Question

How would you prepare a solution containing 50mM NaCl (mwt = 58.5g/mol) and 10mM Na2HPO4 (358.057 g/mol) in a final volume of 250ml?

Answer 1

Solution

1mM= 0.001M

$Molarity=\frac{50mM\times{0.001}}{1mM}=0.05M$

$Moles of NaCl=\frac{250ml\times{0.05mol}}{1000ml}= 0.0125 moles$

$Mass of NaCl = 58.44g/mol\times0.0125= 0.7 g$$Molarity of Na_2HP0_4$

$\frac{10mM\times{0.001M}}{1mM}=0.01M$

$Moles= \frac{250\times{0.01}}{1000}= 0.025moles$

$Mass of Na_2HPO_4= 358.057\times 0.025 moles = 8.95 g$