Biology question - Mgcl2 stock

Question

The MgCl₂ stock we use for PCR is at 25 mM. If the molar mass of magnesium chloride is 95.211 g/mol, how many grams of MgCl₂  would you need to prepare 10. mL of this 25 mM stock?

A. 0.0024 g of MgCl2

B. 0.24 g of MgCl2

C. 2.4 g of MgCl2

D. 0.024 g of MgCl2

Thank you.

Answer 1

Hi Jas A.!

We just need to use dimensional analysis to convert from volume into mass using the concentration as our bridge:

(10. mL MgCl₂)(25 mmol MgCl₂/1000 mL MgCl₂)(1 mol MgCl₂/1000 mmol MgCl₂)(95.211 g MgCl₂/1 mol MgCl₂)

mass = 0.024 g MgCl₂

Hope this helps!

Cheers

Answer 2

D. 0.024 g of MgCl2

Working: Molarity (M) = moles of solute/litres of solution Moles of MgCl2 = Molarity * Litres of solution = 25 mM * 0.010 L = 0.00025 moles Grams of MgCl2 = moles of MgCl2 * molar mass of MgCl2 = 0.00025 moles * 95.211 g/mol = 0.024 g MgCl2

Hope this will help.