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 = [tex]20 g C_{6}H_{12}O_{6}*\frac{1mol C_{6}H_{12}O_{6}}{180.16gC_{6}H_{12}O_{6}} =0.111 molC_{6}H_{12}O_{6}[/tex]
Moles of glucose in 84 mL solution:
[tex]84mL*\frac{0.111 mol C_{6}H_{12}O_{6} }{100mL} =0.0933 molC_{6}H_{12}O_{6}[/tex]
Calculating molarity of 84 mL of 20 % glucose solution:
[tex]\frac{0.0933mol C_{6}H_{12}O_{6}}{84mL}*\frac{1000mL}{1L}=1.11 M[/tex]
Finding the volume of this solution required to prepare 0.0250L of 0.5 M glucose solution:
[tex]M_{1}V_{1}=M_{2}V_{2}[/tex]
[tex]1.11 M(V_{1})=0.5 M(0.0250 L)[/tex]
[tex]V_{1}= 0.01126 L or 11.26 mL[/tex]
