The mass of sodium acetate (solute) in 425 g of solution is Mass or NaC_2H_3O_2
ID: 519547 • Letter: T
Question
The mass of sodium acetate (solute) in 425 g of solution is Mass or NaC_2H_3O_2 = 425 g times 0.0240 = 10.2 g We have the relationship Mass of solution mass NaC_2H_3O_2 + mass H_2O Therefore, the quantity of water in the solution is Mass of H_2O = mass of solution - mass of NaC_2H_3O_2 = 425 g - 10 You would prepare the solution by dissolving 10.2 g of sodium acetate in An experiment calls for 35.0 g of hydrochloric acid that is 20.2% HCÌ by mass. How many grams of HC1 is this?/How many grams of water?Explanation / Answer
12.5)
total mass of solution = 35.0 g
mass of HCl = 20.2 % of total mass
= 20.2 % of 35.0 g
= 20.2 * 35.0 / 100
= 7.07 g
remaining will be water
so,
mass of water = total mass - mass of HCl
= 35.0 g - 7.07 g
=27.93 g
Answer: 27.93 g
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