Question

Calculat molarity of solution containing 16mg glocuse per 300ml solution

Answer 1

Given that  

Amount of glucose = 16 mg or 0.016 g

Number of mole of glucose = amount in g / molar mass

= 0.016 g/ 180.156 g/mol

= 8.88*10^-5 mole

Volume = 300 ml  

300 ml*1.0 L /1000 ml = 0.300 L

Molarity = number of moles / volume in L

= 8.88*10^-5 mole /0.300 L

= 2.96*10^-4 M

= 3.0*10^-4 M