Calculate the weight of water that can be removed from a wet organic phase using 50.0g of magnesium sulfate. Assume that it gives the hydrate listed in table 12.2
I'm not quite sure how I would go about this b/c when drying a organic layer you just scoop a bunch of drying agent into the test tube without measuring.
But I think that this is a simple stoichiometric calculations...is it?
If it is it would be:
50.0g MgSO4(1mol MgSO4/1.19e2g)(7mol water/1mol MgSO4)(1.80e1g/1mol water)=
The answer looks funny but I checked the my work and It looks alright so can someone check both my thinking and work as well?