Answer to Question #153534 in General Chemistry for erica

Question

caffeine has a molecular weight 194.19g how much caffeine is required to make 1L of 0.1w/v caffeine

Answer 1

Solution:

1) We should convert w/v to molarity:

$Molarity(\dfrac{mol}{L})=\dfrac{1000}{M_{solute}}\:\dfrac{\tfrac{w}{v}}{100}=\dfrac{1000}{194.19}\:\dfrac{0.1}{100}$

$C_M=0.00515\:(\tfrac{mole}{L})$

$0.00515=\dfrac{m_{caffeine}}{M_{caffeine}*V_{solution}}=\dfrac{m_{caffeine}}{194.19*1}$

$m_{caffeine}=1\:g$

Answer:

In order to make 1L of 0.1 $(\tfrac{w}{v})$ caffeine, we need 1 g of it.