Question

How many grams of glucose (C6 H12 O6) are needed to prepare 800.0 g of a 5.0% glucose solution?

Answer 1

C_percentage = 5 %

m_solution = 800 g

m_soluble = ?

$C_{percentage}= \cfrac {m_{soluble}}{m_{solution}}* 100$

$m_{soluble}= \cfrac {C_{percentage}~* ~ m_{solution}} {100}$

m_glucose = $\cfrac {5*800}{100}= 40~g$

Answer: 40 g