NPN transistor: collector branch question
When the NPN transistor saturates its voltage between collector and emitter drops to a couple hundred mV. That's far less that the typical 2V for a LED for instance. If the LED voltage is 200mV it won't light. Final.  

But this is not an efficient circuit: you'll have current through the 330\$\Omega\$ resistor, whether the LED is on or not, and because of what I just explained, the current will be even larger when the LED is off, namely 14.5mA vs. 9.1mA.  

The better way is to swap photoresistor and 2k2\$\Omega\$ resistor, and put the LED in series with the 330\$\Omega\$ resistor. Then the base voltage will drop if the photoresistor receives light, and the transistor will be off, and so will the LED.
If it's dark the phototransistor's resistance will rise, and so will the transistor's base voltage. If it's 0.7V the transistor will start to conduct and the LED will go on.  

Measure the photoresistor's (let's call it R1. Better use reference designators on your schematics. Wait until Olin sees this) resistance when it sees light, and use that value to find the 2k2\$\Omega\$ resistor's replacement, let's call it R2:  

\$ \dfrac{R1}{R1 + R2} = \dfrac{0.7V}{5V} \$

or  

\$ R2 = \dfrac{5V - 0.7V}{0.7V}R1  \$