what is the molarity of a solution made by dissolving 8.56 g of sodium acetate in water [Solved]

Molarity is simply just the moles of the solute (whatever youre dissolving) divided by the liters of the entire solution. To find the moles of the sodium acetate, you just divide the grams by the molar mass. 4.00g of sodium acetate divided by 82.0 g of sodium acetate in one mole gives you approximately 0.0488 moles of sodium acetate. Now, change the mL to L (because thats the way it is in the equation) and 750 mL = 0.75 L. Finally, follow the first equation i gave, which was moles divided by liters and you get. 0.0488 moles / 0.75 L = 0.065-molar solution

b) 0.139 M

Molarity is simply just the moles of the solute (whatever youre dissolving) divided by the liters of the entire solution. To find the moles of the sodium acetate, you just divide the grams by the molar mass. 4.00g of sodium acetate divided by 82.0 g of sodium acetate in one mole gives you approximately 0.0488 moles of sodium acetate. Now, change the mL to L (because thats the way it is in the equation) and 750 mL = 0.75 L. Finally, follow the first equation i gave, which was moles divided by liters and you get. 0.0488 moles / 0.75 L = 0.065-molar solution

b) 0.139 M

Molarity is simply just the moles of the solute (whatever youre dissolving) divided by the liters of the entire solution. To find the moles of the sodium acetate, you just divide the grams by the molar mass. 4.00g of sodium acetate divided by 82.0 g of sodium acetate in one mole gives you approximately 0.0488 moles of sodium acetate. Now, change the mL to L (because thats the way it is in the equation) and 750 mL = 0.75 L. Finally, follow the first equation i gave, which was moles divided by liters and you get. 0.0488 moles / 0.75 L = 0.065-molar solution

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Yamakiri

b) 0.139 M

first calculate moles of sodium acetate NaCH3COO = grams / MM = 8.56 / 82.0 = 0.104 mol
then the molarity = n / V = 0.104 / 0.750 = 0.139 mol L-1

This post is last updated on hrtanswers.com at Date : 1st of September – 2022