It's very efficient and it's a great solution overall; the only problem is there are only so many dams we can build, since each one requires drowning a valley.
m = 10^9.10^5 / (10*100) = 10^11 kg of water. = 10^8 m^3
This is not a big reserviour, being 1/20th of the size of Warragamba dam's reserviour (in Sydney). One valley could provide 20GW for a day. Compare that with demand for where I live:
http://www.aemo.com.au/Electricity/Data/Price-and-Demand/Pri...
Peak demand for 7.5M people is 9GW, with an average closer to 7GW. One valley could provide overnight storage for over 20 million people, with first world demands (all of Australia). The limitation is transmission and generating/pumping capacity, rather than storage volume.
The highest summer peak appears to have been just above 13GW. http://www.wattclarity.com.au/2016/02/highest-electricity-de...
However your point still stands I believe.
http://www.fastcodesign.com/1672202/you-deserve-a-house-with...
Could I efficiently use excess solar to pump water in to a house roof storage area and generate electric from that?
The energy in one gallon of gasoline is equivalent to the potential energy stored in 13 tons of water one kilometer up
Gravity potential energy = mgh. Suppose pool is 10 meters up, then you need 239,000 kg of water = 63K gallons or 8400 cu ft. This is a huge pool: 29 ft. x 29 ft x 10 ft.
You could use the city water supply: borrow water through a turbine at night, and return it during the day :-)
Or have pool up very high, for example on a cliff overlooking the Mediterranean:
http://inhabitat.com/crazy-home-carved-into-a-coastal-cliff-...
;)