438 grams.
There are 23.5 grams of solute contained for every 1,000,000 liters of solution that contains 21.7 ppm of SnC12. Therefore, there is 0.00008365 grams of solute in 3.5 liters of solution.Ê
Molarity = moles of solute/Liters of solution ( 75.0 ml = 0.075 Liters ) Get moles NaBr 1.5 M NaBr = moles NaBr/0.075 Liters = 0.1125 moles NaBr (102.89 grams/1 mole NaBr) = 11.575 grams NaBr ( call it 12 grams ) ----------------------------------------------------
30 liters
1 percent = 10 grams 2 % = 20 grams x 3 liters = 60 grams
3.33
88,218 liters.
I suppose that this solution doesn't exist.
897.5 centiliters are contained in 8.975 liters.
0.03696 centiliters are contained in 0.0003696 liters.
0.123 liters are contained in 12.3 centiliters.
109.31 liters are contained in 10,931 centiliters.
386.4 centiliters are contained in 3.864 liters.