethanol (mol.wt. = 78.3). It is a liquid at room temperature with a density of 1.12 g/mL. You decided that you will make up a 50 mL stock of 0.7% (v/v) β-mercaptoethanol. How would you do this and what is the molar concentration of the stock that you’ve made up?

Answer 1

Answer:

7)

For making 50 mL stock of 0.7% (v/v) β-mercaptoethanol.

We need, 0.7 mL of β-mercaptoethanol for 100 mL solution.

Hence, the amount of β-mercaptoethanol for 50 mL solution is = (50*0.7)/100 = 0.35 mL

The density of β-mercaptoethanol is 1.12 g/mL

Hence, the mass of β-mercaptoethanol needed for 50 mL 0.7% (v/v) is = 1.12*0.35 = 0.392 g of β-mercaptoethanol.

Hence, we can make 50 mL solution of 0.7% (v/v) β-mercaptoethanol by adding 0.35 mL of β-mercaptoethanol (0.392 g) in a 50 mL volumetric flask and filled up to the mark by distilled water.

The molecular weight of the β-mercaptoethanol is 78.3 g/mol

Hence, the concentration of the β-mercaptoethanol solution is = (0.392/78.3)*(1000/50) = 0.10013 (M).

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