Question

An experiment requires 0.025 l of a 0.5 m glucose solution. you have 84 ml of a 20 % (m/v) glucose stock solution. how would you prepare the required concentration?

Answer 1

Concentration of the given stock solution of glucose = 20% (m/v)

It means that 20 g glucose is present in 100 mL solution

Volume of the 20 % solution given is 84 mL

Moles of glucose =
`20 g C_(6)H_(12)O_(6)*(1mol C_(6)H_(12)O_(6))/(180.16gC_(6)H_(12)O_(6)) =0.111 molC_(6)H_(12)O_(6)`

Moles of glucose in 84 mL solution:

`84mL*(0.111 mol C_(6)H_(12)O_(6)   )/(100mL) =0.0933 molC_(6)H_(12)O_(6)`

Calculating molarity of 84 mL of 20 % glucose solution:

`(0.0933mol C_(6)H_(12)O_(6))/(84mL)*(1000mL)/(1L)=1.11 M`

Finding the volume of this solution required to prepare 0.0250L of 0.5 M glucose solution:

`M_(1)V_(1)=M_(2)V_(2)`

`1.11 M(V_(1))=0.5 M(0.0250 L)`

`V_(1)= 0.01126 L or 11.26 mL`